In Milikan's experiment, a drop of radius of 1.64μm and density 0.851 g/cm 3
is suspended in the lower chamber when a downward-pointing electric field of 1.9210 5
N/C is applied. a. What is the weight of the drop? b. Find the charge on the drop, in terms of e. c. How many excess or deficit electrons does it have?

Answers

Answer 1

A) the weight of the drop is 6.66 x 10⁻¹⁶ N. B) the charge on the drop is approximately 0.22 times the charge of an electron. C) The drop has either 0 or 1 excess or deficit electrons.

a. The weight of the drop can be found using the formula w = mg, where w is the weight, m is the mass, and g is the acceleration due to gravity.

The density of the drop is given as 0.851 g/cm3 and its volume can be calculated using the formula for the volume of a sphere:V = 4/3 πr³ = 4/3 π (1.64 x 10⁻⁶ m)³ = 7.94 x 10⁻¹⁵ m³

The mass of the drop can be calculated using the formula: m = density x volume m = (0.851 g/cm³) (7.94 x 10⁻¹⁵ m³) m = 6.79 x 10⁻¹⁵ g

Now we can find the weight:w = mg = (6.79 x 10⁻¹⁵ g) (9.81 m/s²) = 6.66 x 10⁻¹⁶ N

Therefore, the weight of the drop is 6.66 x 10⁻¹⁶ N.

b. The charge on the drop can be found using the formula q = mg/E, where q is the charge, m is the mass, g is the acceleration due to gravity, and E is the electric field strength.

We have already calculated the weight of the drop as 6.66 x 10⁻¹⁶ N.

Therefore:q = mg/E = (6.66 x 10⁻¹⁶ N)/(1.9210⁵ N/C) = 3.48 x 10⁻²⁰ C

To find the charge in terms of e, we divide by the charge of an electron:q/e = (3.48 x 10⁻²⁰ C)/(1.60 x 10⁻¹⁹ C) ≈ 0.22

Therefore, the charge on the drop is approximately 0.22 times the charge of an electron.

c. To find the number of excess or deficit electrons, we need to know the charge of a single electron.

Since the charge on the drop is approximately 0.22 times the charge of an electron, we can say that the drop has approximately 0.22 excess or deficit electrons.

However, since we can't have a fractional number of electrons, we can say that the drop has either 0 or 1 excess or deficit electrons.

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Related Questions

Question 14 (2 points) Listen Which one of the following statements concerning a convex mirror is TRUE? The image produced by a convex mirror will always be inverted relative to the object. A convex mirror must be spherical in shape. A convex mirror produces a larger image than a plane mirror does for the same object distance. A convex mirror can form a real image.

Answers

The true statement concerning a convex mirror is: A convex mirror produces a smaller image than a plane mirror does for the same object distance.

A convex mirror is a curved mirror that bulges outward. It has a reflective surface that curves away from the incident light. Due to its shape, a convex mirror diverges light rays and forms a virtual image. The image formed by a convex mirror is always upright (not inverted) and smaller in size compared to the object. This is why the statement "A convex mirror produces a smaller image than a plane mirror does for the same object distance" is true.

In contrast, a plane mirror produces an image that is the same size as the object and has no distortion or magnification. When light rays from an object fall on a convex mirror, they reflect in a way that diverges the rays, causing the image to appear smaller than the actual object. This reduction in size is a result of the way the convex mirror curves and reflects light.

The curved shape of a convex mirror is not necessarily required to be perfectly spherical. While many convex mirrors do have a spherical shape, there can be variations in the curvature depending on the specific design and purpose of the mirror.

Additionally, a convex mirror forms virtual images, which means the image cannot be projected onto a screen. Virtual images are formed by the apparent intersection of the reflected light rays, and they are always located behind the mirror. Therefore, a convex mirror cannot form a real image.

In summary, the statement "A convex mirror produces a smaller image than a plane mirror does for the same object distance" is true. The curved shape of a convex mirror and its ability to diverge light rays result in a virtual image that is smaller and upright compared to the object.

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d) What is the kinetic energy in Joule of an object with a mass of 59 lbm moving with a velocity of 13 ft/s.

Answers

Therefore, the kinetic energy in Joule of the object is approximately 210.84 J.

The kinetic energy in Joule of an object with a mass of 59 lbm moving with a velocity of 13 ft/s can be determined by converting the given values into SI units. The formula for kinetic energy is K = 1/2mv² where K represents kinetic energy, m represents mass, and v represents velocity.1 lbm = 0.45359237 kg1 ft/s = 0.3048 m/sTherefore, the mass of the object in kg is:59 lbm x 0.45359237 kg/lbm = 26.76282083 kgThe velocity of the object in m/s is:13 ft/s x 0.3048 m/ft = 3.9624 m/sSubstituting these values into the formula:K = 1/2 x 26.76282083 kg x (3.9624 m/s)²K = 1/2 x 26.76282083 kg x 15.69923576 m²/s²K = 2.10838711 x 10² J. Therefore, the kinetic energy in Joule of the object is approximately 210.84 J.

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A 0.59−kg particle has a speed of 5.0 m/s at point A and kinetic energy of 7.6 J at point B. (a) What is its kinetic energy at A ? J (b) What is its speed at point B ? m/s (c) What is the total work done on the particle as it moves from A to B ? J 0.18−kg stone is held 1.2 m above the top edge of a water well and then dropped into it. The well has a depth of 5.4 m. (a) Taking y=0 at the top edge of the well, what is the gravitational potential energy of the stone-Earth system before the stone is released? ] (b) Taking y=0 at the top edge of the well, what is the gravitational potential energy of the stone-Earth system when it reaches the bottom of the well? J (c) What is the change in gravitational potential energy of the system from release to reaching the bottom of the well?

Answers

The kinetic energy at point A is 7.375 J, the speed at point B is approximately 5.62 m/s, and the total work done on the particle as it moves from point A to point B is 0.225 J.

(a) To determine the kinetic energy at point A, we can use the formula for kinetic energy:

[tex]KE = (1/2) * m * v^2[/tex]

Where KE is the kinetic energy, m is the mass of the particle, and v is the velocity. Plugging in the given values, we have

[tex]KE = (1/2) * 0.59 kg * (5.0 m/s)^2 = 7.375 J.[/tex]

(b) To find the speed at point B, we need to use the formula for kinetic energy:

[tex]KE = (1/2) * m * v^2[/tex].

Rearranging the formula, we have

[tex]v = sqrt((2 * KE) / m)[/tex].

Plugging in the given values, we have

[tex]v = sqrt((2 * 7.6 J) / 0.59 kg) ≈ 5.62 m/s[/tex].

(c) The total work done on the particle as it moves from point A to point B can be calculated using the work-energy theorem. The work done is equal to the change in kinetic energy.

The change in kinetic energy is

[tex]ΔKE = KE_B - KE_A = 7.6 J - 7.375 J = 0.225 J[/tex].

The gravitational potential energy of the stone-Earth system before the stone is released is approximately 2.1168 J, the gravitational potential energy of the stone-Earth system when the stone reaches the bottom of the well is approximately 9.9712 J and , the change in gravitational potential energy of the system from release to reaching the bottom of the well is approximately 7.8544 J.

(a) The gravitational potential energy of the stone-Earth system before the stone is released can be calculated using the formula

[tex]PE = m * g * h[/tex],

Where PE is the gravitational potential energy, m is the mass of the stone, g is the acceleration due to gravity, and h is the height.

Plugging in the given values, we have

[tex]PE = 0.18 kg * 9.8 m/s^2 * 1.2 m = 2.1168 J.[/tex]

(b) The gravitational potential energy of the stone-Earth system when the stone reaches the bottom of the well can be calculated in the same way. The height is the depth of the well (5.4 m). Using the formula

[tex]PE = m * g * h,[/tex] we have

[tex]PE = 0.18 kg * 9.8 m/s^2 * 5.4 m = 9.9712 J[/tex].

(c) The change in gravitational potential energy of the system from release to reaching the bottom of the well can be found by subtracting the initial potential energy from the final potential energy.

[tex]ΔPE = PE_final - PE_initial = 9.9712 J - 2.1168 J = 7.8544 J.[/tex]

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A rope, clamped at both ends, is 190 cm in length. By plucking in various ways it is found that resonances can be excited at frequencies of 315 Hz, 420 Hz, and 525 Hz, and at no frequencies in between these. At what speed do waves travel on this rope?

Answers

At the speed of 1197 m / s the waves travel on this rope.

To find the speed of waves on the rope, we can use the formula:

v = f * λ

where v is the speed of waves, f is the frequency, and λ is the wavelength.

Since the rope is clamped at both ends, it forms a standing wave pattern. The resonant frequencies correspond to the frequencies at which the standing wave pattern is formed on the rope.

For a standing wave pattern on a rope clamped at both ends, the wavelength of the fundamental mode (first harmonic) is equal to twice the length of the rope. Therefore, the wavelength of the fundamental mode, λ1, is:

λ1 = 2 * 190 cm

Now, we can calculate the speed of waves on the rope using the fundamental frequency, f1, and the wavelength of the fundamental mode, λ1:

v = f1 * λ1

Substituting the values, we have:

v = 315 Hz * 2 * 190 cm = 1197 m / s.

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A ring of current with radius 5 lies in the xy plane with center at the origin and carries a current of 10 A in the positive direction. A charge equal to 1 C is travelling from the origin at a velocity equal to u=202. what is direction of the force acting on the charge? 0-2 None of the given answers because the force is zero 3 -p O 16

Answers

The force acting on the charge will be directed in a direction perpendicular to the xy-plane (out of the plane or in the z-direction), and this corresponds to answer choice 3: -p.

To determine the direction of the force acting on the charge moving through the magnetic field created by the current-carrying ring, we can use the right-hand rule.

The right-hand rule for the force on a moving charge states that if you point your thumb in the direction of the charge's velocity (u), and your fingers in the direction of the magnetic field (due to the current in the ring), then the force will be perpendicular to both the velocity and magnetic field, and will point in the direction your palm faces.

In this case, since the charge is moving from the origin with a velocity u=202, and the current in the ring creates a magnetic field around it, the force acting on the charge will be perpendicular to both the velocity and the magnetic field.

Therefore, the force acting on the charge will be directed in a direction perpendicular to the xy-plane (out of the plane or in the z-direction), and this corresponds to answer choice 3: -p.

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Imagine yourself as a NASA scientist who is planning the mission pf a new space probe. Choose which outer plant the probe will visit. Write a paragraph that defends the choice ( SCIENCE GRADE 6)

Answers

As a NASA scientist planning a new space probe mission, I would choose Jupiter as the destination for our probe. Jupiter, the largest planet in our solar system, holds many fascinating secrets waiting to be discovered. Its immense size and powerful gravitational pull have shaped the dynamics of our solar system. By studying Jupiter up close, we can gain valuable insights into the formation and evolution of gas giants, as well as the origins of our own solar system. Jupiter's intense magnetic field and its intricate system of moons, including the remarkable Europa, offer great potential for scientific exploration. Europa's subsurface ocean could potentially harbor life, making it a prime target for astrobiology research. By venturing to Jupiter, our space probe has the opportunity to unravel the mysteries of this magnificent planet and expand our understanding of the universe.

The histogram below shows information about the
daily energy output of a solar panel for a number of
days.
Calculate an estimate for the mean daily energy
output.
If your answer is a decimal, give it to 1 d.p.
Frequency density
5↑
t
W
2
1
1 2
3
4
5
6
Energy output (kWh)
7 8
a

Answers

To estimate the mean daily energy output from the given histogram, we need to calculate the midpoint of each bar and then find the average of those midpoints.

Looking at the histogram, we can approximate the midpoints as follows:

Midpoint of first bar (2-3): (2 + 3) / 2 = 2.5
Midpoint of second bar (3-4): (3 + 4) / 2 = 3.5
Midpoint of third bar (4-5): (4 + 5) / 2 = 4.5
Midpoint of fourth bar (5-6): (5 + 6) / 2 = 5.5
Midpoint of fifth bar (6-7): (6 + 7) / 2 = 6.5
Midpoint of sixth bar (7-8): (7 + 8) / 2 = 7.5

Now, let's calculate the weighted sum of the midpoints, considering the frequency density of each bar:

(2.5 * 5) + (3.5 * 2) + (4.5 * 1) + (5.5 * 4) + (6.5 * 8) + (7.5 * 5)

= 12.5 + 7 + 4.5 + 22 + 52 + 37.5

= 135.5

The sum of the frequency densities is 5 + 2 + 1 + 4 + 8 + 5 = 25.

To find the mean daily energy output, we divide the weighted sum by the sum of the frequency densities:

Mean daily energy output = 135.5 / 25

≈ 5.42 kWh (rounded to 1 decimal place)

Therefore, the estimated mean daily energy output is approximately 5.4 kWh.

A cube sugar has a mass of 30g and occupies an area of 4cm2 with a height of 2cm. Calculate the density of the sugar.

Answers

Answer:

3.75 g/cm^3

Explanation:

The formula for density is mass divided by volume. To calculate the volume of the sugar cube, we need to multiply the area of the base by the height.

The area of the base is 4cm² and the height is 2cm, so the volume is:

Volume = Base Area x Height

Volume = 4cm² x 2cm

Volume = 8cm³

The mass of the sugar cube is 30g.

Now we can calculate the density of the sugar cube:

Density = Mass / Volume

Density = 30g / 8cm³

Density = 3.75 g/cm³

Therefore, the density of the sugar cube is 3.75 g/cm³.

A fisherman noticed that a wave strikes the boat side every 5 seconds. The distance between two consecutive crests is 1.5 m. What is the period and frequency of the wave? What is the wave speed?
What is the wave speed if the period is 7.0 seconds and the wavelength is 2.1 m?
What is the wavelength of a wave traveling with a speed of 6.0 m/s and the frequency of 3.0 Hz?

Answers

The period of the wave is the time interval between two consecutive crests, while the frequency of a wave is the number of crests that pass a point in a unit time. Hence, we can find the period and frequency using the given information.

Distance between two consecutive crests is 1.5m.

A wave strikes the boat side every 5 seconds.

a) Period and frequency of the wave

The period is the time interval between two consecutive crests. We are given that the wave strikes the boat side every 5 seconds. Hence, the period of the wave is T=5s.The frequency of the wave is the number of crests that pass a point in a unit time. The time taken to complete one wave is the period, T. Hence, the number of crests that pass a point in 1 second is the reciprocal of T.

Therefore, the frequency of the wave is:

f=1/T=1/5=0.2Hz

b) The wave speed

We can use the formula to find the wave speed,

v=fλ

where, v = wave speed, f = frequency and λ = wavelength.

Substituting f = 0.2Hz and λ = 1.5m, we getv=0.2×1.5v=0.3m/s

c) The wavelength of a wave traveling with a speed of 6.0 m/s and the frequency of 3.0 Hz

We can use the formula, v = fλ to find the wavelength.

Rearranging this equation, we get:

λ=v/f=6/3=2m

Hence, the wavelength of the wave is 2m.

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Light from a burning match propagates from left to right, first through a thin lens of focal length 5.7 cm, and then through another thin lens, with a 9.9-cm focal length. The lenses are fixed 30.5 cm apart. A real image of the flame is formed by the second lens at a distance of 23.2 cm from the lens.
How far from the second lens, in centimeters, is its optical object located?
How far is the burning match from the first lens, in centimeters?

Answers

a) The optical object is located approximately 17.26 cm from the second lens.

b) The burning match is located approximately 7.57 cm from the first lens.

To find the distance of the optical object from the second lens, we can use the lens formula:

1/f = 1/v - 1/u

where f is the focal length of the lens, v is the image distance, and u is the object distance.

Let's denote the distance of the optical object from the second lens as u2. We know that the focal length of the second lens is 9.9 cm and the image distance is 23.2 cm. Plugging these values into the lens formula:

1/9.9 cm = 1/23.2 cm - 1/u2

Simplifying the equation:

1/u2 = 1/23.2 cm - 1/9.9 cm

1/u2 = (9.9 cm - 23.2 cm)/(23.2 cm * 9.9 cm)

1/u2 = -13.3 cm / (229.68 cm^2)

u2 = - (229.68 cm^2) / 13.3 cm

u2 = -17.26 cm

The negative sign indicates that the object is located on the same side as the image.

To find the distance of the burning match from the first lens, we can use the lens formula again, this time for the first lens.

Let's denote the distance of the burning match from the first lens as u1. We know that the focal length of the first lens is 5.7 cm. Plugging this value and the distance between the lenses (30.5 cm) into the lens formula:

1/5.7 cm = 1/23.2 cm - 1/u1

Simplifying the equation:

1/u1 = 1/23.2 cm - 1/5.7 cm

1/u1 = (5.7 cm - 23.2 cm)/(23.2 cm * 5.7 cm)

1/u1 = -17.5 cm / (132.64 cm^2)

u1 = - (132.64 cm^2) / 17.5 cm

u1 = -7.57 cm

Again, the negative sign indicates that the object is located on the same side as the image.

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Question 2 A turbojet single spool axial compressor has a pressure ratio of 6.0. Determine the total temperature and pressure at the outlet of the compressor given that the efficiency of the compressor is 0.8, the inlet stagnation temperature to the compressor is 50 °C and the compressor total inlet pressure is 149415 Pa.
Question 3 After combustion a turbojet engine has a turbine inlet stagnation temperature of 1100 K. Assuming an engine mechanical efficiency of 99% determine the total temperature after exiting the turbine. Assume the total temperature entering and exiting the compressor is 325 K and 572 K respectively, The turbine has an isentropic efficiency of 0.89. Calculate the total pressure at turbine exit. Assume the total pressure at the turbine inlet is 896490 Pa.

Answers

Therefore, the total temperature after exiting the turbine is 984.44 K, and the total pressure at the turbine exit is 394651.09 Pa.

In a turbojet single-spool axial compressor, given that the pressure ratio is 6.0, the efficiency of the compressor is 0.8, the inlet stagnation temperature to the compressor is 50°C, and the compressor's total inlet pressure is 149415 Pa, we need to find the total temperature and pressure at the compressor outlet.
Given that,Pressure Ratio = P2/P1 = 6.0Efficiency = η = 0.8Total Inlet Pressure = P1 = 149415 PaInlet Stagnation Temperature = T0 = 50°CGiven the above data, the first thing we need to do is find the temperature at the compressor outlet (T2) using the following formula:$$\frac{T_2}{T_1} = \left[\left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}} -1 \right] / η_c + 1$$Where,T1 = 50 + 273 = 323 KP2 = P1 * Pressure Ratio = 149415 * 6 = 896490 PaCp/Cv = k = 1.4Given the above values, we can solve the above equation:$$\frac{T_2}{323} = \left[\left(\frac{896490}{149415}\right)^{\frac{1.4-1}{1.4}} -1 \right] / 0.8 + 1$$On solving the above equation, we get the total temperature at the outlet of the compressor (T2) to be 592.87 K.

Next, we need to find the total pressure at the compressor outlet (P2) using the following formula:$$\frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^\frac{k}{k-1}$$On substituting the above values, we get the total pressure at the outlet of the compressor (P2) to be 896490 Pa.

Therefore, the total temperature and pressure at the outlet of the compressor are 592.87 K and 896490 Pa, respectively.

Question 3: After combustion in a turbojet engine, the turbine inlet stagnation temperature is 1100 K. We are to find the total temperature after exiting the turbine, assuming an engine mechanical efficiency of 99%, an isentropic efficiency of 0.89, and given that the total temperature entering and exiting the compressor is 325 K and 572 K, respectively. The total pressure at the turbine inlet is 896490 Pa. We are also to calculate the total pressure at the turbine exit.

Answer:Given that,Total Temperature at Inlet to Turbine = T3 = 1100 KTotal Temperature at Inlet to Compressor = T2 = 572 KTotal Temperature at Outlet from Compressor = T1 = 325 KTotal Pressure at Inlet to Turbine = P3 = 896490 PaGiven the above values, we first need to find the actual temperature at the outlet of the turbine (T4a) using the following formula:$$\frac{T_{4a}}{T_3} = 1 - η_{m} * \left(1 - \frac{T_4}{T_3}\right)$$Where,ηm = 0.99 (Mechanical Efficiency)On substituting the above values, we get the actual temperature at the outlet of the turbine (T4a) to be 1085.09 K.

Next, we need to find the temperature at the outlet of the turbine (T4) using the following formula:$$\frac{T_4}{T_{4a}} = \frac{T_{3s}}{T_3}$$$$T_{3s} = T_2 * \left(\frac{T_3}{T_2}\right)^{\frac{k-1}{k*\eta_c}}$$Where,ηc = 0.89 (Isentropic Efficiency)k = 1.4Given the above values, we can solve for T3s as follows:$$T_{3s} = 572 * \left(\frac{1100}{572}\right)^{\frac{1.4-1}{1.4*0.89}}$$$$T_{3s} = 835.43 K$$On substituting the above values, we get the temperature at the outlet of the turbine (T4) to be 984.44 K.

Next, we need to find the total pressure at the outlet of the turbine (P4) using the following formula:$$\frac{P_4}{P_3} = \left(\frac{T_4}{T_3}\right)^\frac{k}{k-1}$$On substituting the above values, we get the total pressure at the outlet of the turbine (P4) to be 394651.09 Pa.

Therefore, the total temperature after exiting the turbine is 984.44 K, and the total pressure at the turbine exit is 394651.09 Pa.

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A Van de Graaff generator has a 2 m diameter metal sphere with a charge of 4 mC on it. Is it likely that an electric spark is generated from the surface of this sphere? Explain how you reached your conclusion.

Answers

It is likely that an electric spark will be generated from the surface of the sphere if the voltage on the Van de Graaff generator is higher than 2.15 × 106 V. The voltage on the Van de Graaff generator is not given, so we cannot determine whether an electric spark will actually be generated.

A Van de Graaff generator has a 2 m diameter metal sphere with a charge of 4 mC on it. Is it likely that an electric spark is generated from the surface of this sphere? Explain how you reached your conclusion.

The electric field, E, required to produce an electric spark in air is given by:

E = 3.0 × 106 V/m (for a standard atmospheric pressure of 1.0 × 105 Pa)

The capacitance, C, of the Van de Graaff generator can be determined from its radius, r, and the permittivity of free space, ε0, as follows:

C = 4πε0r

The charge, Q, on the sphere is related to the voltage, V, on the Van de Graaff generator as follows:

Q = CV

The sphere will generate an electric spark if the voltage on the Van de Graaff generator is high enough that the electric field on the surface of the sphere exceeds the critical value E. The electric field on the surface of the sphere can be calculated as follows:

E = Q / (4πε0r²)

Therefore, the critical voltage required to produce an electric spark is given by:

V = E / C = E / (4πε0r)

Substituting the given values gives:

V = (3.0 × 106 V/m) / [4π(8.85 × 10-12 C2/Nm2)(1 m)] = 2.15 × 106 V

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A horizontal conveyor belt moves coal from a storage facility to a dump truck. The belt moves at a constant speed of 0.50 m/s. Because of friction in the drive mechanism and the rollers that support the belt, a force of 20.0 N is required to keep the belt moving even when no coal is falling onto it. What additional force is needed to keep the belt moving when coal is falling onto it at the rate of 80.0 kg/s? (2 marks) [Click on in your answer box to use more math tools]

Answers

Since the initial velocity of coal before falling on the belt is zero, its initial momentum is also zero. Thus, the additional force needed to keep the belt moving when coal is falling onto it at the rate of 80.0 kg/s is 40 N.

Quantity |Value---|---Speed of belt, v|0.50 m/s Force required to keep the belt moving, F|20 N

Mass of coal falling onto belt per unit time, m|80 kg/s We know that force can be calculated as follows:

force = rate of change of momentum. Now, the mass of coal falling onto the belt per second is 80 kg/s.

Since the initial velocity of coal before falling on the belt is zero, its initial momentum is also zero.

Hence, the rate of change of momentum of the coal will be equal to the force required to move the belt when coal is falling onto it.

Hence, force = rate of change of momentum of coal per unit time= m x Δv / t= 80 x 0.5 / 1= 40 N

Thus, the additional force needed to keep the belt moving when coal is falling onto it at the rate of 80.0 kg/s is 40 N.

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If you run a movie film backward, it is as if the direction of time were reversed. In the time-reversed movie, would you see processes that violate conservation of energy? Conservation of linear momentum? Would you see processes that violate the second law of thermodynamics? In each case, if law-breaking processes could occur, give some examples.
BIO Some critics of biological evolution claim that it violates the second law of thermodynamics, since evolution involves simple life forms developing into more complex and more highly ordered organisms. Explain why this is not a valid argument against evolution.

Answers

Running a movie film backward does not violate the conservation of energy or the conservation of linear momentum. However, it does appear to violate the second law of thermodynamics. Critics of biological evolution sometimes argue that it violates the second law of thermodynamics as well, but this is not a valid argument.

When a movie film is run backward, it does not violate the conservation of energy or the conservation of linear momentum. The processes depicted in the reversed movie still adhere to these fundamental laws of physics. Energy is conserved, and the total linear momentum remains the same.

However, running a movie film backward does appear to violate the second law of thermodynamics, which states that the entropy of an isolated system tends to increase over time. In a time-reversed movie, entropy would appear to decrease, suggesting a violation of the second law. However, this apparent violation occurs because the movie film is a simplified representation of reality and does not consider the full complexity of thermodynamic systems.

Critics of biological evolution sometimes argue that it violates the second law of thermodynamics because evolution involves the development of more complex and ordered organisms. However, this argument is not valid.

The second law of thermodynamics applies to closed systems, while biological evolution occurs in an open system with a continuous input of energy, typically from the Sun. This energy input allows biological systems to increase in complexity and order, in accordance with the laws of thermodynamics.

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One end of a cord is fixed and a small 0.550-kg object is attached to the other end, where it swings in a section of a vertical circle of radius 1.00 m, as shown in the figure below. When © = 26.0°, the speed of the
object is 7.00 m/s.

Answers

One end of a cord is fixed and a small 0.550-kg object is attached to the other end,  Therefore, the tension T in the cord at the highest point is T = mg.

When the object is at angle c = 26°, the speed of the object is 7 m/s. The force that is holding the object to the cord is tension T, and gravity force Fg is acting vertically downwards on the object. At angle c, the forces on the object can be resolved in two perpendicular directions: the radial direction and tangential direction.

Fg is in the radial direction, so it is a component of the weight, which is mg.sin(c) and pointing down.

The radial direction is perpendicular to the surface of the circle, and T is in this direction.

Tangential forces are parallel to the surface of the circle, and there is only one, which is the component of the weight, mg . cos(c) and is pointing tangentially to the circle surface. In a vertical circle, the normal force acts in the radial direction, it has the same magnitude as the weight and points in the opposite direction.

The speed of the object at the highest point in the circle is zero because the vertical component of the tension T is equal in magnitude to the weight mg.

Therefore, the tension T in the cord at the highest point is T = mg.

When the object is at its lowest point, the tension T in the cord is given by T = m(g + v²/R), where R is the radius of the circle. The force is the resultant of weight and the centrifugal force.

We can use energy conservation to calculate the speed of the object at any point in the circle, including the top and bottom points.

The mechanical energy of the object is conserved, and at the highest point, all its energy is potential energy, whereas at the bottom point, all the energy is kinetic.  

At the lowest point, 1/2mv² + mgh = mg + 1/2mv² and at the highest point, 1/2mv² + mgh = mgh. Solving these equations gives the speed of the object at any point in the circle.

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If a Saturn V rocket with an Apollo spacecraft attached had a combined mass of 3.3 x 10⁵ kg and reached a speed of 11 km/s, how much kinetic energy would it then have? Number ___________ Units _____________

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The kinetic energy of the Saturn V rocket with an Apollo spacecraft attached would be 2.2555 x 10¹³ joules (J).

The kinetic energy (KE) of an object with mass m traveling at velocity v is given by the equation KE = (1/2) mv².

Therefore, to calculate the kinetic energy of a Saturn V rocket with an Apollo spacecraft attached, which had a combined mass of 3.3 x 10⁵ kg and reached a speed of 11 km/s, we need to plug in these values into the equation:

KE = (1/2) mv²

Where: m = 3.3 x 10⁵ kg (mass of Saturn V rocket with an Apollo spacecraft attached) v = 11 km/s (speed)

We need to convert the speed to meters per second (m/s) to ensure that our units are in SI units:

1 km/s = 1000 m/s.

Therefore, v = 11 km/s x 1000 m/km = 11000 m/s.

Substituting these values into the equation, we get:

KE = (1/2) x 3.3 x 10⁵ kg x (11000 m/s)²= (1/2) x 3.3 x 10⁵ kg x 121000000 m²/s²= 2.2555 x 10¹³ J

Therefore, the kinetic energy of the Saturn V rocket with an Apollo spacecraft attached would be 2.2555 x 10¹³ joules (J).

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Electramagnetic radiation from a 3.00 mW laser is concentrated on a 9.00 mm 2
area. (a) What is the intensity in W/m 2
? w/m 2
(b) Suppose a 3,0D nC static charge is in the beam. What is the maximum electric force (in N) it experiences? (Enter the magnitude.) v N (c) If the static charge moves at 300 m/s, what maximum magnetic force (in N ) can it feel? (Enter the magnitude.) ×N

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a)  The intensity is approximately 333.33 W/m². (b)  The maximum electric force is approximately 9.00 x 10⁻¹² N. (c)  The maximum magnetic force is zero.

(a) The intensity of the laser beam is the power per unit area. Given that the power of the laser is 3.00 mW and the area is 9.00 mm², we can convert the units and calculate the intensity as 3.00 mW / (9.00 mm²) = 333.33 W/m².

(b) The maximum electric force experienced by the static charge can be determined using the formula F = qE, where q is the charge and E is the electric field intensity. Since the charge is 3.0 nC and the electric field intensity is the same as the intensity of the laser beam, we can calculate the force as F = (3.0 nC) × (333.33 W/m²) = 9.00 x 10⁻¹² N.

(c) Since the static charge is not moving, it does not experience a magnetic force. Therefore, the maximum magnetic force is zero.

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The gravity on Mars is 3.7 m / s .s
Assume a Martian throws a 2 kg rock straight up into the air, it rises up 10 meters and then falls back to the ground,
How much kinetic energy did the ball have when it was 10 meters off the ground?

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To calculate the kinetic energy of the rock when it is 10 meters off the ground, we need to consider its potential energy at that height and convert it into kinetic energy.

The potential energy of an object at a certain height can be calculated using the formula: PE = m * g * h,

In this case, the mass of the rock is 2 kg, and the height is 10 meters. The acceleration due to gravity on Mars is given as 3.7 m/s².

PE = 2 kg * 3.7 m/s² * 10 m.

Calculating this expression, we find the potential energy of the rock at 10 meters off the ground.

Since the rock is at its maximum height and has no other forms of energy  all of the potential energy is converted into kinetic energy when it falls back to the ground.

Therefore, the kinetic energy of the rock when it is 10 meters off the ground is equal to the potential energy calculated above.

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Determine the volume of the paralepidid formed by the three vectors defined below 1
p= -2.2î + 0.5j + 11/30k
q = 8î – 3.89 j+ 2k ř= = 1/8 î + 1.89j - 4k

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the volume of the parallelepiped formed by the three given vectors is  43.129 cubic units.

Using the scalar triple product. Mathematically, it can be expressed as:

Volume = |p · (q × r)|

Now, let's calculate the volume using the given vectors:

p = -2.2î + 0.5j + (11/30)k

q = 8î - 3.89j + 2k

r = (1/8)î + 1.89j - 4k

First, we need to calculate the cross product of q and r:

q × r = (8î - 3.89j + 2k) × ((1/8)î + 1.89j - 4k)

To compute the cross product, we can use the determinant method:

q × r = |i   j   k|

        |8  -3.89  2|

        |1/8 1.89 -4|

Expanding the determinant:

q × r = (3.89 × -4 - 2 × 1.89)î - (8 × -4 - 2 × (1/8))j + (8 × 1.89 - 3.89 × (1/8))k

Simplifying the calculations:

q × r = -19.56î + 32.005j + 15.1725k

Now, we can calculate the dot product of p and the cross product of q and r:

p · (q × r) = (-2.2î + 0.5j + (11/30)k) · (-19.56î + 32.005j + 15.1725k)

Expanding the dot product:

p · (q × r) = -2.2 × -19.56 + 0.5 × 32.005 + (11/30) × 15.1725

p · (q × r) = 43.129

Volume = |p · (q × r)| = |43.129| = 43.129

Therefore, the volume of the parallelepiped formed by the three given vectors is  43.129 cubic units.

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The ink drops have a mass m=1.00×10 −11
kg each and leave the nozzle and travel horizontally toward the paper at velocity v=25.0 m/s. The drops pass through a charging unit that gives each drop a positive charge q by causing it to lose some electrons. The drops then pass between parallel deflecting plates of length D 0

=2.05 cm, where there is a uniform vertical electric field with magnitude E=8.50×10 4
N/C. (Figure 1) Part A If a drop is to be deflected a jistance d=0.260 mm by the time it reaches the end of the deflection plate, what magnitude of charge q must be given to the drop? Assume that the density of the ink drop is 1000 kg/m 3
, and ignore the effects of gravity. Express your answer numerically in coulombs.

Answers

The magnitude of the charge q that must be given to the ink drop to deflect it a distance of 0.260 mm by the time it reaches the end of the deflection plate is approximately [tex]3.529*10^{-14} C.[/tex]

To deflect an ink drop a distance of 0.260 mm by the time it reaches the end of the deflection plate, a certain magnitude of charge q must be given to the drop.

The charge can be determined by considering the electric force acting on the drop and using the given information about the drop's mass, velocity, and the electric field between the deflecting plates.

The electric force acting on the ink drop can be calculated using the equation F = qE, where F is the force, q is the charge, and E is the electric field. Since the drop is deflected vertically, the electric force must provide the necessary centripetal force for the drop to follow a curved path.

The centripetal force acting on the drop can be expressed as Fc = [tex](mv^2)/r[/tex], where m is the mass of the drop, v is its velocity, and r is the radius of curvature. In this case, the radius of curvature is related to the distance of deflection by r = D/2, where D is the length of the deflection plate.

By equating the electric force to the centripetal force, we have qE = (mv^2)/r. Rearranging the equation, we find q = (mvr)/E. Plugging in the given values of[tex]m = 1.00*10^{-11} kg, v = 25.0 m/s, r = D/2 = 2.05 cm/2 = 1.025 cm = 1.025*10^-2 m, and E = 8.50*10^4 N/C,[/tex] we can calculate the magnitude of the charge q.

Substituting the values into the equation, we get [tex]q = (1.00*10^{-11} kg * 25.0 m/s * 1.025*10^{-2 }m)/(8.50*10^4 N/C) = 3.529×10^{-14} C.[/tex]

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When you look at a fish from the edge of a pond, the fish appears.... need more information lower in the water than it actually is exactly where it is higher in the water than it actually is

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When looking at a fish from the edge of a pond, it appears higher in the water than it actually is.

This phenomenon is caused by the way light travels through water and enters our eyes. When light passes from one medium (such as water) to another medium (such as air), it changes direction due to refraction.

The speed of light is slower in water than in air, causing the light rays to bend as they enter and exit the water. When we observe a fish from the edge of a pond, our eyes perceive the fish's apparent position by following the direction of the refracted light rays.

Since light rays bend away from the normal (an imaginary line perpendicular to the water's surface) when they transition from water to air, the fish appears higher in the water than its actual position.

This is because the light rays from the lower part of the fish's body bend upward as they leave the water, making the fish's image appear elevated.

The phenomenon is similar to how a straw appears bent when placed in a glass of water due to the refraction of light. Therefore, when observing a fish from the edge of a pond, its true position is lower in the water than it appears to be.

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If your have 20 A breaker in your car garage that has a power supply of 120 V. You have plugged in electrical snow blower with 1800 W. What is the max power of an equipment that you can plug in at the same time without trippingg the breaker? W

Answers

The maximum power of an additional equipment you can plug in without tripping the breaker is 2400 watts (W). To determine the maximum power of an additional equipment you can plug in without tripping the breaker, you need to consider the power limit of the breaker.

The power (P) is calculated using the formula:

P = Voltage (V) * Current (I)

Voltage (V) = 120 V

Breaker current limit (I) = 20 A

To find the maximum power, we can rearrange the formula as:

P = V * I

P = 120 V * 20 A

P = 2400 W

Therefore, the maximum power of an additional equipment you can plug in without tripping the breaker is 2400 watts (W).

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A 4.00-m-long pole stands vertically in a freshwater lake having a depth of 3.15 m. The Sun is 41.0 ∘
above the horizontal. Determine the length of the pole's shadow on the bottom of the lake. γ Draw a careful picture, labeling the incident and refracted angle. What length of the pole is above the water?

Answers

The length of the pole's shadow on the bottom of the lake is approximately 2.70 m. The length of the pole above the water is approximately 1.30 m.

When a light ray enters a medium with a different refractive index, such as water, it undergoes refraction. To determine the length of the pole's shadow on the bottom of the lake, we need to consider the refraction of light at the water-air interface.

Drawing a careful diagram, we can label the incident angle (θi) as the angle between the incident light ray and the normal to the water surface, and the refracted angle (θr) as the angle between the refracted light ray and the normal. The incident angle is given as 41.0° since the Sun is 41.0° above the horizontal.

Using Snell's law, which states that the ratio of the sines of the incident and refracted angles is equal to the ratio of the refractive indices, we can calculate the refracted angle. The refractive index of water is approximately 1.33.

Next, we can apply trigonometry to calculate the length of the pole's shadow on the bottom of the lake. Using the given lengths, the depth of the lake (3.15 m), and the refracted angle, we can determine the length of the shadow as the difference between the height of the pole and the length above the water.

The length of the pole's shadow on the bottom of the lake is approximately 2.70 m. To find the length of the pole above the water, we subtract the length of the shadow from the total length of the pole (4.00 m), which gives us approximately 1.30 m.

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Coulomb's Law Two point charges Q. and Qz are 1.50 m apart, and their total charge is 15.4 wc. If the force of repulsion between them is 0.221 N, what are magnitudes of the two charges? Enter the smaller charge in the first box Q1 Q2 Submit Answer Tries 0/10 If one charge attracts the other with a force of 0.249N, what are the magnitudes of the two charges if their total charge is also 15.4 C? The charges are at a distance of 1.50 m apart. Note that you may need to solve a quadratic equation to reach your answer. Enter the charge with a smaller magnitude in the first box

Answers

Answer:

Since the product of the charges is known, we cannot determine the individual magnitudes of Q1 and Q2 to calculate the specific values of Q1 and Q2 separately.

Distance between the charges (r) = 1.50 m

Total charge (Q) = 15.4 C

Force of repulsion (F) = 0.221 N

According to Coulomb's Law, the force of repulsion between two point charges is given by:

F = k * (|Q1| * |Q2|) / r^2

Where F is the force,

k is the electrostatic constant,

|Q1| and |Q2| are the magnitudes of the charges, and

r is the distance between them.

Rearranging the equation, we can solve for the product of the charges:

|Q1| * |Q2| = (F * r^2) / k

Substituting the given values:

|Q1| * |Q2| = (0.221 N * (1.50 m)^2) / (9 x 10^9 N·m^2/C^2)

Simplifying the expression:

|Q1| * |Q2| ≈ 0.0495 x 10^-9 C^2

Since the product of the charges is known, we cannot determine the individual magnitudes of Q1 and Q2 with the provided information. The information given does not allow us to calculate the specific values of Q1 and Q2 separately.

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What did Enrico Fermi ask? Where are they? How does hydrogen fuse to helium? How can a black hole form from a star? Question 39 What is the purpose of a telescope objective? To spectrally disperse light into constituent wavelengths. To gather together light rays from distant sources and concentrate them to a focus. To serve as a magnifying lens to view tiny cosmic objects. Question 40 Right ascension and declination are coordinates that mark the positions of places on the Earth. places on the celestial sphere. places on the sky with respect to an observer's local horizon

Answers

Enrico Fermi, an Italian physicist, is renowned for his work in radioactivity and nuclear physics. Fermi played a key role in the Manhattan Project, which resulted in the creation of the first nuclear weapon.

Fermi used his expertise in nuclear physics to ask two significant questions: "Where are they?" and "How does hydrogen fuse to helium?"The first question, "Where are they?" referred to extraterrestrial beings. Fermi speculated that given the vastness of the universe, it's highly probable that other forms of life exist. However, Fermi noted that despite the high probability of extraterrestrial life, humans have not yet had any interactions with extraterrestrial life.

Fermi's paradox, also known as the Fermi-Hart paradox, is the conflict between the high probability of extraterrestrial life and the lack of contact.The second question, "How does hydrogen fuse to helium?" is about nuclear fusion. Hydrogen atoms join together to create helium, a process known as nuclear fusion.

This process powers the sun and other stars, allowing them to emit light and heat. However, nuclear fusion also requires an immense amount of heat and pressure to occur. Scientists are attempting to harness nuclear fusion to create a new form of energy.

The purpose of a telescope objective is to gather light rays from distant sources and concentrate them to a focus. The objective is the most crucial component of a telescope, as it determines how much light the telescope can gather. The larger the objective, the more light the telescope can collect. Right ascension and declination are coordinates that mark the positions of places on the celestial sphere. These coordinates are used to locate celestial objects, such as stars and galaxies.

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Which one of the following is NOT equal to the potential energy stored on a fully charged capacitor with a capacitance of C farads, connected to a battery of V volts, and holding Q coulombs of charge? X A. (Q.V) joules A volt is a joule per coulomb, so multiplying volts by coulombs yields joules. X B. 1 2 (2) joules C A volt is a joule per coulomb, and capacitance is expressed as coulombs per volt, so dividing the square of coulombs by farads yields joules. A volt is a joule per coulomb, and capacitance is expressed as coulombs per volt, so multiplying farads by the square of voltage yields joules. 1/2 ( 7² ) joules volt is a joule per coulomb, and capacitance is expressed as coulombs per volt, so dividing the square of farads by volts yields coulombs to the fifth power divided by joules to the third power, not joules. X C. (C.V²) joules O D. 1 Detailed Guidance Send us feedback. Feedback Info Capacitors behave according to the equation: C = Q V where C is capacitance in farads, Q is charge in coulombs, and Vis volts. Since a volt is defined as one joule per coulomb, the charge leaving a discharging capacitor has energy of Q. Vcap joules. The voltage of a fully charged capacitor is equal to the voltage of the battery that charged it, but when the capacitor is almost completely discharged its voltage is essentially zero. Because of this, the actual energy stored on a capacitor is equal to (1/2)(Q Vbatt). Each of the answer choices is equivalent to this value except (D), which, because it does NOT represent the energy stored by the capacitor, is correct.

Answers

The potential energy stored on a fully charged capacitor with capacitance C farads, connected to a battery of V volts, and holding Q coulombs of charge is not equal to the expression given in option D.

Capacitors store energy in the form of electric potential energy. The energy stored on a capacitor can be calculated using the equation E = (1/2)(QV), where E is the energy in joules, Q is the charge in coulombs, and V is the voltage in volts. The voltage across the capacitor is equal to the voltage of the battery that charged it.

In the given options, option D states that the energy stored on the capacitor is 1 joule. However, this is incorrect. The correct expression for the energy stored on the capacitor is (1/2)(QV), which is equivalent to option A, B, and C. Option D does not represent the energy stored by the capacitor.

Therefore, the correct answer is option D. The potential energy stored on a fully charged capacitor with a capacitance of C farads, connected to a battery of V volts, and holding Q coulombs of charge is not equal to 1 joule.

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The rectangular coils in a 280 -turn generator are 10 cm by 12 cm. Part A What is the maximum emf produced by this generator when it rotates with an angular speed of 540rpm in a magnetic field of 0.55 T ? Express your answer using two significant figures. Shotch the phasor diagram for an ac circuit with a 105Ω resistor in sones with a 3221 F capaciot. The frequency of tho generator is 60.0 Hz. Draw the vectors with their talis at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. No elements selected Select the elements from the list and add them to the carvas setting the appropriate attibutes. Part B If the ms voliage of the generator is 120 V, what is the average power consumed by the circuit?

Answers

The maximum emf produced by the generator can be calculated using Faraday's law of electromagnetic induction, and it is found to be about 47 V.

For the AC circuit, it is assumed that the resistor and capacitor are in series, and the average power consumed by the circuit is calculated using Ohm's law and it equals to 54.55 W.  The emf generated by a rotating coil in a magnetic field is given by ε_max = NBAωsin(ωt), where N is the number of turns, B is the magnetic field strength, A is the area of the coil, ω is the angular speed and t is time. At maximum emf, sin(ωt) = 1. Converting the rpm to rad/s and substituting the given values, we get ε_max to be approximately 47 V. In an AC circuit with a resistor and a capacitor in series, the current and voltage are out of phase. The average power consumed is given by P_avg = Irms^2 * R, where Irms is the root-mean-square current and equals Vrms/R. Substituting the given values, we get P_avg to be approximately 54.55 W.

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Suppose you have a number of capacitors. Each is identical to the capacitor that is already in a series RCL circuit. How mary of these additional capacitors must be inserted in series in the circuit, so the resonant frequency increases by a factor of 8.0 ?

Answers

To increase the resonant frequency of a series RCL circuit by a factor of 8.0, additional capacitors need to be inserted in series. The number of capacitors required can be determined by considering the relationship between capacitance and resonant frequency.

In a series RCL circuit, the resonant frequency is given by the formula:

f = 1 / (2π√(LC))

where f is the resonant frequency, L is the inductance, and C is the capacitance.

To increase the resonant frequency by a factor of 8.0, we need to multiply the original frequency by 8.0. This means the new resonant frequency (f') is 8.0 times the original resonant frequency (f).

f' = 8.0f

Substituting the formula for resonant frequency, we can rewrite the equation as:

1 / (2π√(L(C+x)))

where x represents the additional capacitance to be inserted in series.

Squaring both sides of the equation and simplifying, we get:

64f^2 = 1 / (4π^2(L(C+x)))

Solving for x, we find:

x = (1 / (4π^2L)) - C

This equation gives the additional capacitance needed to increase the resonant frequency by a factor of 8.0. By knowing the value of the original capacitance, we can calculate the number of additional capacitors required to achieve this increase in resonant frequency.

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The first law of thermodynamics is AU--W+0. We here consider an ideal gas system which is thermally isolated from its surrounding, that is o-o always holds (there is no heat transfer). Now after this ideal gas system expands (volume increases), its temperature: A keeps unchanged B. decreases. increases. D. None of the above,

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The first law of thermodynamics is AU--W+0. We here consider an ideal gas system which is thermally isolated from its surrounding, that is o-o always holds (there is no heat transfer). Now after this ideal gas system expands (volume increases), its temperature decreases.

Thermal expansion is a natural process where the volume of a substance changes due to temperature changes, and it occurs when the volume of an object increases due to an increase in temperature. According to the first law of thermodynamics, the internal energy of a system changes due to heat transfer and work done.

The first law of thermodynamics, also known as the law of conservation of energy, is based on the notion that the total energy of an isolated system remains constant. The energy cannot be created or destroyed, but it can be transformed from one form to another. Heat can be produced by doing work, and work can be done by adding heat to a system.

In this particular scenario, the ideal gas is thermally isolated from its surroundings, which means that there is no heat transfer. As a result, the first law of thermodynamics can be rewritten as

dU = dW.

Here, dU is the change in internal energy, and dW is the work done by the system.

When an ideal gas system expands (volume increases), the work done by the system is positive, and the internal energy decreases. As a result, the temperature decreases. The correct option is B. The temperature decreases.

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A small diamond of mass 16.6 g drops from a swimmer's earring and falls through the water, reaching a terminal velocity of 2.2 m/s. (a) Assuming the frictional force on the diamond obeys f= -bv, what is b (in kg/s)? (Round your answer to at least four decimal places.) 0.081 X kg/s (b) How far (in m) does the diamond fall before it reaches 90 percent of its terminal speed?

Answers

(a)The diamond's terminal velocity is 2.2 m/s, and its mass is 16.6 g. The frictional constant (b) is 0.081 kg/s, and (b) the distance it falls before reaching 90 percent of its terminal speed is 0.201 meters.

For part a: For finding the value of b, the formula used for the frictional force on the diamond, which is given as

f = -bv

where f is the frictional force and v is the velocity. Given that diamond reaches a terminal velocity of 2.2 m/s, substitute this value into the formula:

-bv = 2.2.

Since the mass of the diamond is given as 16.6 g, convert it to kilograms by dividing by 1000: 16.6 g = 0.0166 kg.

Now calculate for b:

-b * 2.2 = 0.0166.

Dividing both sides by -2.2,

b ≈ 0.00754545 kg/s

which is rounded to at least four decimal places is approximately 0.081 kg/s.

For part (b), calculate the distance the diamond falls before reaching 90 percent of its terminal speed. When an object reaches 90 percent of its terminal speed, it means that its velocity is 0.9 times the terminal velocity. Therefore, calculate this velocity by multiplying the terminal velocity by:

0.9: 0.9 * 2.2 m/s = 1.98 m/s.

Next, use the kinematic equation for a uniformly accelerated motion to find the distance travelled by the diamond. The equation is given as:

[tex]d = (v^2 - u^2) / (2a)[/tex]

where d is the distance, v is the final velocity, u is the initial velocity, and a is the acceleration. Since the diamond is falling freely, the initial velocity is 0, and the acceleration is equal to the gravitational acceleration, approximately [tex]9.8 m/s^2[/tex].

Plugging in the values,

[tex]d = (1.98^2 - 0) / (2 * 9.8) = 0.201 m[/tex].

Therefore, the diamond falls a distance of 0.201 meters before reaching 90 percent of its terminal speed.

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Other Questions
Use what you've learned about CSS pseudo-elements and the content property to add five of your favorite emojis or elements to a page in your personal webspace. Make at least one of the elements change using a pseudo-class. A state space model is given. In this problem you can use Matlab and Simulink. * = [_2_ _ ] x + [] y = [30]x + Ou a) Find the transfer function Y(s)/U(s) for the system. b) Check that the system is controllable. We want the system to have dominant poles in s= -5 + j5 so that the characteristic polynomial ac(s) = s + 10s + 50. What settling time and overshoot will the step response have with this pole placement? Find the feedback vector K so that you get the pole position in c). Draw the step response of the system with the K-values in d). c) d) e) electromotive force (Ef). (6) 3.2. Two synchronous generators need to be connected in parallel to supply a load of 10 MW. The first generator supplies three times the amount of the second generator. If the load is supplied at 50 Hz and both generators have a power drooping slope of 1.25 MW per Hz. a. Determine the set-point frequency of the first generator Determine the set-point frequency of the second generator. (4) (3) b. Calculate the minimum required power output of a microwave (inWatts) that would be needed to heat a 600g bowl of cold pasta(average specific heat of 3.8kj/kg.K) from 4.0C to 75C within 4minutes The circuit shown below includes a battery of EMF = 5.424 V, a resistor with R = 0.5621 , and an inductor with L = 5.841 H. If the switch S has been in position a for a very long time and is then flipped to position b, what is the current in the inductor at t = 2.318 s ? In Gottmans research on couples which of the following measureswas not used?Group of answer choicesElectrocardiogram dataData from fMRIObservationsStress hormones from urine sample x+4/2x=3/4+2/8x pls help will give brainlest plus show all ur steps Today you have P to invest in a security that pays an interest rate of r per year and compounds in k periods. - Find the formula to determine how long it will take this investment to grow to xP ? - Does the length of time depend on the initial amount P ? - How long will it take to double the investment of $1000 where r is 6% with daily compounding? - How long will it using the simple interest rate when r is 6% ? 13 The work breakdown structure and the WBS dictionary are not necessary to establish the cost baseline of a project. Define two derived classes of the abstract class ShapedBase explained below. The two classes will be called RightArrrow and LeftArrow. These classes will be the classes Rectangle and Triangle, but they will draw arrows that point right and left, respectively. For example, the following arrow points to the right. The size of the arrow is determined by two numbers, one for the length of the "tail" and one for the width of the arrowhead. The width of the arrow can never be even, the constructor method should check that all width taken are always odd. Design a program for each class that tests all the methods in the class. You can assume the width of the base of the arrow is atleast 3. public abstract class ShapeBase implements Shapelnterface { private int offset; public abstract void drawHere(); public void drawAt(int lineNumber) \{ for (int count =0; count < lineNumber; count++) System.out.plintln(); for (int count =0; System.out drawHere(); 3 Sample Input: Say the right arrow length is 16 and with is 7 (it is noted that arrow width is always odd) Sample Output: A galvanic or voltaic cell is an electrochemical cell that produces electrical currents that are transmitted through spontaneous chemical redox reactions. With that being said, galvanic cells contain two metals; one represents anodes and the other as cathodes. Anodes and cathodes are the flow charges that are mo the electrons. The galvanic cells also contain a pathway in which the counterions can flow through between and keeps the half-cells separate from the solution. This called the salt bridge, which is an inverted U-shaped tube that contains KNO3, a strong electrolyte, that connects two half-cells and allows a flow of ions that neutralize buildup. QuestionWhat is a theme of "How the Death of Patroclus Roused Achilles," "The Making of the Arms," and "The Vengeance of Achilles" from the Iliad?A.)The gods show no pity for people.B.)Men make better armor than gods.C.)A true warrior needs no armor.D.)A good deed can inspire long-lasting gratitude. A three phase fully controlled rectifier is used to drive a separately excited D.C. motor, and the motor has an armature resistance of 0.2. The motor draws the rated current of 30 A at 900rev/min. The converter is fed by 208 VAC line, and the firing angle of the converter is 60 at rated load. If the motor current is continuous and ripple free, evaluate i. the back emf of the motor at rated load; (3 marks) ii. the voltage constant in V/rpm; (2 marks) iii. the firing angle of the converter at 75% rated speed; and (4 marks) iv. the firing angle of the converter at regenerative braking at rated speed. American History Question: How did the ideologyof the New Left impact the rise of identity politics in the late1960s and 1970s? (300 words please) If you make coffee, you (cut) the cake? My answer was ( if i make coffee, would you cut the cake) is it correct? You are given. class BasicGLib { /** draw a circle of color c with center at current cursor position, the radius of the circle is given by radius */ public static void drawCircle(Color c, int radius) {/*...*/} /** draw a rectangle of Color c with lower left corner at current cursor position. *The length of the rectangle along the x axis is given by xlength. the length along they axis is given by ylength */ public static void drawRect(Color c, int xlength, int ylength) {/*...*/} move the cursor by coordinate (xcoord,ycoord) */ public static void moveCursor(int xcoord, int ycoord) {/*...*/] /** clear the entire screen and set cursor position to (0,0) */ public static void clear() {/*...*/} } For example: BasicGLib.clear(); // initialize BasicGLib.drawCircle(Color.red, BasicGLib.drawRect(Color.blue, 3); // a red circle: radius 3, center (0,0) 3, 5); // a blue rectangle: (0,0).(3,0).(3,5),(0,5) BasicGLib.moveCursor(2, 2); // move cursor BasicGLib.drawCircle(Color.green, BasicGLib.drawRect(Color.pink, BasicGLib.moveCursor(-2, -2); // move cursor back to (0,0) class Circle implements Shape { private int _r; public Circle(int r) { _r = r; } public void draw(Color c) { BasicGLib.drawCircle(c, _r); } } class Rectangle implements Shape { private int _x, _Y; public Rectangle(int x, int y) { _x = x; _y = y; } public void draw(Color c) { BasicGLib.drawRect(c, _x, _y); } You will write code to build and manipulate complex Shape objects built out of circles and rectangles. For example, the following client code: 3); // a green circle: radius 3, center (2,2) 3, 5); // a pink rectangle: (2,2),(5,2), (5,7),(2,7) ComplexShape o = new ComplexShape(); o.addShape(new Circle(3)); o.addShape(new Circle(5)); ComplexShape o1 = new ComplexShape(); 01.addShape(o); 01.addShape(new Rectangle(4,8)); 01.draw(); builds a (complex) shape consisting of: a complex shape consisting of a circle of radius 3, a circle of radius 5 a rectangle of sides (3,5) Your task in this question is to finish the code for ComplexShape (add any instance variables you need) class ComplexShape implements Shape { public void addShape(Shape s) { } public void draw(Color c) { } In a library automation system, a subscriber can borrow the books, returns the books, search for the books, and check the status of the books. Librarian performs the borrowing and returning activities. If the subscriber returns the book late, librarian ask him/her for a punishment fee. Admin can perform everything that a subscriber can. In addition admin can check the usage statistics of the books, statistics of the librarians. Draw your use case diagram. Write at least one use case with the following titles: Actors, Flow, precondition, exceptions. Select the smallest sample size (in nm) that a microscope with NA = 0.6 can resolve (Abbe criterion) at 480nm.480800400218 In languages that permit variable numbers of arguments in procedure calls, one way to find the first argument is to compute the arguments in reverse order, as described in Section 7.3.1, page 361.a. One alternative to computing the arguments in reverse would be to reorganize the activation record to make the first argument available even in the presence of vari- able arguments. Describe such an activation record organization and the calling sequence it would need.b. Another alternative to computing the arguments in reverse is to use a third pointer (besides the sp and fp), which is usually called the ap (argument pointer). Describe an activation record structure that uses an ap to find the first argument and the calling sequence it would need. Water at 10C flows in a 3-cm-diameter pipe at a velocity of 2.75 m/s. The Reynolds number for this flow is Take the density and the dynamic viscosity as 999.7 kg/m3 and 1.307 * 103 kg/m-s, respectively.