a. The dog's displacement after 5 seconds is equal to the change in position. To determine this, we must determine the distance between the final position and the initial position.
The dog's initial position was zero, and its final position was 1 meter west (negative direction), so the displacement is equal to 1 meter in the negative direction.
Displacement = final position - initial position = -1 m - 0 m = -1 m.
b.
The distance traveled is the total distance covered by the dog. We must determine the sum of the magnitudes of each vector quantity in this case. The displacement from the previous part was equal to 1 m, but we must now account for the distance that the dog covered in the positive direction (east) before moving back west. 2 m + 1 m = 3 m total distance covered. The dog's distance traveled after 5 seconds is equal to 3 meters.
c. The dog is motionless when its position remains constant. The dog is stationary between 2 and 3 seconds because the graph is flat. The dog is not in any position when it is stopped.
d.
Velocity is defined as the rate at which the position changes over time. If the position increases over time, the velocity is positive, whereas if the position decreases over time, the velocity is negative.
When the position remains constant, the velocity is zero. The graph is flat between 2 and 3 seconds, so the velocity is zero. When the dog is at a position of 1 meter west of the origin at 5 seconds, the dog's velocity is calculated as follows:
Velocity = displacement/time = (-1 m - 0 m) / (5 s - 0 s) = -1/5 m/s.
The dog's velocity at t = 4 s is -1/5 m/s.
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73. A small soap factory in Laguna supplies soap containing 30% water to a local hotel at P373 per 100 kilos FOB. During a stock out, a different batch of soap containing 5% water was offered instead.
The new cost of the soap containing 5% water would be cheaper. However, it is important to note that the new batch of soap may not have the same quality as the original batch containing 30% water. The hotel may also not be satisfied with the quality of the new batch and may choose to switch suppliers.
In the case of a small soap factory in Laguna that supplies soap containing 30% water to a local hotel at P373 per 100 kilos FOB and then experiencing a stock out, the factory may provide a different batch of soap containing 5% water. This will change the cost of the soap. The cost of the soap containing 30% water is calculated using:P373 per 100 kilos = (30% x 100 kilos) water + (70% x 100 kilos) soap= 30 kilos water + 70 kilos soap Therefore, the cost of the soap component is:P373/70 kilos soap = P5.33/kilo soapOn the other hand, if the soap contains 5% water, the cost of the soap will change. The cost of the soap component in this case would be:P373/95 kilos soap = P3.93/kilo soap. Therefore, the new cost of the soap containing 5% water would be cheaper. However, it is important to note that the new batch of soap may not have the same quality as the original batch containing 30% water. The hotel may also not be satisfied with the quality of the new batch and may choose to switch suppliers.
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An X-ray photon is scattered at an angle of θ=180.0 ∘
from an electron that is initially at rest. After scattering, the electron has a speed of 5.40×10 6
m/s. Find the wavelength of the incident X-ray photon.
The wavelength of the incident X-ray photon is 3.57 × 10-11 m.
A wavelength of the incident X-ray photon is required if an X-ray photon is scattered at an angle of θ = 180.0∘ from an electron that is initially at rest and the electron has a speed of 5.40 × 10 6 m/s after scattering.
The momentum conservation law holds for the electron and photon before and after scattering because the interaction is a collision. Before scattering: p i = h/λ... (1)
After scattering:p f = h/λ′... (2)
The momentum conservation law can be stated as follows:p i = p f + p e... (3), where pe is the momentum of the electron after scattering and can be calculated using the following equations:
Kinetic energy = 1/2 mv2Pe = mv... (4), where m is the mass of the electron, and v is the velocity of the electron, which is given in the problem as 5.40 × 10 6 m/s.
The momentum of the photon can be calculated using the following equations: E = pc... (5), where E is the energy of the photon, c is the speed of light, and p is the momentum of the photon.
The energy of the photon before scattering is equal to the energy of the photon after scattering because the scattering is elastic. Therefore, E i = E f... (6), where Ei is the energy of the incident photon and Ef is the energy of the scattered photon.
The energy of a photon can be expressed as E = hc/λ... (7), where h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon.
Substituting equations (1) through (7) into equation (3) and solving for λ gives: λ = h/(mc)(1−cosθ)+(h/mcλ′)
Substituting the given values into the above equation:λ = [(6.63 × 10-34)/(9.11 × 10-31 × 3 × 108)](1 - cos 180°) + [(6.63 × 10-34)/(9.11 × 10-31 × 5.40 × 106)]λ = 1.03 × 10-11 + 2.54 × 10-11λ = 3.57 × 10-11 m
Therefore, the wavelength of the incident X-ray photon is 3.57 × 10-11 m.
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a. Lights enters an unknown material from air at 47 degrees and is refracted to 28.1 degrees. Find the index of refraction of the new material
b. (Image is for b) This image shows two mirrors with a 120 degree angle between them. The incident angle to the first mirror is 65 degrees. What is the angle of reflection off of the second mirror?
a. The index of refraction of the new material is approximately 1.51.
b. The angle of reflection off the second mirror is 55 degrees.
a. To find the index of refraction of the new material, use the formula:n1sinθ1 = n2sinθ2, where n1 is the index of refraction of the original material (in this case, air), θ1 is the angle of incidence, n2 is the index of refraction of the new material, and θ2 is the angle of refraction. Substitute the given values: n1 = 1, θ1 = 47°, θ2 = 28.1°.
Then, n1sinθ1 = n2sinθ2 => 1(sin47°) = n2(sin28.1°) => n2 = sin47°/sin28.1°.
This evaluates to n2 = 1.51 (rounded to two decimal places).
Therefore, the index of refraction of the new material is approximately 1.51.
b. According to the laws of reflection, the angle of incidence equals the angle of reflection.
Therefore, the angle of reflection off the first mirror is 65 degrees.
Subsequently, we need to find the angle of incidence on the second mirror to calculate the angle of reflection of it.
We know that the two mirrors are at an angle of 120 degrees between them and that the angle of incidence is equal to the angle of reflection.
Thus, the angle of incidence on the second mirror will be 120 - 65 = 55 degrees. Then, the angle of reflection off the second mirror will be the same as the angle of incidence, so it will be 55 degrees.
a. The index of refraction of the new material is approximately 1.51.
b. The angle of reflection off the second mirror is 55 degrees.
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Alternating current have voltages and currents through
the circuit elements that vary as a function of time. Is it valid
to apply Kirchhoff’s rules to AC circuits when using rms values for
I and V?
Kirchhoff's rules can be applied to AC circuits using rms values for current and voltage. RMS values represent the effective values, allowing analysis of current distribution and voltage drops in AC circuits.
It is valid to apply Kirchhoff's rules to AC circuits when using rms (root mean square) values for current (I) and voltage (V). Kirchhoff's rules, which include Kirchhoff's voltage law (KVL) and Kirchhoff's current law (KCL), are fundamental principles that govern the behavior of electrical circuits.
The rms values of current and voltage represent the effective values of the alternating current or voltage. They are calculated as the square root of the average of the squares of the instantaneous values over a complete cycle. By using rms values, we can treat AC circuits in a similar manner to DC circuits.
Kirchhoff's rules state that the algebraic sum of currents at any node in a circuit is zero (KCL), and the algebraic sum of voltages in any closed loop of a circuit is zero (KVL). These rules hold true for AC circuits because they are based on the conservation of charge and energy.
By using rms values, we can effectively analyze and solve AC circuits using Kirchhoff's rules, allowing us to determine current distribution, voltage drops, and power calculations in AC circuits.
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A 200 g metal container, insulated on the outside, holds 100 g of water in thermal equilibrium at 22.0°C. A 9 g ice cube, at temperature -20.0°C, is dropped into the water, and when thermal equilibrium is reached the temperature is 12.0°C. Assume there is no heat exchange with the surroundings. Find the specific heat of the metal the container is made from. cwater = 4190 J/kg∙C°
cice = 2090 J/kg∙C°
Lf = 3.34×105 J/kg
The specific heat of the metal container is approximately 2095 J/kg∙C°.
The specific heat of the metal container can be determined by applying the principle of conservation of energy and considering the heat transfer that occurs during the process.
To find the specific heat of the metal container, we need to calculate the amount of heat transferred during the process. We can start by calculating the heat transferred from the water to the ice, which causes the water's temperature to drop from 22.0°C to 12.0°C.
The heat transferred from the water to the ice can be calculated using the formula:
Qwater → ice = mcΔT
where:
m is the mass of the water (100 g),
c is the specific heat of water (4190 J/kg∙C°), and
ΔT is the change in temperature (22.0°C - 12.0°C = 10.0°C).
Substituting the given values into the equation, we have:
Qwater → ice = (0.1 kg) * (4190 J/kg∙C°) * (10.0°C) = 4190 J
The heat transferred from the water to the ice is equal to the heat gained by the ice, causing it to melt. The heat required to melt the ice can be calculated using the formula:
Qmelting = mLf
where:
m is the mass of the ice (9 g),
Lf is the latent heat of fusion for ice (3.34×10^5 J/kg).
Substituting the given values into the equation, we have:
Qmelting = (0.009 kg) * (3.34×10^5 J/kg) = 3010 J
Since the metal container is insulated and there is no heat exchange with the surroundings, the heat transferred from the water to the ice and the heat required to melt the ice must be equal. Therefore, we can equate the two equations:
Qwater → ice = Qmelting
4190 J = 3010 J
Now, we can solve for the specific heat of the metal container (cm) by rearranging the equation:
cm = Qwater → ice / (mwater * ΔTwater)
Substituting the known values, we get:
cm = (4190 J) / ((0.2 kg) * (10.0°C)) ≈ 2095 J/kg∙C°
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Earth is about 150 million kilometers from the Sun, and the apparent brightness of the Sun in our sky is about 1300 watts /m 2
. Using these two facts and the inverse square law for light, determine the apparent brightness that we would measure for the Sun if we were located at the following positions. Half Earth's distance from the Sun. Express your answer in watts per meter squared to two significant figures. Part B Twice Earth's distance from the Sun. Express your answer in watts per meter squared to two significant figures. watts /m 2
8 times Earth's distance from the Sun. Express your answer in watts per meter squared to two significant figures. Sirius A has a luminosity of 26L Sun and a surface temperature of about 9400 K. What is its radius? (Hint. See Mathematical Insight Calculating Stellar Radii.) Express your answer in meters to two significant figures.
The Sun is 5200 watts/m² (approx), at twice Earth's distance from the Sun is 325 watts/m² (approx), and at 8 times Earth's distance from the Sun is 20.3125 watts/m² (approx).The radius of Sirius A is 1.71 × 10⁹ meters (approx).
Earth is about 150 million kilometers from the Sun, and the apparent brightness of the Sun in our sky is about 1300 watts /m2.
We have to determine the apparent brightness that we would measure for the Sun if we were located at half of the Earth's distance from the Sun. Using the inverse square law for light, we know that the apparent brightness of the Sun is inversely proportional to the square of the distance from the Sun. Suppose the distance between the Sun and us is r, then the brightness is proportional to r⁻².
According to the question, when the distance between Earth and the Sun is r, the brightness is 1300 watts/m².
So, when the distance between Earth and the Sun is half of that, i.e., 0.5r, the brightness is proportional to (0.5r) ⁻² = 4r⁻². Therefore, the brightness is 4 × 1300 watts/m² = 5200 watts/m² (approx) at half of the Earth's distance from the Sun.
We have to determine the apparent brightness that we would measure for the Sun if we were located at twice Earth's distance from the Sun.
Using the inverse square law for light, we know that the apparent brightness of the Sun is inversely proportional to the square of the distance from the Sun. Suppose the distance between the Sun and us is r, then the brightness is proportional to r⁻².
According to the question, when the distance between Earth and the Sun is r, the brightness is 1300 watts/m². So, when the distance between Earth and the Sun is twice of that, i.e., 2r, the brightness is proportional to (2r)⁻² = 0.25r⁻². Therefore, the brightness is 0.25 × 1300 watts/m² = 325 watts/m² (approx) at twice Earth's distance from the Sun.
8 times Earth's distance from the Sun. Using the inverse square law for light, we know that the apparent brightness of the Sun is inversely proportional to the square of the distance from the Sun. Suppose the distance between the Sun and us is r, then the brightness is proportional to r⁻². According to the question, when the distance between Earth and the Sun is r, the brightness is 1300 watts/m². So, when the distance between Earth and the Sun is eight times of that, i.e., 8r, the brightness is proportional to (8r) ⁻² = 0.015625r⁻².
Therefore, the brightness is 0.015625 × 1300 watts/m² = 20.3125 watts/m² (approx) at 8 times Earth's distance from the Sun.
Sirius A has a luminosity of 26LSun and a surface temperature of about 9400 K. To calculate its radius, we use the following formula:
L = 4πR²σT⁴where L is the luminosity, R is the radius, T is the surface temperature, and σ is the Stefan-Boltzmann constant. Rearranging the formula to solve for R, we get: R = √(L/4πσT⁴)
Substituting the given values, we get:
R = √(26 × LSun / (4 × π × (5.67 × 10⁻⁸) × (9400)⁴) - 1.71 × 10⁹ meters (approx)
Therefore, the radius of Sirius A is 1.71 × 10⁹ meters.
Therefore, the apparent brightness that we would measure for the Sun if we were located at half of the Earth's distance from the Sun is 5200 watts/m² (approx), at twice Earth's distance from the Sun is 325 watts/m² (approx), and at 8 times Earth's distance from the Sun is 20.3125 watts/m² (approx).The radius of Sirius A is 1.71 × 10⁹ meters (approx).
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A car with a mass of 750 kg moving at a speed of 23 m/s rear-ends a truck with a mass of 1250 kg and a speed of 15 m/s. (The two vehicles are initially traveling in the same direction.) If the collision is elastic, find the final velocities of the two vehicles. (This is a 1-dimensional collision.)
The final velocities of the two vehicles, if the collision is elastic, then v₁ = 18 m/s and v₂ = 48 m/s.
It is given that, Mass of car, m₁ = 750 kg, Initial velocity of car, u₁ = 23 m/s, Mass of truck, m₂ = 1250 kg, Initial velocity of truck, u₂ = 15 m/s and the collision is elastic. Therefore, the total momentum of the system is conserved, i.e.,
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Putting the values, we get,
750 × 23 + 1250 × 15 = 750v₁ + 1250v₂
(17250 + 18750) = (750v₁ + 1250v₂)
36000 = 750v₁ + 1250v₂
(6 × 6000) = 750v₁+ 1250v₂
Now, we have two variables and only one equation. We need another equation. We can use the conservation of kinetic energy to get another equation.
Since the collision is elastic, the total kinetic energy of the system is conserved, i.e.,
(1/2)m₁*2u₁ + (1/2)m₂*2u₂ = (1/2)m₁*2v₁ + (1/2)m₂*2v₂
Putting the values, we get,
(1/2) × 750 × (23)2 + (1/2) × 1250 × (15)2 = (1/2) × 750 × 2v₁ + (1/2) × 1250 × 2v₂
Solving further, we get,
195375 = 375v₁ + 937.5v₂(195375 / 375) = v₁ + (937.5 / 375)v₂(521 / 5) = v₁ + (25 / 2)v₂
Multiplying the first equation by 25 and subtracting the second equation, we get,
15000 = (625/2)v₂
v₂ = 48 m/s
Putting the value of v₂ in the first equation, we get,
6 × 6000 = 750v1 + 1250(48)
v₁ = 18 m/s
Therefore, the final velocities of the two vehicles are:v₁ = 18 m/s , v₂= 48 m/s.
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The Starship Enterprise returns from warp drive to ordinary space with a forward speed of 60 km/s. To the crew's great surprise, a Klingon ship is 150 km directly ahead, traveling in the same direction at a mere 22 km/s. Without evasive action, the Enterprise will overtake and collide with the Klingons in just about 3.9 s . The Enterprise's computers react instantly to brake the ship. 6 of 6 Review | Constants Part A What magnitude acceleration does the Enterprise need to just barely avoid a collision with the Klingon ship? Assume the acceleration is constant.
Hint: Draw a position-versus-time graph showing the motions of both the Enterprise and the Klingon ship. Let zo = 0km be the location of the Enterprise as it returns from warp drive. How do you show graphically the situation in which the collision is "barely avoided"? Once you decide what it looks like graphically, express that situation mathematically.
The Enterprise needs to come to a stop just as it reaches position of Klingon ship. Therefore position-versus-time graph for Enterprise would be a straight line with a positive slope initially, representing its initial velocity of 60 km/s.
At the moment of collision avoidance, the Enterprise's position should match that of the Klingon ship. This means the two lines on the graph should intersect at the same point.
Mathematically, this can be expressed by setting the equations for the positions of the Enterprise and the Klingon ship equal to each other:
60t = 22t + 150
By rearranging the equation, we have: 60t - 22t = 150
38t = 150
t ≈ 3.95 seconds
Therefore, to just barely avoid a collision with the Klingon ship, the Enterprise needs to achieve an acceleration that brings it to a stop within approximately 3.95 seconds.
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) Fourier Transform of Signals a) Obtain the Fourier Transform of the signal: x(t) = e-alt where "a" is a positive real number. (4 Marks) b) Obtain the Fourier Transform of the signal: x(t) = 8(t) + sin(wot) + 3. Where 8(t) is a unit impulse function.
The Fourier Transform of the given signal is 8(ω) + (1/2j) [δ(w-w0) - δ(w+w0)] + 3δ(w) is the answer. The notation used here assumes a two-sided Fourier Transform, where the frequencies can be positive or negative.
a) Obtain the Fourier Transform of the signal x(t) = e^-at where "a" is a positive real number. A Fourier Transform is defined as the mathematical technique that decomposes a time-domain signal into its corresponding frequency-domain spectrum.
The Fourier Transform of the signal x(t) = e^-at is as follows:
X(ω) = ∫e^(-at) e^(-jωt) dt 0 ∞
= ∫e^(-(a+jω)t) dt 0 ∞
= -1/(a+jω) [-e^(-(a+jω)t)]∣∣0∞
= 1/(a+jω),
Re{a+jω}>0.
b) Obtain the Fourier Transform of the signal x(t) = 8(t) + sin(wot) + 3.
Where 8(t) is a unit impulse function.
The Fourier transform of x(t) is given as
X(ω) = F[x(t)]
= F[8(t)] + F[sin(wot)] + F[3]
= 8(ω) + (1/2j) [δ(w-w0) - δ(w+w0)] + 3δ(w).
Hence, the Fourier Transform of the given signal is 8(ω) + (1/2j) [δ(w-w0) - δ(w+w0)] + 3δ(w).
Please note that the notation used here assumes a two-sided Fourier Transform, where the frequencies can be positive or negative. If you are working with a one-sided Fourier Transform, you may need to adjust the representation accordingly.
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Your spaceship lands on an unknown planet. To determine the characteristics of this planet, you drop a wrench from 5.00 m above the ground and measure that it hits the ground 0.811 s later. (a) What is the acceleration of gravity near the surface of this planet? (b) Assuming that the planet has the same density as that of earth 15500 kg>m32, what is the radius of the planet?
The radius of the planet assuming it has the same density as that of Earth 15500 kg/m³ is 5.58 × 10³ km.
(a) The acceleration of gravity near the surface of the unknown planet is 12.3 m/s². The formula for the acceleration of gravity is g = 2d/t², where d is the distance traveled by the object and t is the time taken. Using this formula, we have: g = 2d/t² = 2(5.00 m) / (0.811 s)² = 12.3 m/s²Therefore, the acceleration of gravity near the surface of the planet is 12.3 m/s².(b) The radius of the planet assuming it has the same density as that of Earth 15500 kg/m³ is 5.58 × 10³ km. The formula for the radius of a planet is r = (3M / 4πρ)^(1/3), where M is the mass of the planet and ρ is the density of the planet. Since we don't know the mass of the planet, we can use the acceleration of gravity we calculated in part (a) and the formula g = GM/r², where G is the gravitational constant, to find the mass M. We have:G = 6.67 × 10^-11 Nm²/kg²g = GM/r²M = gr²/G = (12.3 m/s²)(5.00 m)² / (6.67 × 10^-11 Nm²/kg²) = 2.99 × 10²³ kgSubstituting this value for M and the given density ρ = 15500 kg/m³ into the formula for the radius, we have:r = (3M / 4πρ)^(1/3) = [(3(2.99 × 10²³ kg) / (4π(15500 kg/m³))]^(1/3) = 5.58 × 10³ km. Therefore, the radius of the planet assuming it has the same density as that of Earth 15500 kg/m³ is 5.58 × 10³ km.
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A boat whose velocity through the water is 14 km/h is moving in a river whose current is 6 km/in relative to the riverbed. The velocity of the boat relative to the riverbed must be between O 6 and 14 km/h 6 and 20 km/h and 14 km/h 8 and 20 km/h
A boat whose velocity through the water is 14 km/h is moving in a river whose current is 6 km/h.
To determine the velocity of the boat relative to the riverbed, we need to calculate the resultant velocity of the boat. The velocity of the boat relative to the riverbed must be between 8 km/h and 20 km/h.Resolution of the velocities can be used to determine the resultant velocity. It refers to the separation of a vector quantity into two or more components. By definition, these components are scalar components.
A velocity vector's resolution into two perpendicular components is known as a rectangular resolution.
Let’s find the resultant velocity by using the formula of the Pythagorean theorem.
Velocity of the boat relative to the riverbed = Velocity of the boat in still water + velocity of the rivercurrent
= 14 km/h + 6 km/h= 20 km/h
Using the Pythagorean theorem, the resultant velocity is determined as follows:
Resolving the velocity in the x and y directions:
Velocity in the x-direction (upstream) = V × cos θ= 20 × cos 30°
= 17.32 km/h
Velocity in the y-direction (downstream) = V × sin θ= 20 × sin 30°= 10 km/h
Therefore, the boat's velocity relative to the riverbed is between 8 km/h and 20 km/h.
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a.) a golf ball rolls off a cliff horizontally with a speed of 15.9 m/s. the cliff is a vertical distance of 14.8 m above the surface of a lake below. find how long the ball was in the air.
b.) what is the impact speed of the ball just as it reaches the surface of the water?
(a) The ball was in the air for approximately [tex]\sqrt{3}[/tex] seconds.
(b) The impact speed of the ball as it reaches the surface of the water is 15.9 m/s.
a) To find how long the ball was in the air, we can use the equation of motion for vertical motion:
Δy = v₀y × t + (1/2) × g × t²
Where:
Δy is the vertical distance (14.8 m),
v₀y is the initial vertical velocity (0 m/s since the ball is rolling horizontally),
t is the time,
g is the acceleration due to gravity (-9.8 m/s²).
Since the initial vertical velocity is 0 m/s, the equation simplifies to:
Δy = (1/2) × g × t²
Plugging in the values, we have:
14.8 = (1/2) × (-9.8) × t²
Simplifying the equation:
14.8 = -4.9 × t²
Dividing both sides by -4.9:
t² = -14.8 / -4.9
t² = 3
Taking the square root of both sides:
t = [tex]\sqrt{3}[/tex]
So, the ball was in the air for approximately [tex]\sqrt{3}[/tex] seconds.
b) To find the impact speed of the ball just as it reaches the surface of the water, we can use the equation of motion for horizontal motion:
Δx = v₀x × t
Where:
Δx is the horizontal distance (which we assume to be the same as the initial speed, 15.9 m/s),
v₀x is the initial horizontal velocity (also 15.9 m/s),
t is the time.
Plugging in the values, we have:
15.9 = 15.9 × t
Solving for t:
t = 1
So, the time taken for the ball to reach the surface of the water is 1 second.
Since the horizontal velocity remains constant, the impact speed of the ball is equal to the initial horizontal velocity. Therefore, the impact speed of the ball as it reaches the surface of the water is 15.9 m/s.
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material and energy balance equations for an unsteady
compressible flow in Cartesian coordinates
In an unsteady compressible flow in Cartesian coordinates, material and energy balance equations are used to describe the conservation of mass and energy within the system. These equations are commonly referred to as the continuity equation and the energy equation, respectively.
Continuity Equation:
The continuity equation for an unsteady compressible flow in Cartesian coordinates can be expressed as:
∂ρ/∂t + ∂(ρu)/∂x + ∂(ρv)/∂y + ∂(ρw)/∂z = 0
where:
ρ is the density of the fluid,
t is the time,
u, v, and w are the components of velocity in the x, y, and z directions, respectively, and
x, y, and z are the Cartesian coordinates.
This equation represents the conservation of mass and states that the rate of change of density within a control volume, together with the divergence of the mass flux in each direction, must sum to zero.
Energy Equation:
The energy equation for an unsteady compressible flow in Cartesian coordinates can be expressed as:
∂(ρe)/∂t + ∂(ρeu)/∂x + ∂(ρev)/∂y + ∂(ρew)/∂z = -∇•(pu) + ∂(τu)/∂x + ∂(τv)/∂y + ∂(τw)/∂z + Q
where:
e is the specific internal energy of the fluid,
p is the pressure,
τ is the stress tensor,
Q is the heat transfer per unit volume,
and other terms have the same meaning as in the continuity equation.
This equation represents the conservation of energy and states that the rate of change of energy within a control volume, together with the work done by pressure forces, heat transfer, and viscous effects, must sum to zero.
These material and energy balance equations provide a mathematical framework for analyzing and predicting the behavior of unsteady compressible flows in Cartesian coordinates. They are essential tools in fields such as fluid dynamics, aerodynamics, and thermodynamics for understanding the flow and energy exchange processes within a system.
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As shown in the figure, where V = 0 at infinity, what is the net electric potential at P due to the q1= 3.8, q2 = 3.8, q3 = 2.5, q4 = 6, q5 = 4.6, q6 = 8.6 with d =9.1.
The net electric potential at P due to charges q1, q2, q3, q4, q5, q6 is 13.47 x 10⁹ V
Given, q1= 3.8 μC, q2 = 3.8 μC, q3 = 2.5 μC, q4 = 6 μC, q5 = 4.6 μC, q6 = 8.6 μC and d =9.1. We have to find the net electric potential at P due to these charges.Let V1, V2, V3, V4, V5, V6 be the electric potentials at point P due to charges q1, q2, q3, q4, q5, q6 respectively.
Also, let VP be the resultant potential at P due to all charges.We know that the electric potential at any point due to a point charge q at a distance d from it is given by,V = (1/4πε) (q/d) ...........(1)Where ε is the permittivity of free space and has a constant value of 8.85 x 10⁻¹² C²/Nm².
Therefore, the electric potential at P due to charges q1, q2, q3, q4, q5, q6 can be given by,V1 = (1/4πε) (q1/d) ...........(2)V2 = (1/4πε) (q2/d) ...........(3)V3 = (1/4πε) (q3/d) ...........(4)V4 = (1/4πε) (q4/d) ...........(5)V5 = (1/4πε) (q5/d) ...........(6)V6 = (1/4πε) (q6/d) ...........(7)The net electric potential at P is given by the sum of all the potentials.
Therefore,VP = V1 + V2 + V3 + V4 + V5 + V6 ...........(8)Substituting the given values in equations (2) to (7), we get,V1 = (1/4πε) (3.8 x 10⁻⁶/9.1) = 1.35 x 10⁹ VV2 = (1/4πε) (3.8 x 10⁻⁶/9.1) = 1.35 x 10⁹ VV3 = (1/4πε) (2.5 x 10⁻⁶/9.1) = 8.85 x 10⁸ VV4 = (1/4πε) (6 x 10⁻⁶/9.1) = 2.12 x 10⁹ VV5 = (1/4πε) (4.6 x 10⁻⁶/9.1) = 1.64 x 10⁹ VV6 = (1/4πε) (8.6 x 10⁻⁶/9.1) = 3.06 x 10⁹ V.
Substituting these values in equation (8), we get,VP = 1.35 x 10⁹ + 1.35 x 10⁹ + 8.85 x 10⁸ + 2.12 x 10⁹ + 1.64 x 10⁹ + 3.06 x 10⁹= 13.47 x 10⁹ VTherefore, the net electric potential at P due to charges q1, q2, q3, q4, q5, q6 is 13.47 x 10⁹ V when V = 0 at infinity and d = 9.1 m. Answer: 13.47 x 10⁹ V.equations
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If your heart rate is 150 beats per minute during strenuous exercise, what is the time per beat in units of seconds? Answer 14. A stroboscope is set to flash every 9.00×10 −5
s. What is the frequency of the flashes? Answer 15. When an 90.0-kg man stands on a pogo stick, the spring is compressed 0.150 m. What is the force constant of the spring? Answer 16. What is the period of a 1.00−m-long pendulum?
The period of the 1.00-meter-long pendulum is approximately 2.01 seconds. The period represents the time it takes for the pendulum to complete one full swing, moving back and forth from one extreme to the other.
The period of a pendulum is the time it takes to complete one full swing. For a 1.00-meter-long pendulum, the period can be calculated using the formula T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
To find the period of a pendulum, we can use the formula T = 2π√(L/g), where T represents the period, L is the length of the pendulum, and g is the acceleration due to gravity. In this case, we have a 1.00-meter-long pendulum. The acceleration due to gravity on Earth is approximately 9.8 m/s². Plugging these values into the formula, we get:
T = 2π√(1.00/9.8)
≈ 2π√(0.102)
≈ 2π × 0.320
≈ 2.01 seconds
Therefore, the period of the 1.00-meter-long pendulum is approximately 2.01 seconds. The period represents the time it takes for the pendulum to complete one full swing, moving back and forth from one extreme to the other. This value is influenced by the length of the pendulum and the acceleration due to gravity, and it remains constant as long as these factors remain unchanged.
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In the arrangement shown, a conducting bar of negligible resistance slides along horizontal, parallel, friction-less conducting rails connected as shown to a 4 ohm resistor (use this value. Ignore the 2.0 ohm mentioned in the figure for the resistance). A uniform 1.8-T magnetic field is perpendicular to the plane of the paper. If L=40 cm, at what rate is thermal energy being generated (in terms of joules/second) in the resistor at the instant the speed of the bar is equal to 2.7 m/s ? Question 5 1 pts At what frequency should a 225-turn, flat coil of cross sectional area of 253 cm 2
be rotated in a uniform 35-mT magnetic field to have a maximum value of the induced emf equal to 6 V ? Write your answer in hertz.
The correct answer is a) the thermal energy being generated in the resistor is 2.02 J/s. and b) the frequency of rotation of the coil should be 27.68 Hz.
Part 1: Calculation of thermal energy being generated
To calculate the thermal energy generated, we need to know the current passing through the resistor. Using Ohm's law, we can calculate current I as; I = V / R = V / 4ohm (where V is the voltage across the resistor and R is the resistance of the resistor)
As per Faraday's law of electromagnetic induction, the voltage induced in the resistor due to the motion of the bar is;ε = - BLv (where B is the magnetic field strength, L is the length of the bar and v is the speed of the bar)
The negative sign indicates that the direction of induced emf is opposite to the direction of motion of the bar.
Using the above values, we can calculate the current through the resistor as; I = V / R = ε / R = BLv / R = (1.8T)(0.4m)(2.7m/s) / 4 ohm = 0.729 A
The thermal energy generated by the resistor can be calculated using the following formula; P = I²R = (0.729 A)²(4 ohm) = 2.02 W
Therefore, the thermal energy being generated in the resistor is 2.02 J/s.
Part 2: Calculation of frequency
The maximum value of the induced emf can be given by the formula;ε = NBA w sin ωt(where ε is the induced emf, N is the number of turns in the coil, B is the magnetic field strength, A is the area of the coil, w is the angular velocity, ωt is the angular displacement)If ε is maximum and sin ωt = 1;ε = NBA w = 6 VN = 225, A = 253cm² = 253 x 10⁻⁴ m²B = 35 x 10⁻³ TW =?
Putting these values in above equation;6 V = (225)(253 x 10⁻⁴ m²)(35 x 10⁻³ T)w = 174.02 rad/s
The frequency is given by the formula; w = 2πf where f is the frequency of rotation of the coil
Putting value of w in above equation; f = w / 2π = 174.02 rad/s / 2π = 27.68 Hz
Therefore, the frequency of rotation of the coil should be 27.68 Hz.
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A concave mirror is to form an image of the filament of a headlight Part A lamp on a screen 8.50 m from the mirror. The filament is 8.00 mm tall, and the image is to be 26.0 cm tall. How far in front of the vertex of the mirror should the filament be placed? Express your answer in meters. Part B What should be the radius of curvature of the mirror? Express your answer in meters.
A)The filament should be placed approximately 0.325 meters (or 32.5 cm) in front of the vertex of the mirror. B)The radius of curvature of the mirror should be approximately 0.156 meters (or 15.6 cm).
Part A: The magnification formula for a concave mirror is given by:
magnification (m) = -image height ([tex]h_i[/tex]) / object height ([tex]h_o[/tex])
Given that the image height ([tex]h_i[/tex]) is 26.0 cm and the object height ([tex]h_o[/tex]) is 8.00 mm. Converting the object height to centimetres,
object height = 0.80 cm.
Rearranging the formula, solve for the object distance:
[tex]d_o = -h_i / (m * h_o)[/tex]
Since the mirror forms a real and inverted image, the magnification (m) is negative. Substituting the given values,
[tex]d_o = -26.0 cm / (-1 * 0.80 cm) \approx 32.5 cm[/tex]
Converting the object distance to meters, the filament should be placed approximately 0.325 meters (or 32.5 cm) in front of the vertex of the mirror.
Part B: The mirror equation for a concave mirror is given by:
[tex]1 / d_o + 1 / d_i = 1 / f[/tex]
It's already determined that the object distance ([tex]d_o[/tex]) is approximately 0.325 meters. The image distance ([tex]d_i[/tex]) is the distance between the mirror and the screen, which is given as 8.50 m.
Substituting these values into the mirror equation, focal length (f):
1 / 0.325 + 1 / 8.50 = 1 / f
Simplifying the equation,
f ≈ 0.1556 m
Therefore, the radius of curvature of the mirror should be approximately 0.156 meters (or 15.6 cm).
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220V, 50Hz, n=1400rpm, equivalent circuit parameters of one phase induction motor R1=2.9ohm, R2'=3ohm, X1=X2'=3.3ohm, Xm=56ohm When operating at the rated power of the engine;
a) current drawn from the network
b) induced rotating field strength
c) If P friction = 43W, the induced mechanical power
d) calculate efficiency.
220V, 50Hz, n=1400rpm, equivalent circuit parameters of one phase induction motor R1=2.9ohm, R2'=3ohm, X1=X2'=3.3ohm, Xm=56ohm When operating at the rated power of the engine;(a)I= 7.88 amps + j8.85 amps(b)he induced rotating field strength is: 446.8 volts(c)he induced mechanical power is Pmech =1217 watts(d)η =70.4%
a) Current drawn from the network
The current drawn from the network can be calculated using the following equation:
I = V / Z
where V is the applied voltage and Z is the impedance of the motor.
The applied voltage is 220 volts, and the impedance of the motor is:
Z = R1 + jX1 = 2.9 ohms + j3.3 ohms
Therefore, the current drawn from the network is:
I = 220 volts / (2.9 ohms + j3.3 ohms) = 7.88 amps + j8.85 amps
b) Induced rotating field strength
The induced rotating field strength can be calculated using the following equation:
E = I ×Xm
where I is the current flowing through the motor and Xm is the magnetizing reactance of the motor.
The current flowing through the motor is 7.88 amps, and the magnetizing reactance of the motor is:
Xm = 56 ohms
Therefore, the induced rotating field strength is:
E = 7.88 amps ×56 ohms = 446.8 volts
c) If P friction = 43W, the induced mechanical power
The induced mechanical power can be calculated using the following equation:
Pmech = V × I × cos(phi) - Pfric
where V is the applied voltage, I is the current flowing through the motor, phi is the power factor, and Pfric is the frictional power.
The applied voltage is 220 volts, the current flowing through the motor is 7.88 amps, the power factor is 0.8, and the frictional power is 43 watts.
Therefore, the induced mechanical power is:
Pmech = 220 volts × 7.88 amps × 0.8 - 43 watts = 1260 watts - 43 watts = 1217 watts
d) Calculate efficiency
The efficiency of the motor can be calculated using the following equation:
η = Pmech / Pinput
where Pmech is the induced mechanical power and Pinput is the input power.
The input power is the power supplied to the motor by the network, which is 220 volts × 7.88 amps = 1739 watts.
Therefore, the efficiency of the motor is:
η = 1217 watts / 1739 watts = 0.704 = 70.4%
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A 33.70-kg object is moved through 2.00 m by a 1.80-N force acting in the same
direction as the distance it moves through. How much work is on the object during this
process?
A 33.70-kg object is moved through 2.00 m by a 1.80-N force acting in the same direction as the distance it moves through. the work done on the object during this process is 3.60 joules (J).
To determine the work done on the object, we can use the formula:
Work = Force × Distance × cos(θ)
Where the force and distance are given, and θ is the angle between the force and the direction of motion. In this case, the force and distance are in the same direction, so the angle θ is 0 degrees.
Given:
Force = 1.80 N
Distance = 2.00 m
θ = 0 degrees
Pllgging these values into the formula, we have:
Work = 1.80 N × 2.00 m × cos(0 degrees)
Since cos(0 degrees) = 1, the equation simplifies to:
Work = 1.80 N × 2.00 m × 1
Work = 3.60 N·m
Therefore, the work done on the object during this process is 3.60 joules (J).
The work done represents the energy transferred to the object as a result of the applied force over a given distance. In this case, a force of 1.80 N is exerted over a distance of 2.00 m in the same direction. As a result, the object gains 3.60 J of energy. This work can be used to change the object's speed, increase its potential energy, or perform other forms of mechanical work.
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lamp and a 30 02 lamp are connected in series with a 10 V battery. Calculate the following: a. the equivalent resistance Question 18 1 pts A 20 02 lamp and a 30 02 lamp are connected in series with a 10 V battery. Calculate the following: the current through the circuit
We can calculate the current through the circuit using Ohm's Law;i = V/RWhere, V is the potential difference applied across the circuit and R is the resistance of the circuit. Substituting the given values;i = 10 V / 50 Ω = 0.2 ATherefore, the current through the circuit is 0.2 A.
Given information:Two lamps - a 20 Ω lamp and a 30 Ω lamp are connected in series with a 10 V battery.To calculate: The equivalent resistance and current through the circuit.The equivalent resistance of the circuit is given by;Req = R1 + R2Where, R1 and R2 are the resistances of the lamps in the circuit.Substituting the given values;Req = 20 Ω + 30 Ω = 50 ΩThe equivalent resistance of the circuit is 50 Ω.Now, we can calculate the current through the circuit using Ohm's Law;i = V/RWhere, V is the potential difference applied across the circuit and R is the resistance of the circuit. Substituting the given values;i = 10 V / 50 Ω = 0.2 ATherefore, the current through the circuit is 0.2 A.
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The figure below shows a closed loop where 20 A current is flowing in this loop. A uniform magnetic field of 3.0 T in the -x axis direction. The loop is in a plane that is 30 degrees with the yz−plane. Find: a. The y-component of the magnetic force on the segment AB of the loop. N b. The torque magnitude that the magnetic field exerts on the loop. N.m
a. The y-component of the magnetic force on the segment AB of the loop is zero. b. The torque magnitude that the magnetic field exerts on the loop can be calculated using the formula τ = IAB × B × sin(θ), where I is the current, AB is the area of the loop, B is the magnetic field, and θ is the angle between the loop and the magnetic field.
a. The y-component of the magnetic force on the segment AB of the loop is zero because the magnetic field is directed in the -x axis direction, perpendicular to the y-axis. The magnetic force experienced by a current-carrying segment is given by the equation F = I × L × B × sin(θ), where I is the current, L is the length of the segment, B is the magnetic field, and θ is the angle between the segment and the magnetic field.
In this case, the segment AB is parallel to the magnetic field (θ = 90°), resulting in sin(90°) = 1, but the y-component of the force is zero because the force is in the x-direction.
b. The torque magnitude that the magnetic field exerts on the loop can be calculated using the formula τ = IAB × B × sin(θ), where I is the current, AB is the area of the loop, B is the magnetic field, and θ is the angle between the loop and the magnetic field.
The torque acts to rotate the loop around an axis perpendicular to the plane of the loop. To calculate the torque, we need to determine the area of the loop and the angle θ. Once these values are known, we can plug them into the formula to find the torque magnitude.
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Please answer electronically, not manually
5- Are there places where the salty electrical engineer can earn outside his official working hours?
As an Electrical Engineer, you can find several ways to earn extra money outside your official working hours by working as Online tutor, Freelancer, part time teacher etc.
1. Online Tutoring: You can use your engineering degree and expertise to tutor students online. There are several online tutoring websites available where you can register yourself and start teaching students in your free time.
2. Freelancing: Several freelancing websites are available that provide opportunities for Engineers to work on projects. You can register yourself and find work in your domain and complete projects in your free time.
3. Part-time teaching: If you are interested in teaching, you can work as a part-time lecturer or tutor in educational institutions.
4. Content creation: You can use your technical knowledge to create content for technical websites or blogs. You can also start your own blog and earn money through ads.
5. Consulting: As an engineer, you can provide consultancy services to companies or individuals. You can use your expertise to solve their technical problems and earn some extra cash.
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Vector A points in the negative z direction. Vector points at an angle of 31.0" above the positive z axis. Vector C has a magnitude of 16 m and points in a direction 42.0* below the positive x axis. Part B Express your answer using two significant figures. |B|= ________ m
Vector A points in the negative z direction. Vector points at an angle of 31.0" above the positive z axis. Vector C has a magnitude of 16 m and points in a direction 42.0* below the positive x axis.
Vector A, A = {0, 0, -a}
Vector C, C = {16 cos 42.0°, 0, - 16 sin 42.0°}
Let B = A + B + C. Hence, B = {0, 0, -a} + {B sin 31.0° cos θ, B sin 31.0° sin θ, B cos 31.0°} + {16 cos 42.0°, 0, - 16 sin 42.0°}
Then, equating the x, y, and z components of the above equation separately, we get:
B sin 31.0° cos θ = - 16 cos 42.0°B sin 31.0° sin θ = 0
B cos 31.0° = a - 16 sin 42.0°
From the second equation, we have B = 0 or sin θ = 0, we have B = 0. But, B = 0 doesn't satisfy the third equation. Hence, sin θ = 0. So, θ = 0° or θ = 180°.When θ = 0°, we get,
B sin 31.0° cos θ = - 16 cos 42.0°B sin 31.0° (1) = - 16 cos 42.0°
B = - 16 cos 42.0° / sin 31.0°
Then, |B| = 22 m (approx.)
So, the required value of |B| is 22 m (approx.)
Note: You can also solve it by using the dot product of the vectors.
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You are given a lens that is thinner in the center than at the edges. Questions 23-24 refer to this lens. This lens is: Concave Diverging Convex Converging You are given a lens that is thinner in the center than at the edges. Questions 23-24 refer to this lens. This lens could be used to remedy: Glaucoma Cataracts Nearsightedness Farsightedness
A lens that is thinner in the center than at the edges is called a concave or diverging lens. It is denoted by a negative sign (-).So the correct option is (B) diverging lens and it is used to remedy myopia or nearsightedness.
Nearsightedness occurs when the light rays entering the eye are focused in front of the retina instead of directly on it. As a result, the individual can see nearby objects more clearly than distant objects.A diverging lens is used to correct nearsightedness by spreading out the light rays entering the eye so that they are focused directly on the retina. This results in the distant objects appearing clearer to the individual.
The other options, such as glaucoma, cataracts, and farsightedness are not corrected by a diverging lens.
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. Consider the signal x = cos((2π/3)n). The signal is downsampled by a factor of two. Indicate the frequency of the resulting output, normalized by 27. (E.g., if the frequency is π/2, write 1/4)
Frequency of the resulting output, normalized by 27 is 1/3.
To determine the frequency of the resulting output after downsampling, we need to consider the original signal and the downsampling factor.
The original signal is given by x = cos((2π/3)n), where n represents the discrete time index.
When downsampling by a factor of two, every other sample of the original signal is selected, effectively reducing the sampling rate by half.
Since the original signal has a frequency of (2π/3) radians per sample, downsampling by a factor of two reduces the frequency by half as well.
Therefore, the frequency of the resulting output, normalized by 27, would be (2π/3) / 2π = 1/3.
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Photoelectric effect is observed on two metal surfaces.
Light of wavelength 300.0 nm is incident on a metal that has a work function of 2.80 eV. What is the maximum speed of the emitted electrons?
...m/s
Therefore, the maximum speed of the emitted electrons is 1.62 x 10⁶ m/s.
The photoelectric effect is observed on two metal surfaces. If light of wavelength 300.0 nm is incident on a metal that has a work function of 2.80 eV, the maximum speed of the emitted electrons is 1.62 x 10⁶ m/s. What is the photoelectric effect? The photoelectric effect, also known as the Hertz–Lenard effect, is a phenomenon in which electrons are emitted from a metal surface when light is shone on it. The photoelectric effect was initially studied by Heinrich Hertz in 1887 and later by Philipp Lenard in 1902.Latex-free answer: To calculate the maximum speed of emitted electrons using the photoelectric effect equation, we can use the following formula: KEmax = hν - φwhere KE max is the maximum kinetic energy of the ejected electron, h is Planck's constant, ν is the frequency of the incident light, and φ is the work function of the metal. Using the equation, we can convert the given wavelength of 300.0 nm to frequency by using the formula c = λν where c is the speed of light and λ is the wavelength. c = λνν = c/λν = (3.0 x 10⁸ m/s) / (300.0 x 10⁻⁹ m)ν = 1.0 x 10¹⁵ Hz, Now we can plug in the values in the equation: KE max = (6.626 x 10⁻³⁴ J s) (1.0 x 10¹⁵ Hz) - (2.80 eV)(1.60 x 10⁻¹⁹ J/eV)KE max = 1.06 x 10⁻¹⁹ J - 4.48 x 10⁻¹⁹ JKE max = -3.42 x 10⁻¹⁹ J. Since KE max is a positive value, we can convert the value to speed using the equation KE = 1/2mv² where m is the mass of the electron and v is the velocity of the electron: v = √(2KE/m)v = √[(2)(3.42 x 10⁻¹⁹ J)/(9.11 x 10⁻³¹ kg)]v = 1.62 x 10⁶ m/s. Therefore, the maximum speed of the emitted electrons is 1.62 x 10⁶ m/s.
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A typical wall outlet in a place of residence in North America is RATED 120V, 60Hz. Knowing that the voltage is a sinusoidal waveform, calculate its: a. PERIOD b. PEAK VOLTAGE Sketch: c. one cycle of this waveform (using appropriate x-y axes: show the period on the y-axis and the peak voltage on the x-axis)
The typical wall outlet in North America has a rated voltage of 120V and operates at a frequency of 60Hz. The period of the voltage waveform is 1/60 seconds, and the peak voltage is ±170V.
The frequency of the voltage waveform represents the number of complete cycles per second, which is given as 60Hz. The period of the waveform can be calculated by taking the reciprocal of the frequency: 1/60 seconds. This means that the waveform completes one cycle every 1/60 seconds.
The peak voltage refers to the maximum voltage value reached by the waveform. In this case, the rated voltage is 120V, which represents the RMS voltage. Since the waveform is sinusoidal, the peak voltage can be both positive and negative. The [tex]V_{peak} = \sqrt{2} V_{RMS} = \sqrt{2} * 120 V = 170V[/tex]. Therefore, the peak voltage is ±170V, indicating that the voltage swings from positive 170V to negative 170V during each cycle.
The cycle of wave form is given below.
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Just then, you realized something---the wavelength of this man's butt beam is 525 nm. Didn't your pendulum have the print "project 525?" Was that a coincidence? When you confronted him, he said "I was just funding project 525. I was told to produce and sell as many free electrons as possible. Muons and antimuons have mean life (not half life) of 2.20 us, so it didn't take me a long time to produce 600 electrons from 1000 muons/antimuons that I was given." How long did it actually take him to do that? O 1.62 us 02.91 us 01.12 us 2.02 us
It took the man 880 μs (or 0.88 μs) to produce 600 electrons from the given 1000 muons/antimuons.
The man funded Project 525, which involved producing and selling free electrons. He was given 1000 muons/antimuons, and he managed to produce 600 electrons. Since muons and antimuons have a mean life (not half-life) of 2.20 μs, we can calculate the time it took for him to produce 600 electrons.
The mean life (τ) of a particle is related to its decay rate (λ) by the equation τ = 1/λ. In this case, the mean life of muons/antimuons is given as 2.20 μs.
The decay rate can be calculated using the formula λ = N/t, where N is the number of decays and t is the time interval. In this case, the number of decays is 1000 - 600 = 400, as 600 electrons were produced from the given 1000 muons/antimuons.
We can rearrange the formula to find the time interval: t = N/λ. Substituting the values, we have t = 400 / (1/2.20 μs) = 400 * (2.20 μs) = 880 μs.
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The drag force of a projectile in air is proportional to the square of the velocity: D=bv² Which of the following options correctly represents the dimensions of the constant b? a. m² = kg/s² b. kg/m c. m³kg d. Ns/m² e. kg/s²
The dimensions of the constant b is Ns/m². The correct option is d
The drag force of a projectile in air is proportional to the square of the velocity.
This means that D= bv²
where
D is the drag force,
v is the velocity,
b is a constant.
Therefore, the dimensions of the constant b can be obtained as follows:
Dimension of force F = MLT−2
Dimension of velocity v = LT−1
Dimension of drag coefficient b = D/F = [MLT−2]/[L2T−2] = [M/T] [1/L]2
The above is the dimensional formula for b.
To make this dimensionless constant into SI units we need to do some conversions to get the right combination of dimensions that give the correct unit.
Now, mass is in kilograms (kg), length is in meters (m), and time is in seconds (s).
Therefore, we have,
Dimension of b = [M/T] [1/L]2
= kg/s . 1/m2
= Ns/m²
Hence, the correct option is d. Ns/m².
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(a) A 0.552 kg particle has a speed of 2.17 m/s at point A and kinetic energy of 7.64 ) at point B. What is its kinetic energy at A? Submit Answer Tries 0/10 (b) What is its speed at point B? Submit Answer Tries 0/10 (c) What is the total work done on the particle as it moves from A to B?
The total work done on the particle as it moves from A to B is 11.32 J.
(a) A 0.552 kg particle has a speed of 2.17 m/s at point A and kinetic energy of 7.64 J at point B. What is its kinetic energy at A?The kinetic energy of an object is given by the formula KE = (1/2)mv², where m is the mass of the object and v is its velocity.Therefore, at point A,KE = (1/2)mv²= (1/2)(0.552 kg)(2.17 m/s)²= 1.44 J(b) What is its speed at point B?Given that, the particle has kinetic energy of 7.64 J at point B. KE = (1/2)mv², where m is the mass of the object and v is its velocity.Therefore, at point B, 7.64 J = (1/2)(0.552 kg)v²Therefore, v² = (2 x 7.64 J) / (0.552 kg) v² = 27.71
The speed of the object at point B is given by the formula, v = √(27.71) = 5.26 m/s(c) What is the total work done on the particle as it moves from A to B?
The work done on the particle as it moves from A to B is given by the difference in kinetic energy, W = ΔKEKE(B) - KE(A) = (1/2)mv(B)² - (1/2)mv(A)²= (1/2)m(v(B)² - v(A)²) = (1/2)(0.552 kg)(5.26 m/s)² - (1/2)(0.552 kg)(2.17 m/s)²= 11.32 JTherefore, the total work done on the particle as it moves from A to B is 11.32 J.
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