The runner's kinetic energy at that instant is 857.30 J, and the net work required to double the runner's speed is 2574.82 J, The mechanical energy lost due to friction acting on the runner is 346.8 J, and the base runner slides approximately 0.849 meters.
To calculate the runner's kinetic energy at the given instant, we use the formula for kinetic energy:
KE = (1/2) * m * v^2
Where KE is the kinetic energy, m is the mass of the runner, and v is the velocity. Plugging in the given values, we have
KE = (1/2) * 66.1 kg * (5.10 m/s)^2 = 857.30 J.
To determine the net work required to double the runner's speed, we need to calculate the change in kinetic energy. Doubling the speed will result in a new velocity of
2 * 5.10 m/s = 10.20 m/s.
The initial kinetic energy is
KE1 = (1/2) * 66.1 kg * (5.10 m/s)^2 = 857.30 J.
The final kinetic energy is
KE2 = (1/2) * 66.1 kg * (10.20 m/s)^2 = 3432.12 J.
The change in kinetic energy is
ΔKE = KE2 - KE1 = 3432.12 J - 857.30 J = 2574.82 J.
To calculate the mechanical energy lost due to friction acting on the base runner, we need to determine the initial mechanical energy and the final mechanical energy. Mechanical energy is the sum of kinetic energy and potential energy.
The initial kinetic energy is
KE1 = (1/2) * 60 kg * (3.4 m/s)^2 = 346.8 J.
The initial potential energy is
PE1 = 60 kg * 9.8 m/s^2 * 0 = 0 J (assuming the base is at ground level).
The initial mechanical energy is
E1 = KE1 + PE1 = 346.8 J.
The final kinetic energy is
KE2 = (1/2) * 60 kg * (0 m/s)^2 = 0 J (since the speed is zero).
The final potential energy is
PE2 = 60 kg * 9.8 m/s^2 * 0 = 0 J.
The final mechanical energy is
E2 = KE2 + PE2 = 0 J.
The mechanical energy lost is
ΔE = E2 - E1 = 0 J - 346.8 J = -346.8 J
(negative sign indicates energy loss).
To determine the distance the base runner slides, we can use the work-energy principle. The work done by friction is equal to the change in mechanical energy. The work done by friction is
W = -ΔE = -(-346.8 J) = 346.8 J.
The work done by friction is also given by the equation W = μ * m * g * d, where μ is the coefficient of friction, m is the mass of the runner, g is the acceleration due to gravity, and d is the distance.Solving for d, we have
d = W / (μ * m * g) = 346.8 J / (0.70 * 60 kg * 9.8 m/s^2)
≈ 0.849 m.
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A photon with a wavelength of 3.50×10 −13
m strikes a deuteron, splitting it into a proton and a neutron. Calculate the released kinetic energy in the unit of MeV.
The released kinetic energy in the unit of MeV is 12.48 MeV (rounded off to two decimal places).Hence, the required solution.
The given photon strikes a deuteron and splits it into a proton and a neutron. We need to calculate the released kinetic energy in the unit of MeV.Given, wavelength of the photon, λ = 3.50 × 10^-13 mSpeed of light, c = 3 × 10^8 m/sPlanck’s constant, h = 6.63 × 10^-34 J.sThe energy of a photon, E = hc/λThe energy of the photon is calculated as follows:E = hc/λ= (6.63 × 10^-34 J.s × 3 × 10^8 m/s)/ 3.50 × 10^-13 m= 5.68 × 10^-19 J
The above energy of the photon is used to split the deuteron into proton and neutron. As the deuteron is split into two particles, the total mass of the two particles is equal to the mass of the deuteron, m. The mass of the proton is 1.00728 amu, and the mass of the neutron is 1.00866 amu.
Thus, the total mass of the two particles is m = 2.01594 amu. (amu is the atomic mass unit)The mass of 1 amu is 1.66054 × 10^-27 kg.The total mass, m = 2.01594 amu = 2.01594 × 1.66054 × 10^-27 kg = 3.34402 × 10^-27 kgAs the deuteron splits into proton and neutron, there is a decrease in the mass of the particles by an amount Δm.Δm = 2m(1 - mp/m)
Where mp is the mass of the proton and m is the mass of the deuteron.Substituting the values,Δm = 2 × 3.34402 × 10^-27 (1 - 1.00728/2.01594)= 2.22557 × 10^-29 kgThe kinetic energy released in this reaction is given by E = Δmc^2Substituting the values,E = Δmc^2= (2.22557 × 10^-29 kg) × (3 × 10^8 m/s)^2= 2.00301 × 10^-12 JConverting this to MeV,1 eV = 1.602 × 10^-19 J1 MeV = 10^6 eVThus, E = 2.00301 × 10^-12 J= (2.00301 × 10^-12 J)/(1.602 × 10^-19 J/MeV)= 12.48 MeV
The released kinetic energy in the unit of MeV is 12.48 MeV (rounded off to two decimal places).Hence, the required solution.
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An electron moves along the z-axis with v. = 5.5 x 107 m/s. As it passes the origin, what are the strength and direction of the magnetic field at the following (x, y, z) positions? (0 cm, 2.0 cm , 1.0 cm) Express your answers in teslas separated by commas.
At the position (0 cm, 2.0 cm, 1.0 cm), the magnetic field strength is approximately -8.22 × 10^-13 T in the x-direction, and the magnetic field is zero in the y and z-directions.
To calculate the strength and direction of the magnetic field at a given point due to the motion of an electron, we can use the Biot-Savart law. The Biot-Savart law states that the magnetic field at a point due to a moving charged particle is given by:
B = (μ₀ / 4π) * (q * v × r) / r³
Where:
B is the magnetic field
μ₀ is the permeability of free space (4π × 10^-7 T·m/A)
q is the charge of the electron (-1.6 × 10^-19 C)
v is the velocity vector of the electron
r is the vector pointing from the electron to the point of interest
Let's calculate the magnetic field at the given point (0 cm, 2.0 cm, 1.0 cm):
Position vector r = (0 cm, 2.0 cm, 1.0 cm)
First, let's convert the position vector from centimeters to meters:
r = (0.00 m, 0.02 m, 0.01 m)
Now we can calculate the magnetic field using the given velocity vector:
v = 5.5 × 10^7 m/s in the z-direction
Plugging the values into the Biot-Savart law equation:
B = (μ₀ / 4π) * (q * v × r) / r³
B = (4π × 10^-7 T·m/A / 4π) * (-1.6 × 10^-19 C * (0, 0, 5.5 × 10^7 m/s) × (0.00, 0.02, 0.01) / (0.00² + 0.02² + 0.01²)^(3/2)
B = (-1.6 × 10^-19 C * (0, 0, 5.5 × 10^7 m/s) × (0.00, 0.02, 0.01) / (0.0005)^(3/2)
B = (-1.6 × 10^-19 C * (0, 0, 5.5 × 10^7 m/s) × (0.00, 0.02, 0.01) / 0.00353553
B = (-1.6 × 10^-19 C * (0, 0, 5.5 × 10^7 m/s) × (0.00, 0.02, 0.01) / 0.00353553
B ≈ (-8.22 × 10^-13 T, 0 T, 0 T)
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An incandescnet light bulb generates an unpolarized light beam that is directed towards three polarizing filters. The first one is oriented with a horizontal transmission axis. The second and third filters have transmission axis 20.0° and 40.0°from the horizontal, respectively. What percent of the light gets through this combination of filters?
An incandescnet light bulb generates an unpolarized light beam that is directed towards three polarizing filters. Percent transmitted =is given by (Total intensity / I₀) * 100
When an unpolarized light beam passes through a polarizing filter, it becomes partially polarized, meaning its electric field vectors align in a specific direction. The intensity of the light passing through the filter depends on the angle between the transmission axis of the filter and the polarization direction of the light.
In this case, we have three polarizing filters:
1. First filter: Transmission axis is horizontal (0° from the horizontal).
2. Second filter: Transmission axis is 20.0° from the horizontal.
3. Third filter: Transmission axis is 40.0° from the horizontal.
The intensity of light passing through each filter is given by Malus' Law:
I = I₀ * cos²(θ)
Where I₀ is the initial intensity of the light, and θ is the angle between the polarization direction of the light and the transmission axis of the filter.
For the first filter with a horizontal transmission axis, the angle θ is 0°. Therefore, the intensity remains unchanged: I₁ = I₀.
For the second filter with a transmission axis 20.0° from the horizontal, the angle θ is 20.0°. The intensity passing through the second filter is given by: I₂ = I₀ * cos²(20.0°).
For the third filter with a transmission axis 40.0° from the horizontal, the angle θ is 40.0°. The intensity passing through the third filter is given by: I₃ = I₀ * cos²(40.0°).
To find the total intensity of light passing through the combination of filters, we multiply the intensities of each filter together:
Total intensity = I₁ * I₂ * I₃ = I₀ * cos²(20.0°) * cos²(40.0°)
To find the percentage of light transmitted, we divide the total intensity by the initial intensity I₀ and multiply by 100:
Percent transmitted = (Total intensity / I₀) * 100
By substituting the values and calculating, we can determine the percentage of light that gets through the combination of filters.
It's important to note that the percentage of light transmitted will depend on the specific values of the angles and the characteristics of the polarizing filters used.
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Calculate the angular momenta of the earth due to its rotational motion about its own axis (effect days and nights) and due to its rotational motion around the sun (effect season change).
The angular momenta about its own axis is7.2 *[tex]10^{33}[/tex] kg[tex]ms^{2}[/tex][tex]s^{-1}[/tex].The angular momenta of earth around the sun is 2.663x[tex]10^{40}[/tex] kg[tex]m^{2} s^{-1}[/tex]
To calculate the angular momenta of the Earth, we need to consider two components: Angular momentum due to the Earth's rotational motion about its own axis (causing day and night).
Angular momentum due to the Earth's rotational motion around the Sun (causing season change).Let's calculate each component separately:
Angular momentum due to the Earth's rotational motion about its own axis:The formula for angular momentum is given by L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
The moment of inertia for a solid sphere rotating about its axis is given by I = (2/5) * M * R^2, where M is the mass of the Earth and R is the radius of the Earth.
The angular velocity of the Earth's rotation about its own axis is approximately ω = 2π/T, where T is the period of rotation. The period of rotation for the Earth is approximately 24 hours, which is equivalent to 86,400 seconds.
Substituting the values into the formula, we have:
L1 = I * ω = (2/5) * M * R^2 * (2π / T)=7.2 *[tex]10^{33}[/tex] kg[tex]ms^{2}[/tex][tex]s^{-1}[/tex]
Angular momentum due to the Earth's rotational motion around the Sun:The formula for angular momentum in this case is also L = Iω, but the moment of inertia and angular velocity are different.
The moment of inertia for a planet rotating around an axis passing through its center and perpendicular to its orbital plane is given by I = M * R^2, where M is the mass of the Earth and R is the average distance from the Earth to the Sun (approximately 149.6 million kilometers).
The angular velocity for the Earth's rotation around the Sun is approximately ω = 2π / T', where T' is the period of revolution. The period of revolution for the Earth around the Sun is approximately 365.25 days, which is equivalent to approximately 31,557,600 seconds.
Substituting the values into the formula, we have:
L2 = I * ω = M * R^2 * (2π / T')=2.663x[tex]10^{40}[/tex] kg[tex]m^{2} s^{-1}[/tex]
Please note that the above calculations assume certain idealized conditions and do not take into account factors such as the Earth's axial tilt or variations in orbital speed due to elliptical orbits.
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A canoe has a velocity of 0.33 m/s south relative to the river. The canoe is on a river that is flowing 0.57 m/s east relative to the earth. a) What is the magnitude and direction of the velocity of the canoe relative to the ground? (Circle one.) i. 0.66 m/s at 30° south of east ii. 0.66 m/s at 60° south of east iii. 0.66 m/s at 50° south of east iv. 0.46 m/s at 36° south of east v. 0.46 m/s at 51° south of east b) Sketch a velocity vector diagram showing the velocity of the river with respect to the ground, the velocity of the canoe with respect to the ground and the velocity of the canoe with respect to the river.
Tthe magnitude and direction of the velocity of the canoe relative to the ground is 0.66 m/s at 30° south of east, which corresponds to option ( i ).
The magnitude and direction of the velocity of the canoe relative to the ground is 0.66 m/s at 30° south of east.
To find the velocity of the canoe relative to the ground, we can add the velocities of the canoe relative to the river and the river relative to the ground. The velocity of the canoe relative to the river is given as 0.33 m/s south. The velocity of the river relative to the ground is given as 0.57 m/s east.
To add these velocities, we can use vector addition. The magnitude of the resultant velocity is found by taking the square root of the sum of the squares of the individual velocities. So, √((0.33)^2 + (0.57)^2) = 0.66 m/s.
The direction of the resultant velocity can be found using trigonometry. The angle can be calculated as arctan(0.33/0.57) = 30°. Since the canoe's velocity is south relative to the river and the river's velocity is east relative to the ground, the resultant velocity will be south of east.
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A heat engine does 180 JJ of work per cycle while exhausting 610 JJ of heat to the cold reservoir.
Part A: What is the engine's thermal efficiency? Express your answer using two significant figures.
Part B: A Carnot engine with a hot-reservoir temperature of 390 ∘C∘C has the same thermal efficiency. What is the cold-reservoir temperature in ∘C∘C?
Express your answer using two significant figures.
The thermal efficiency of the engine is 23% and the cold reservoir temperature of the carrot engine is approx. 511 °C.
Part A: The thermal efficiency of an engine can be defined as the ratio of work done by the engine to the heat energy supplied to it. It is given as: thermal efficiency = work done by the engine/heat energy supplied to the engine. From the question, work done by the engine = 180 J and heat energy exhausted to the cold reservoir = 610 J. Hence, the thermal efficiency of the engine = work done by the engine/heat energy supplied to the engine= (work done by the engine)/(heat energy supplied - heat energy exhausted to the cold reservoir)= (180 J)/(Q_h - 610 J) ... equation (1)Now, to calculate the value of Q_h, we can use the first law of thermodynamics, which states that the energy supplied to the engine is equal to the sum of work done by the engine and heat energy exhausted to the cold reservoir. Mathematically, it can be represented as: energy supplied to the engine = work done by the engine + heat energy exhausted to the cold reservoir Q_h = work done by the engine + heat energy exhausted to the cold reservoir= 180 J + 610 J= 790 J. Now, substituting this value in equation (1), we get: thermal efficiency = (180 J)/(790 J)= 0.23 or 23% (approx). Hence, the thermal efficiency of the engine is 23% (approx).
Part B: Let T_h and T_c be the hot and cold reservoir temperatures of the Carnot engine, respectively. Then, the thermal efficiency of a Carnot engine is given by: thermal efficiency = (T_h - T_c)/T_h= (T_h/T_h) - (T_c/T_h)= 1 - (T_c/T_h)Since the Carnot engine has the same thermal efficiency as the given engine, we can equate the two expressions and solve for T_c. That is,0.23 = 1 - (T_c/T_h)T_c/T_h = 1 - 0.23 = 0.77T_c = 0.77 × T_h. Now, given that T_h = 390 °C (note that the temperature must be converted to Kelvin), we can calculate the value of T_c as:T_c = 0.77 × T_h= 0.77 × (390 + 273) K= 0.77 × 663 K= 511 K (approx)Thus, the cold-reservoir temperature of the Carnot engine is approximately 511 °C.
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A particle m=0.0020kg, is moving (v=2.0m/s) in a direction that is perpendicular to a magnetic field (B=3.0T). The particle moves in a circular path with radius 0.12m. How much charge is on the particle? Please show your work.
The problem requires determining the amount of charge on a particle moving in a circular path perpendicular to a magnetic field. The charge on the particle is approximately 0.0111 Coulombs.
When a charged particle moves in a magnetic field perpendicular to its velocity, it experiences a force that causes it to move in a circular path. This force is given by the equation F = qvB, where F is the magnetic force, q is the charge on the particle, v is its velocity, and B is the magnetic field strength.
In this case, the mass of the particle (m = 0.0020 kg), its velocity (v = 2.0 m/s), and the magnetic field strength (B = 3.0 T) is given. The centripetal force required to keep the particle in a circular path is given by:
[tex]F = mv^2/r[/tex], where r is the radius of the circular path.
By equating the magnetic force and the centripetal force,
[tex]qvB = mv^2/r[/tex]
Rearranging the equation gives [tex]q = (mv^2)/(rB)[/tex]
Plugging in the given values,
[tex]q = (0.0020 kg * (2.0 m/s)^2) / (0.12 m * 3.0 T)[/tex].
Calculating the expression yields q ≈ 0.0111 C.
Therefore, the charge on the particle is approximately 0.0111 Coulombs.
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A 225kg floor safe is being moved by thief-cats 8.5 m from its initial location. One thief pushes 12.0N at an angle of 30 ° downward and another pulls with 10.0N at an angle of 40 ° upward. What is the net work done by the thieves on the safe? How much work is done by the gravitational force and the normal force? If the safe was initially at rest, what is the speed at the end of the 8.5 m displacement?
The net work done by the thieves on the safe is 173.644 Joules, the work done by the gravitational force is -17364 Joules, and the normal force does no work.
The final speed of the safe at the end of the 8.5 m displacement is approximately 2.29 m/s.
To solve this problem, we need to calculate the net work done by the thieves, the work done by the gravitational force, and the work done by the normal force. We can then use the work-energy theorem to find the final speed of the safe.
1. Net Work Done by the Thieves:
The net work done by the thieves can be calculated by adding the work done by each thief. The work done by a force is given by the equation: work = force * displacement * cos(angle).
Thief 1:
Force = 12.0 N
Displacement = 8.5 m
Angle = 30°
Work1 = 12.0 N * 8.5 m * cos(30°)
Thief 2:
Force = 10.0 N
Displacement = 8.5 m
Angle = 40°
Work2 = 10.0 N * 8.5 m * cos(40°)
Net Work Done by the Thieves = Work1 + Work2
2. Work Done by the Gravitational Force:
The work done by the gravitational force can be calculated using the equation: work = force * displacement * cos(angle).
Force (weight) = mass * gravitational acceleration
mass = 225 kg
gravitational acceleration = 9.8 m/s² (approximate value on Earth)
Displacement = 8.5 m
Angle = 180° (opposite direction of displacement)
Work done by the gravitational force = (225 kg * 9.8 m/s²) * 8.5 m * cos(180°)
3. Work Done by the Normal Force:
Since the safe is on a flat surface and not accelerating vertically, the normal force does no work. The normal force is perpendicular to the displacement, so the angle between them is 90°, and cos(90°) = 0.
Work done by the normal force = 0
4. Final Speed of the Safe:
We can use the work-energy theorem to find the final speed of the safe. The work-energy theorem states that the net work done on an object is equal to its change in kinetic energy.
Net Work Done by the Thieves = Change in Kinetic Energy
Since the safe was initially at rest, the initial kinetic energy is zero. Therefore, the net work done by the thieves is equal to the final kinetic energy.
Net Work Done by the Thieves = (1/2) * mass * final speed^2
We can solve this equation for the final speed:
(1/2) * mass * final speed² = Net Work Done by the Thieves
final speed² = (2 * Net Work Done by the Thieves) / mass
final speed = √((2 * Net Work Done by the Thieves) / mass)
Now, let's calculate the values:
1. Net Work Done by the Thieves:
Work1 = 12.0 N * 8.5 m * cos(30°)
Work2 = 10.0 N * 8.5 m * cos(40°)
Net Work Done by the Thieves = Work1 + Work2
2. Work Done by the Gravitational Force:
Work done by the gravitational force = (225 kg * 9.8 m/s²) * 8.5 m * cos(180°)
3. Work Done by the Normal Force:
Work done by the normal force = 0
4. Final Speed of the Safe:
final speed = √((2 * Net Work Done by the Thieves) / mass)
Now, let's calculate these values:
Calculations:
Work1 = 12.0 N * 8.5 m * cos(30°) = 102.180 J
Work2 = 10.0 N * 8.5 m * cos(40°) = 71.464 J
Net Work Done by the Thieves = Work1 + Work2 = 173.644 J
Work done by the gravitational force = (225 kg * 9.8 m/s^2) * 8.5 m * cos(180°) = -17364 J (negative sign indicates work done against the gravitational force)
Work done by the normal force = 0 J
final speed = √((2 * Net Work Done by the Thieves) / mass) = sqrt((2 * 173.644 J) / 225 kg) = 2.29 m/s (approximately)
Therefore, the net work done by the thieves on the safe is 173.644 Joules, the work done by the gravitational force is -17364 Joules, and the normal force does no work. The final speed of the safe at the end of the 8.5 m displacement is approximately 2.29 m/s.
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Calculate the capacitance of the capacitor (pF) in the given scenario.
There are two plates in a parallel-plate capacitor with A=3cm² with a separation of d=0.5mm. A wedge with insulating material is placed between the plates and provides capacitor with max voltage of 35000V. Provide the answer two places right of the decimal. Must be in pF
The capacitance of the capacitor is 53.12 pF.
The formula to calculate the capacitance of the capacitor is given as;
C = ε * A/d Where,
C is capacitance of the capacitor,
ε is the permittivity of the insulating material placed between the plates,
A is the area of the plates of the capacitor,
d is the separation between the plates of the capacitor.
The given area A = 3cm² = 3 × 10⁻⁴ m²
The given separation between the plates d = 0.5 mm = 0.5 × 10⁻³ m
Now, the permittivity of air is taken as 8.854 × 10⁻¹² F/m
C = ε * A/d
C = (8.854 × 10⁻¹² F/m) * (3 × 10⁻⁴ m²) / (0.5 × 10⁻³ m) = 53.124 × 10⁻¹² F = 53.12 pF
Therefore, the capacitance of the capacitor is 53.12 pF.
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A 30-kg boy puts his entire weight on the small plunger of a hydraulic press. What weight can the larger piston lift if the diameters of both pistons are 1 cm and 12 cm?
The larger piston can lift a weight of 1686.42 N
The ratio of the diameter of the larger piston to the diameter of the smaller piston is 12: 1. So the ratio of the area of the larger piston to the area of the smaller piston will be (12/1)² : 1² = 144:1.
Therefore, the larger piston can lift a weight that is 144 times heavier than the weight placed on the smaller piston. Now, the smaller piston has a surface area of: (1/2)²π = 0.785 sq cm. So, if the 30 kg boy puts his entire weight on the small plunger, then the force exerted on the small plunger will be 30 kg x 9.8 m/s² = 294 N. And, this force will act over the surface area of the small plunger.
Thus, the pressure in the system will be: Pressure = Force / Area of the small piston = 294 N / 0.785 sq cm = 374.52 N/sq cm. And, this pressure will be transmitted uniformly throughout the hydraulic system.
Finally, using the formula: Pressure = Force / Area of the large piston, we can calculate the weight that the larger piston can lift.
So, the weight that the larger piston can lift will be:
Force = Pressure x Area of the large piston = 374.52 N/sq cm x (6 cm)²π / 4 = 1686.42 N.
So, the larger piston can lift a weight of 1686.42 N if the diameters of both pistons are 1 cm and 12 cm.
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An ocean-going research submarine has a 20-cm-diameter window 8.0 cm thick. The manufacturer says the window can withstand forces up to 1.0 X 100 N. What is the submarine's maximum safe depth? The pressure inside the submarine is maintained at 1.0 atm.
The maximum safe depth of the submarine is approximately 10,317 meters can be determined by calculating the pressure exerted on the window and comparing it to the manufacturer's stated limit.
To calculate the maximum safe depth of the submarine, we need to consider the pressure exerted on the window. The pressure exerted by a fluid is given by the equation P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth. In this case, the fluid is seawater.
First, we need to determine the pressure exerted on the window at the maximum safe depth. The pressure inside the submarine is maintained at 1.0 atm, which is equivalent to 101,325 Pa. We can assume that the density of seawater is approximately [tex]1,030 kg/m^3[/tex] and the acceleration due to gravity is [tex]9.8 m/s^2[/tex].
Using the equation P = ρgh, we can rearrange it to solve for h: h = P / (ρg). Plugging in the values, we have h = [tex]101,325 Pa / (1,030 kg/m^3 * 9.8 m/s^2)[/tex], which gives us the maximum safe depth of the submarine.
To find out the numerical value, we need to evaluate the expression. The maximum safe depth of the submarine is approximately 10,317 meters.
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An insulated beaker with negligible mass contains liquid water with a mass of 0.230 kg and a temperature of 83.7°C. Part A
How much ice at a temperature of −10.2°C must be dropped into the water so that the final temperature of the system will be 29.0°C ? Take the specific heat of liquid water to be 4190 J/kg. K, the specific heat of ice to be 2100 J/kg−K, and the heat of fusion for water to be 3.34×10⁵/kg.
The mass of ice to be added is 0.0685 kg.
Mass of water in the beaker = m1 = 0.230 kg
Temperature of water = T1 = 83.7 °C
Specific heat of liquid water = c1 = 4190 J/kg. K
Mass of ice to be added = m2
Temperature of ice = T2 = −10.2 °C
Specific heat of ice = c2 = 2100 J/kg. K
Heat of fusion for water = L = 3.34 × 10⁵ /kg
Final temperature of the system = T3 = 29.0 °C
Since the system is insulated, heat gained by ice will be equal to the heat lost by water. So,
m1c1(T1 - T3) = mL + m2c2(T3 - T2)
{Let L be the heat of fusion for water.}
m1c1T1 - m1c1T3 = mL + m2c2T3 - m2c2
T2m2 = [m1c1(T1 - T3) - mL] / [c2(T3 - T2)]
m2 = [(0.230 kg) × (4190 J/kg. K) × (83.7 - 29.0) °C - (0.230 kg) × (3.34 × 10⁵ /kg)] / [(2100 J/kg. K) × (29.0 - (-10.2)) °C)]≈ 0.0685 kg
Therefore, the mass of ice to be added is 0.0685 kg.
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The force on a particle is directed along an x axis and given by F= F₀(x/x₀ - 1) where x is in meters and F is in Newtons. If F₀ = 2.2 N and x₀ = 5.9 m, find the work done by the force in moving the particle from x = 0 to x = 2x₀ m. Number ______________ Units ________________
A 80 kg block is pulled at a constant speed of 3.8 m/s across a horizontal floor by an applied force of 120 N directed 43° above the horizontal. What is the rate at which the force does work on the block? Number ______________ Units ________________
Answer: 1. The work done by the force in moving the particle from x = 0 to x = 2x₀ m is 3.92 J.
2.The rate at which the force does work on the block is 334 W.
1. Finding the work done by the force in moving the particle from x = 0 to x = 2x₀ m
Using the formula, F= F₀(x/x₀ - 1) where x is in meters and F is in Newtons, and given that F₀ = 2.2 N and x₀ = 5.9 m, we can find the work done by the force in moving the particle from x = 0 to x = 2x₀ m.
The work done by the force is equal to the change in kinetic energy. Therefore, the work done by the force in moving the particle from x = 0 to x = 2x₀ m is given by,
W = K₂ - K₁ = (1/2) mv₂² - (1/2) mv₁²
where v₂ and v₁ are the final and initial velocities, respectively, and m is the mass of the particle. In this case, since the force is in the x-direction, we know that the velocity is in the x-direction as well. Therefore, we can use the kinematic equation:
v² - u² = 2as
where v and u are the final and initial velocities, respectively, a is the acceleration, and s is the displacement. We can solve for the final velocity:v = √(u² + 2as)
Using this equation, we can find the final velocity of the particle at
x = 2x₀ m.
We know that the initial velocity is zero since the particle starts from rest. Therefore,
v₂ = √(0 + 2a(2x₀)) = √(4ax₀)
Using the force equation, we can find the acceleration of the particle:
a = F/m = F₀(x/x₀ - 1)/m
Substituting the values of F₀, x₀, and m, we get
a = (2.2 N)(x/5.9 m - 1)/(1 kg) = (2.2 N/m)(x/5.9 - 1)
v₂ = √(4ax₀)
= √(4(2.2 N/m)(2x₀/5.9 - 1)(5.9 m))
= √(17.6(2x₀/5.9 - 1))
= 2.8 m/s
Now, we can find the work done by the force in moving the particle from x = 0 to x = 2x₀ m. We know that the initial velocity is zero, so the initial kinetic energy is zero.
Therefore, W = (1/2) mv₂² = (1/2)(1 kg)(2.8 m/s)² = 3.92 J.
The work done by the force in moving the particle from x = 0 to x = 2x₀ m is 3.92 J.
2. Given that a 80 kg block is pulled at a constant speed of 3.8 m/s across a horizontal floor by an applied force of 120 N directed 43° above the horizontal.
The rate at which the force does work on the block is given by:
P = Fv cosθ
where P is the power, F is the force, v is the velocity, and θ is the angle between F and v. Substituting the values given, we get
P = (120 N)(3.8 m/s) cos 43°
= (120 N)(3.8 m/s)(0.731)
= 334 W.
The rate at which the force does work on the block is 334 W.
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A wire of unknown composition has a resistance of R 0
=36.5Ω when immersed in water at 26.2 ∘
C. When the wire is placed in boiling water, its resistance rises to 71.3Ω. What is the temperature when the wire has a resistance of 41.6Ω ? Number Units
Therefore, we cannot use the temperature coefficient of resistance for copper wire, which is 0.00428/°C. We would need to know the temperature coefficient of resistance for the specific wire we are using.
The temperature when the wire has a resistance of 41.6Ω is 45.7 ∘C.What is the resistance-temperature characteristic of the wire?The equation used to solve this problem isR = R0 (1 + αΔT)where R is the resistance at temperature T, R0 is the resistance at a reference temperature T0, α is the temperature coefficient of resistance, and ΔT is the difference between T and T0.Rearranging the equation givesΔT = (R - R0) / (R0α)The temperature coefficient of resistance α for a wire of unknown composition is not given. However, the resistance-temperature characteristic for most materials is known, and the temperature coefficient of resistance can be determined from it. For a copper wire, for example, α = 0.00428/°C.Substituting the given values,R0 = 36.5ΩR = 41.6ΩT0 = 26.2°CΔT = (41.6Ω - 36.5Ω) / (36.5Ω × α)For the copper wire, ΔT = (41.6Ω - 36.5Ω) / (36.5Ω × 0.00428/°C) = 28.5°C.Therefore, the temperature when the wire has a resistance of 41.6Ω is T = T0 + ΔT = 26.2°C + 28.5°C = 54.7°C.However, we were not given the material composition of the wire. Therefore, we cannot use the temperature coefficient of resistance for copper wire, which is 0.00428/°C. We would need to know the temperature coefficient of resistance for the specific wire we are using.
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The work function of a metal surface is 4.80 x 10⁻¹⁹ J. The maximum speed of the emitted electrons is va = 730 km/s when the wavelength of the light is λA. However, a maximum speed of vB = 500 km/s is observed when the wavelength is λB. Find the wavelengths.
The wavelengths of the electrons at maximum speed 730km/s and 500 km/s are 1.008 × 10^-12 km and 6.9× 10^-13 km respectively.
What is wavelength?Wavelength is the distance between identical points (adjacent crests) in the adjacent cycles of a waveform signal propagated in space or along a wire.
Wavelength can also be defined as the distance between two successive crest or trough.
Work function of a surface is the minimum energy required to a free electrons to come out of the metal surface.
W = h( v/λ)
where h is the Planck constant = 6.63 × 10^-34 J/s
Therefore;
4.80 × 10^-19 = 6.63 × 10^-34 × 730/λ
λ = 6.63 × 10^-34 × 730)/4.80 × 10^-19
λ = 1.008 × 10^-12 km
Also
4.80 × 10^-19 = 6.63 × 10^-34 × 500/λ
λ = 6.63 × 10^-34 × 500)/4.80 × 10^-19
λ = 6.9× 10^-13 km
Therefore the wavelengths are 1.008 × 10^-12 km and 6.9× 10^-13 km
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A point source that is stationary on an x axis emits a sinusoidal sound wave at a frequency of 874 Hz and speed 343 m/s. The wave travels radially outward from the source, causing air molecules to oscillate radially inward and outward. Let us define a wavefront as a line that connects points where the air molecules have the maximum, radially outward displacement. At any given instant, the wavefronts are concentric circles that are centered on the source. (a) Along x, what is the adjacent wavefront separation? Next, the source moves along x at a speed of 134 m/s. Along x, what are the wavefront separations (b) in front of and (c) behind the source?
The adjacent wavefront separation is 39.24 centimeters. The spacetime submanifolds whose normals n annul the characteristic determinant are the wave fronts of a differential system. Wave fronts are used to propagate discontinuities.
(a) The adjacent wavefront separation along the x-axis can be determined using the formula:
λ = v/f
where λ is the wavelength, v is the speed of the wave, and f is the frequency.
Given that the frequency is 874 Hz and the speed is 343 m/s, we can calculate the wavelength:
λ = 343 m/s / 874 Hz = 39.24 centimeters
(b) When the source is moving along the x-axis at a speed of 134 m/s, the wavefront separation in front of the source can be calculated by considering the relative motion between the source and the wavefront. In this case, the source is moving towards the wavefront, which causes a Doppler shift.
The formula for the Doppler shift in frequency when the source is moving towards the observer is:
f' = (v + v_s) / (v + v_o) * f
where f' is the observed frequency, v is the speed of the wave, v_s is the speed of the source, v_o is the speed of the observer, and f is the original frequency.
In this case, the observer is stationary, so v_o = 0. We can substitute the given values into the formula to find the observed frequency. Then, we can use the observed frequency and the speed of the wave to calculate the wavefront separation.
(c) Similarly, when the source is moving along the x-axis at a speed of 134 m/s, the wavefront separation behind the source can be calculated using the same method as in part (b). The only difference is that the source is moving away from the observer, which will cause a Doppler shift in the opposite direction.
By considering the Doppler shift, we can calculate the observed frequency and then use it with the speed of the wave to determine the wavefront separation behind the source.
Note: The specific values of wavefront separations in front of and behind the source would require numerical calculations using the given values for the speed of the source, speed of the wave, and original frequency.
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A vector A is defined as: A=−2.62i^+−5.91j^. What is θA, the direction of A ? Give your answer as an angle in degrees and in standard form. Round your answer to one (1) decimal place. If there is no solution or if the solution cannot be found with the information provided, give your answer as: −1000
Answer: the answer is 67.8.
The given vector A is A = -2.62i - 5.91j.
The direction of vector A can be found using the formula θA = tan⁻¹(y/x),
where x is the horizontal component and y is the vertical component of vector A.
In this case, x = -2.62 and y = -5.91. So,
θA = tan⁻¹(-5.91/-2.62)
θA = tan⁻¹(2.25)
θA = 67.8 degrees.
Therefore, the direction of vector A is 67.8 degrees in standard form.
Thus, the answer is 67.8.
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Objective: Go through a few problems involving Newton's Laws and friction! Tasks (10 points) 1. Find the mass of a 745 N person and find the weight of an 8.20 kg mass. Use metric units! What is known? What is unknown? What is the basic equation? What is the working equation? Plug in your values. 2. A 2000 kg car is slowed down uniformly from 20.0 m/s to 5.00 m/s in 4.00 seconds. a. What average force acted on the car during that time? What is known? What is unknown? What is the basic equation? What is the working equation? Plug in your values. What is the answer? b. How far did the car travel during that time? What is known? What is unknown? What is the basic equation? What is the working equation? Plug in your values. What is the answer? 3. A 38.4-pound block sits on a level surface, and a horizontal 21.3-pound force is applied to the block. If the coefficient of static friction between the block and the surface is 0.75, does the block start to move? Hint: it may help to draw a force diagram to visualize where everything is happening. What is known? What is unknown? What is the basic equation? What is the working equation? Plug in your values. What is the answer?
The average force acted on the car during the deceleration is 7500 N.The car traveled a distance of 60 meters during the deceleration.The block does not start to move because the applied force is not sufficient to overcome the static friction.
To find the mass of a person given their weight, we use the equation weight = mass × gravity, where weight is given as 745 N. Solving for mass, we have mass = weight / gravity. Assuming standard gravity of 9.8 m/s², the mass is approximately 75.7 kg. To find the weight of a mass, we use the equation weight = mass × gravity, where mass is given as 8.20 kg. Plugging in the values, we have weight = 8.20 kg × 9.8 m/s², which gives a weight of approximately 80.2 N.
2a. To find the average force acting on the car during deceleration, we use Newton's second law, which states that force = mass × acceleration. The change in velocity is 20.0 m/s - 5.00 m/s = 15.0 m/s, and the time is given as 4.00 seconds. The acceleration is calculated as change in velocity / time, which is 15.0 m/s / 4.00 s = 3.75 m/s². Plugging in the mass of 2000 kg and the acceleration, we have force = 2000 kg × 3.75 m/s² = 7500 N.
2b. To determine the distance the car traveled during deceleration, we can use the equation of motion x = x₀ + v₀t + 0.5at². Since the car is slowing down, the final velocity is 5.00 m/s, the initial velocity is 20.0 m/s, and the time is 4.00 seconds. Plugging in these values and using the equation, we get x = 0 + 20.0 m/s × 4.00 s + 0.5 × (-3.75 m/s²) × (4.00 s)² = 60 meters.
To determine if the block starts to move, we need to compare the applied force to the maximum static friction. The equation for static friction is fs ≤ μs × N, where fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force. The normal force is equal to the weight of the block, which is given as 38.4 pounds. Converting the weight to Newtons, we have N = 38.4 lb × 4.45 N/lb = 171.12 N. Plugging in the values, we have fs ≤ 0.75 × 171.12 N. Since the applied force is 21.3 pounds, which is less than the maximum static friction, the block does not start to move.
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a gravitational wave signal. - Evaluate what characteristics of a gravitational wave signal give us information about the source. How does the total mass of the merging black hole system affect the amplitude (height) of the gravitational wave signal? How does the distance to the merging black hole pair affect the amplitude of the gravitational wave signal? How does the total mass affect the period of the gravitational wave signal? How does the distance to the merging black hole pair affect the period of the gravitational wave signal? What is the best estimate for the distance to the merging black holes? What is the best estimate for the total mass of the merging black holes? Reflection - In the box below, describe how theoretical models can be used to determine the properties of merging black holes in galaxies very far from our own.
Gravitational wave signals provide information about the source, such as total mass and distance of merging black holes. Theoretical models and observations of gravitational waves help determine properties of merging black holes and their impact on the surrounding environment.
Gravitational wave signals have certain characteristics that provide valuable information about their source. The strength of a gravitational wave signal is dependent on the total mass of the black hole system undergoing a merger. By studying the amplitude of the signal, researchers can gather insights into the source. Additionally, the period of the gravitational wave signal is influenced by both the total mass of the merging black hole system and the distance to the black hole pair.
The amplitude (height) of a gravitational wave signal is affected by the total mass of the merging black hole system. A larger total mass results in a greater amplitude of the gravitational wave signal. Furthermore, the distance to the merging black hole pair also impacts the amplitude. If the black hole pair is closer, the amplitude of the gravitational wave signal will be higher.
Similarly, the period of the gravitational wave signal is influenced by the total mass of the merging black hole system. A larger total mass leads to a shorter period of the gravitational wave signal. The distance to the merging black hole pair also plays a role in determining the period. If the black hole pair is further away, the period of the gravitational wave signal will be longer.
In the case of the merging black holes with an estimated distance of 1.3 billion light-years and a total mass of 62 solar masses, these values provide the best estimate for their properties.
Theoretical models are utilized to understand the characteristics of merging black holes in galaxies located far from our own. These models enable scientists to make predictions about the properties of gravitational waves emitted by merging black hole systems. By comparing these predictions to actual observations of gravitational waves, scientists can gain valuable insights into the properties of merging black holes, such as their mass, spin, and distance. Theoretical models also help in studying the impact of black hole mergers on their surrounding environment, including the emission of high-energy particle jets.
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A shoe sits on a ramp without moving. As the angle of the ramp is increased, the shoe starts to move. This is because A) the component of gravity acting along the plane of the ramp has increased. B) the component of the normal force along the ramp has increased. C) the normal force has increased. D) the coefficient of static friction has decreased.
The correct answer is A) the component of gravity acting along the plane of the ramp has increased.
When an object sits on a ramp, its weight (which is the force due to gravity) can be resolved into two components: one perpendicular to the ramp (the normal force) and one parallel to the ramp. The parallel component of the weight, often referred to as the force of gravity acting along the ramp, determines the frictional force between the shoe and the ramp. For the shoe to remain at rest, the force of static friction between the shoe and the ramp must be equal to or greater than the parallel component of the weight. This static friction counteracts the tendency of the shoe to slide down the ramp.
As the angle of the ramp is increased, the ramp becomes steeper, and the angle between the ramp and the vertical direction increases. Consequently, the parallel component of the weight, which is responsible for the frictional force, increases. This increase in the parallel component of the weight provides a greater force to overcome static friction, allowing the shoe to start moving. Therefore, the shoe starts to move because the component of gravity acting along the plane of the ramp (parallel to the ramp) has increased.
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A perfectly elastic collision conserves Select all that apply. mass mechanical energy momentum
In a perfectly elastic collision, mass, mechanical energy, and momentum are conserved.
In a perfectly elastic collision, two objects collide and then separate without any loss of kinetic energy. This means that the total mechanical energy of the system remains constant before and after the collision. The conservation of mechanical energy implies that no energy is lost to other forms, such as heat or sound, during the collision.
Additionally, the law of conservation of momentum holds true in a perfectly elastic collision. Momentum, which is the product of an object's mass and velocity, is conserved before and after the collision. This means that the total momentum of the system remains constant, even though the individual objects involved in the collision may experience changes in their velocities.
Lastly, the conservation of mass is another important aspect of a perfectly elastic collision. The total mass of the system, which includes all the objects involved in the collision, remains constant throughout the collision. This principle holds true as long as there is no external force acting on the system that could change the mass.
In conclusion, a perfectly elastic collision conserves mass, mechanical energy, and momentum. These principles are fundamental to understanding the behavior of objects interacting through collisions, and they provide valuable insights into the dynamics of physical systems.
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An electron moves across Earth's equator at a speed of 2.52×10 6
m/s and in a direction 33.5 ∘
N of E. At this point, Earth's magnetic field has a direction due north, is parallel to the surface, and has a magnitude of 0.253×10 −4
T. (a) What is the magnitude of the force acting on the electron due to its interaction with Earth's magnetic field? N (b) Is the force toward, away from, or parallel to the Earth's surface? toward the Earth's surface away from the Earth's surface parallel to the Earth's surface
The magnitude of the force acting on the electron due to its interaction with Earth's magnetic field is 1.61 × [tex]10^{-17}[/tex] N and force on the electron is perpendicular to both the velocity and the magnetic field direction. Since the force is perpendicular to the Earth's surface, it is parallel to the Earth's surface.
(a) To calculate the magnitude of the force acting on the electron due to its interaction with Earth's magnetic field, we can use the formula:
F = q * v * B * sin(θ)
where:
F is the magnitude of the force,
q is the charge of the electron (1.6 × 10^-19 C),
v is the velocity of the electron (2.52 × 10^6 m/s),
B is the magnitude of Earth's magnetic field (0.253 × 10^-4 T),
θ is the angle between the velocity and the magnetic field (90° since the velocity is perpendicular to the magnetic field).
Plugging in the values, we have:
F = (1.6 × 10^-19 C) * (2.52 × 10^6 m/s) * (0.253 × 10^-4 T) * sin(90°)
Simplifying the expression, we get:
F = 1.61 × [tex]10^{-17}[/tex] N
Therefore, the magnitude of the force acting on the electron is 1.61 × [tex]10^{-17}[/tex] N.
(b) The force on the electron is perpendicular to both the velocity and the magnetic field direction.
Since the force is perpendicular to the Earth's surface, it is parallel to the Earth's surface.
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A hot wire radiates heat at 100 Watts. If its temperature measured in degrees Kelvin is doubled then the power radiated wit be what? Select one: 1. Draw a free body diagram of a hanging mass before it is submerged in water. Make sure to label your forces.
If the temperature of a hot wire measured in degrees Kelvin is doubled, the power radiated will increase by a factor of 16.
The power radiated by a hot wire is given by the Stefan-Boltzmann law:
P = σ * A * ε * T^4
where P is the power radiated, σ is the Stefan-Boltzmann constant, A is the surface area of the wire, ε is the emissivity (a measure of how effectively the wire radiates heat), and T is the temperature in Kelvin.
If the temperature T is doubled, the power radiated P' can be calculated by substituting 2T for T:
P' = σ * A * ε * (2T)^4 = σ * A * ε * 16T^4
Comparing P' to the original power P, we find that P' is 16 times greater than P:
P' = 16P
Therefore, if the temperature of the hot wire is doubled (measured in degrees Kelvin), the power radiated by the wire will increase by a factor of 16.
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43. What is precipitation hardening? 44. Diffusion is driven by two things, what are they? 45. Diffusion processes can be in two states, what are they? 46. Which Laws pertain to each type of Diffusion
43. Precipitation hardening is a heat treatment technique used to strengthen certain alloys by creating a fine dispersion of precipitates within the material, increasing its strength and hardness.
44. Diffusion is driven by two things: concentration gradient (difference in concentration) and temperature gradient (difference in temperature).
45. Diffusion processes can be in two states: Fickian diffusion and Non-Fickian diffusion.
46. Fick's first law and Fick's second law pertain to Fickian diffusion, which is the diffusion process governed by concentration gradients and follows Fick's laws.
Heat is a form of energy that is transferred between objects or systems due to temperature difference. It flows from hotter regions to colder regions until thermal equilibrium is reached. Heat can be transferred through conduction, or radiation. It is measured in units of joules (J) or calories (cal) and plays crucial role in thermodynamics and understanding thermal processes.
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6. An airplane heads from Calgary, Alberta to Sante Fe, New Mexico at [S 28.0° E] with an airspeed of 662 km/hr (relative to the air). The wind at the altitude of the plane is 77.5 km/hr [S 75 W) relative to the ground. Use a trigonometric approach to answer the following. (4 marks) a. What is the resultant velocity of the plane, relative to the ground (groundspeed)?
The resultant velocity of the plane, relative to the ground (groundspeed) is approximately 315.82 km/hr which is calculated using a trigonometric approach.
To find the groundspeed of the plane, we need to calculate the resultant velocity by considering the vector addition of the plane's airspeed and the wind velocity.
First, we decompose the airspeed into its components. The southward component of the airspeed can be found by multiplying the airspeed (662 km/hr) by the sine of the angle between the direction of the airspeed and the south direction ([tex]28.0^0[/tex]). This gives us a southward airspeed component of approximately 309.81 km/hr.
Next, we decompose the wind velocity into its components. The westward component of the wind velocity is obtained by multiplying the wind velocity (77.5 km/hr) by the cosine of the angle between the wind direction and the east direction ([tex]180^0 - 75^0 = 105^0[/tex]). This gives us a westward wind component of approximately 31.59 km/hr.
Now, we can find the resultant velocity by adding the components. The groundspeed is the magnitude of the resultant velocity and can be calculated using the Pythagorean theorem. The groundspeed is approximately 315.82 km/hr.
To summarize, the resultant velocity of the plane, relative to the ground, is approximately 315.82 km/hr. This is obtained by considering the vector addition of the plane's airspeed and the wind velocity.
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Sound level of fireworks At a fireworks show, a mortar shell explodes 25 m above the ground, momentarily radiating 75 kW of power as sound. The sound radiates from the explosion equally efficiently in all directions. You are on the ground, directly below the explosion. Calculate the sound level produced by the explosion, at your location.
The sound level produced by the fireworks explosion at your location is approximately 104.8 dB that can be calculated using the given information of power and distance.
To calculate the sound level produced by the fireworks explosion, we can use the formula for sound intensity level (L), which is given by L = 10 log(I/I0), where I is the sound intensity and I0 is the reference intensity [tex](10^{(-12)} W/m^2)[/tex].
First, we need to calculate the sound intensity (I) at the location directly below the explosion. Since the sound radiates equally in all directions, we can assume that the sound energy is spread over the surface of a sphere with a radius equal to the distance from the explosion.
The power (P) of the sound is given as 75 kW. We can use the formula [tex]P = 4\pi r^2I[/tex], where r is the distance from the explosion (25 m in this case), to calculate the sound intensity (I). Rearranging the formula, we have [tex]I = P / (4\pi r^2)[/tex].
Substituting the values into the formula, we get [tex]I = 75,000 / (4\pi(25^2)) = 75,000 / (4\pi(625)) = 0.03 W/m^2.[/tex]
Now, we can calculate the sound level (L) using the formula L = 10 log(I/I0). Substituting the values, we have[tex]L = 10 log(0.03 / 10^{(-12)}) = 10 log(3 * 10^1^0) ≈ 10 * 10.48 = 104.8 dB.[/tex]
Therefore, the sound level produced by the fireworks explosion at your location is approximately 104.8 dB.
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A motorcycle rounds a banked turn of 7% with a radius of 85m. If the friction coefficient between the tires and the road surface is 1.2 and the mass of the motorcycle with a rider is 260 kg, how fast can the motorcycle round the turn? Assume g=9.8m/s2.
please provide a detailed answer with a free body diagram. thank you (the answer is 34m/s)
The motorcycle can round the banked turn with a speed of 34 m/s.
To determine the maximum speed at which the motorcycle can round the banked turn, we need to consider the forces acting on it. A free body diagram can help visualize these forces. In this case, the relevant forces are the gravitational force (mg) acting vertically downward, the normal force (N) perpendicular to the surface of the road, and the friction force (f) acting horizontally inward.
Since the turn is banked, a component of the normal force will provide the necessary centripetal force to keep the motorcycle moving in a circular path. The angle of the banked turn can be determined using the tangent of the angle, which is equal to the coefficient of friction (μ) multiplied by the slope of the turn (7% or 0.07). Therefore, tanθ = μ = 0.07.
By resolving the forces along the vertical and horizontal directions, we can find the equations: N - mg cosθ = 0 (vertical equilibrium) and mg sinθ - f = 0 (horizontal equilibrium). Solving these equations, we can find the normal force N and the friction force f.
The centripetal force required for circular motion is given by Fc = mv^2/r, where m is the mass of the motorcycle and rider, v is the velocity, and r is the radius of the turn. Equating Fc to the horizontal force f, we can solve for v.
Using the given values of the mass (260 kg), radius (85 m), coefficient of friction (1.2), and gravitational acceleration (9.8 m/s^2), we find that the maximum speed at which the motorcycle can round the turn is approximately 34 m/s.
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A 25-m dianteter wheel accelerates uniformly about its center from 149rpm to 270rpm in 2.35. Determine the angular velocity (rad/s) of the wheel 50 s after it has started accelerating.
Given data: Diameter of the wheel (D) = 25 mInitial angular velocity (ω₁) = 149 rpmFinal angular velocity (ω₂) = 270 rpmTime taken (t) = 2.35 s.
Formula used:We know that acceleration of an object is given bya = (ω₂ - ω₁) / tThe angular velocity of an object is given byω = 2πn / 60where,n = number of rotations in 1 second.
Therefore, the angular velocity (ω) of the wheel can be calculated as:ω₁ = 2πn₁ / 60 => n₁ = ω₁ * 60 / 2πω₂ = 2πn₂ / 60 => n₂ = ω₂ * 60 / 2πa = (ω₂ - ω₁) / ta = (270 - 149) / 2.35a = 94.468 rad/s²Let the angular velocity of the wheel after 50 s be ω₃Number of rotations in 1 second = 1 / 60Total number of rotations after 50 s = 50 / 60 = 5 / 6s = ω₁t + (1/2)at²s = 149 * 2.35 + (1/2) * 94.468 * (2.35)²s = 451.50 m.
After 5 / 6 rotations, the distance covered by the wheel can be calculated as follows: Distance covered in 1 rotation = πD = 3.14 * 25 mDistance covered in 5 / 6 rotations = (5 / 6) * 3.14 * 25 m = 130.90 mThe time taken to cover this distance can be calculated as:t = s / vt = 130.90 / (25 * ω₃)t = 5.236 / ω₃Now, we can write the equation for angular velocity as:50 / 60 = ω₁ * 50 + (1/2) * 94.468 * (50)² + (1/2) * 94.468 * (5.236 / ω₃)² + ω₃ * (5.236 / ω₃)ω₃² - 10.472ω₃ + 143.245 = 0Using the quadratic formula, we get,ω₃ = [ 10.472 ± sqrt((10.472)² - 4(143.245)(1)) ] / 2ω₃ = [ 10.472 ± 42.348 ] / 2ω₃ = 26.410 rad/s (approx)Therefore, the angular velocity of the wheel 50 s after it has started accelerating is approximately 26.410 rad/s. Answer: 26.410
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Learning Goal: Photoelectric Effect The work function of calcium metal is W 0
=2.71eV. 1 electron volt (eV)=1.6×10 −19
J. Use h=6.626×10 −34
J⋅s for Planck's constant and c= 3.00×10 8
m/s for the speed of light in a vacuum. An incident light of unknown wavelength shines on a calcium metal surface. The max kinetic energy of the photoelectrons is 9.518×10 −20
J. Part A - What is the energy of each photon in the incident light? Use scientific notations, format 1.234 ∗
10 n,⋅
unit is Joules photon energy = J Part B - What is the wavelength of the incident light? Enter a regular number with 1 digit after the decimal point, in nm.1 nm=10 −9
m
A.The energy of each photon in the incident light is approximately 1.1854 × 10^-19 J
B.The wavelength of the incident light is approximately 1993 nm.
Work function (W₀) = 2.71 eV
1 electron volt (eV) = 1.6 × 10^−19 J
Max kinetic energy of photoelectrons = 9.518 × 10^−20 J
Planck's constant (h) = 6.626 × 10^−34 J·s
Speed of light in a vacuum (c) = 3.00 × 10^8 m/s
Part A: Calculating the energy of each photon in the incident light.
We know that the maximum kinetic energy (K.E.) of the photoelectrons is given by the equation:
K.E. = Energy of incident photon - Work function
Let's denote the energy of each photon as E and rearrange the equation:
E = K.E. + Work function
Substituting the given values:
E = 9.518 × 10^−20 J + 2.71 eV × 1.6 × 10^−19 J/eV
Converting eV to joules:
E = 9.518 × 10^−20 J + (2.71 eV × 1.6 × 10^−19 J/eV)
E = 9.518 × 10^−20 J + 4.336 × 10^−20 J
E = 1.1854 × 10^−19 J
So, the energy of each photon in the incident light is approximately 1.1854 × 10^−19 J.
Now, let's move on to Part B: Calculating the wavelength of the incident light.
We can use the equation E = hc/λ, where λ represents the wavelength.
Rearranging the equation, we have:
λ = hc/E
Substituting the given values:
λ = (6.626 × 10^−34 J·s × 3.00 × 10^8 m/s) / (1.1854 × 10^−19 J)
Calculating the value:
λ = 1.993 × 10^−6 m
Converting meters to nanometers:
λ = 1.993 × 10^−6 m × 10^9 nm/m
λ ≈ 1993 nm
Rounding to one decimal place, the wavelength of the incident light is approximately 1993 nm.
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An automobile manufacturer claims that its
product will, starting from rest, travel 267
min 11.0 s. What is the magnitude of the
constant acceleration required to do this?
The magnitude of the constant acceleration required to do this is `0.0000248 m/s^2`.
Initial velocity u = 0
Distance travelled from rest, s = 267 min 11.0 s=267.1833 m
Time taken t = 267 min 11.0 s=16031 s
The equation for calculating acceleration is given by the relation;`
s = ut + 1/2at^2`
Substituting the given values we get;
267.1833=0+1/2a(16031)^2
=>`a=(267.1833)/(1/2*16031^2)`=`0.0000248 m/s^2
`Therefore, the magnitude of the constant acceleration required to do this is `0.0000248 m/s^2`.
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