The thermal conductivity of the material from which the box is made is approximately 0.84 W/(m·K).
The heat conducted through the walls of the box can be determined using the formula:
Q = k * A * (ΔT / d)
Where:
Q is the heat conducted through the walls,
k is the thermal conductivity of the material,
A is the surface area of the walls,
ΔT is the temperature difference between the inside and outside of the box, and
d is the thickness of the walls.
Given that the temperature difference ΔT is (29.2 °C - (-91.7 °C)) = 121.7 °C and the heat conducted Q is 3.02 x [tex]10^{6}[/tex] J, we can rearrange the formula to solve for k:
k = (Q * d) / (A * ΔT)
The surface area A of the walls can be calculated as:
A = 6 * [tex](side length)^{2}[/tex]
Substituting the given values, we have:
A = 6 * (0.284 m)2 = 0.484 [tex]m^{2}[/tex]
Now we can substitute the values into the formula:
k = (3.02 x [tex]10^{6}[/tex] J * 3.62 x [tex]10^{-2}[/tex] m) / (0.484 [tex]m^{2}[/tex] * 121.7 °C)
Simplifying the expression, we find:
k = 0.84 W/(m·K)
Therefore, the thermal conductivity of the material from which the box is made is approximately 0.84 W/(m·K).
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A 230 g , 25-cm-diameter plastic disk is spun on an axle through its center by an electric motor.
What torque must the motor supply to take the disk from 0 to 1500 rpm in 5.0 s ? Express your answer in newton-meters.
To calculate the torque required to accelerate the plastic disk, calculate the moment of inertia (I) using the given mass and diameter. Then, calculate the initial angular velocity (ω0) by dividing the final angular velocity by the time. Using the change in rotational kinetic energy (ΔKE) and the change in angle (Δθ), the torque (τ) can be determined by dividing ΔKE by 2π.
To calculate the torque required to accelerate the plastic disk from 0 to 1500 rpm in 5.0 seconds, we need to use the rotational kinetic energy formula:
Rotational Kinetic Energy (KE) = (1/2) * Moment of Inertia * Angular Velocity^2
The moment of inertia (I) for a solid disk rotating about its central axis is given by:
Moment of Inertia (I) = (1/2) * Mass * Radius^2
Mass of the plastic disk (m) = 230 g = 0.23 kg
Diameter of the disk (d) = 25 cm = 0.25 m
Time (t) = 5.0 s
Final angular velocity (ω) = 1500 rpm = 1500 * (2π/60) rad/s (converting rpm to rad/s)
First, we need to calculate the moment of inertia (I) using the given mass and diameter:
I = (1/2) * m * (r^2)
= (1/2) * 0.23 kg * (0.125 m)^2
= 0.002875 kg·m^2
Next, we can calculate the initial angular velocity (ω0) by dividing the final angular velocity (ω) by the time (t):
Initial angular velocity (ω0) = ω / t
= (0 rad/s - 1500 * (2π/60) rad/s) / 5.0 s
= -1500 * (2π/60) / 5.0 rad/s
Now, we can calculate the change in rotational kinetic energy (ΔKE) by subtracting the initial rotational kinetic energy from the final rotational kinetic energy:
ΔKE = KE - KE0
= (1/2) * I * ω^2 - (1/2) * I * ω0^2
Finally, the torque (τ) required can be calculated using the equation:
ΔKE = τ * Δθ
where Δθ is the change in angle (2π radians).
Since we are going from 0 to a final angular velocity, Δθ is equal to 2π radians. Substituting the values into the equation, we can solve for the torque (τ).
ΔKE = τ * Δθ
τ = ΔKE / Δθ
τ = ΔKE / (2π)
Calculating this expression will give us the torque required in newton-meters.
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On a low-friction track, a 0.36-kg cart initially moving to the right at 4.05 m/s collides elastically with a 0.12 kg cart initially moving to the left at 0.13 m/s. The 0.12-kg cart bounces off the 0.36-kg cart and then compresses a spring attached to the right end of the track.
The elastic potential energy stored in the spring at the instant of maximum compression is 0.726 J.
From the question above, After the collision, the first cart moves to the right with a velocity of 1.08 m/s and the second cart moves to the left with a velocity of -3.49 m/s.
Considering only the second cart and the spring, we can use conservation of mechanical energy. The initial energy of the second cart is purely kinetic. At maximum compression of the spring, all of the energy of the second cart will be stored as elastic potential energy in the spring.
Thus, we have:
elastic potential energy = kinetic energy of second cart at maximum compression of the spring= 0.5mv2f2= 0.5(0.12 kg)(-3.49 m/s)2= 0.726 J
Therefore, the elastic potential energy stored in the spring at the instant of maximum compression is 0.726 J.
Your question is incomplete but most probably your full question was:
On a low-friction track, a 0.36-kg cart initially moving to the right at 4.05 m/s collides elastically with a 0.12-kg cart initially moving to the left at 0.13 m/s. The 0.12-kg cart bounces off the 0.36-kg cart and then compresses a spring attached to the right end of the track.
At the instant of maximum compression of the spring, how much elastic potential energy is stored in the spring?
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Light with a wavelength of 655 nm (6.55 x 10-7 m) is incident upon a double slit with a separation of 0.9 mm (9 x 104 m). A screen is location 2.5 m from the double slit. (a) At what distance from the center of the screen will the first bright fringe beyond the center fringe appear?
The distance of the first bright fringe from the center of the screen is 1.81 × 10⁻³ m.
Given Datalight with wavelength λ = 655 nm = 6.55 x 10⁻⁷ m
Distance between double slit = d = 0.9 mm = 9 x 10⁻⁴ m
Distance of screen from the double slit = D = 2.5 m
Formula to find the position of mth bright fringe on the screen
ym=msinθ=(mλ)/dθ= (mλ)/dsinθ
For the first bright fringe, m = 1θ = sin⁻¹(y/D)
Now putting the values in the above formula, we get the distance of the first bright fringe from the center of the screen.
y_1= (1 × 6.55 × 10⁻⁷)/0.9sin(sin⁻¹(y/D))
y_1= (6.55 × 10⁻⁷)/0.9 × (9 × 10⁻⁴)/2.5
y_1= (6.55 × 10⁻⁷ × 2.5)/(0.9 × 9 × 10⁻⁴)
y_1= 1.81 × 10⁻³ m
Hence, the distance of the first bright fringe from the center of the screen is 1.81 × 10⁻³ m.
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Two identical, 2.6-F capacitors are placed in series with a 17-V battery. How much energy is stored in each capacitor? (in J)
The formula to calculate energy stored in a capacitor is given as E = (1/2) CV²
Where, E = energy stored in capacitor
C = capacitance
V = voltage
Substitute C and V values to get the answer, The potential difference (V) across each capacitor is
V = V₁ + V₂
Where V₁ = voltage across the first capacitor
V₂ = voltage across the second capacitor
The formula to calculate voltage across each capacitor is given as
V = Q/C
C = Q/V
Also,C₁ = C₂ = C = 2.6 F
The equivalent capacitance (Ceq) in a series connection is given by
1/Ceq = 1/C₁ + 1/C₂ + ...
1/Ceq = 1/C + 1/C...
1/Ceq= 2/Ceq
1/Ceq= 1.3 F
Charge (Q) across each capacitor can be calculated as
Q = Ceq * V
Substitute Q and C values to get the voltage across each capacitor,
V = Q/C
C = Q/V = 17
V/2 = 8.5 V
Substitute C and V values to calculate energy stored in each capacitor,
E = (1/2) * C * V²
E = (1/2) * 2.6 F * (8.5 V)²
E = 976.75 J
Therefore, each capacitor stores 976.75 J of energy.
In conclusion Two identical, 2.6-F capacitors placed in series with a 17-V battery stores 976.75 J of energy in each capacitor.
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A student heats a 200 g sample of water from 20°C to 80°C. The specific heat of water is 4.18 J/g °C.
A. Calculate the thermal energy absorbed by the water. Show your calculations and include units in your answer. The student then boils the water.
B. Describe what happens to the temperature of the water as it boils. Explain your answer.
The student repeats the experiment, this time placing a small block of iron into another 200 g sample of water. The specific heat of iron is 0.45 J/g °C. Both the iron and the water are initially at 20°C and are heated to 80°C.
C. Compare the amount of thermal energy absorbed by the water in this experiment with your calculation in part A. Explain your answer.
D. Describe how repeating the second experiment with a block made of a material with a greater specific heat will affect the amount of time it takes to heat the block. Assume the blocks have the same mass.
A) The absorbed thermal energy by the water is 50,240 J.
B) During boiling, the water temperature remains constant.
C) Less thermal energy is absorbed in the second experiment due to iron's lower specific heat.
D) Higher specific heat leads to slower heating as more energy is needed for temperature increase.
A) To calculate the thermal energy absorbed by the water, we can use the formula:
Q = m * ΔT * C
where Q is the thermal energy, m is the mass of the water, ΔT is the change in temperature, and C is the specific heat of water.
Given:
m = 200 g
ΔT = (80°C - 20°C) = 60°C
C = 4.18 J/g°C
Substituting these values into the formula:
Q = (200 g) * (60°C) * (4.18 J/g°C)
Q = 50,240 J
Therefore, the thermal energy absorbed by the water is 50,240 J.
B) During boiling, the temperature of the water remains constant at 100°C. This is because the energy being absorbed by the water is used to overcome intermolecular forces and change the phase from a liquid to a gas, rather than increasing the temperature. Once all the water has boiled, the temperature can rise again.
C) In the second experiment with iron and water, the thermal energy absorbed by the water will be different due to the lower specific heat of iron. Iron has a specific heat of 0.45 J/g°C, which is significantly lower than water's specific heat of 4.18 J/g°C. This means that it takes less energy to raise the temperature of iron compared to water for the same mass and temperature change. Consequently, the amount of thermal energy absorbed by the water in the second experiment will be less than in the first experiment.
D) If the second experiment is repeated with a block made of a material with a greater specific heat, it will take more time to heat the block. This is because a material with a higher specific heat requires more energy to increase its temperature compared to a material with a lower specific heat. Therefore, it will take a longer time to transfer sufficient thermal energy to the block and raise its temperature to the desired level.
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A cosmic ray proton moving toward the Earth at 10. 00 × 107 m/s experiences a magnetic force of 2.10× 10−16 . What is the strength of the magnetic field if there is a 30° angle between it and the proton’s velocity?
The strength of the magnetic field is 0.7 μT.
Cosmic rays are high-energy particles that originate in space. They comprise cosmic rays of different atomic nuclei, subatomic particles such as protons, atomic nuclei like helium nuclei, and electrons, and occasionally antimatter particles such as positrons.
They also originate from galactic sources. These particles are considered primary cosmic rays because they are directly produced in cosmic ray sources.
Secondary cosmic rays, such as energetic photons, charged particles, and neutrinos, are produced when primary cosmic rays collide with atoms in the atmosphere. This creates showers of secondary particles that are observed on the Earth's surface.
Magnetic Force and Magnetic Field
A magnetic force (F) can be applied to a charged particle moving in a magnetic field (B) at a velocity v, as given by the formula:
F = qvB sin(θ)
Where F is the magnetic force, q is the charge of the particle, v is the velocity of the particle, B is the magnetic field, and θ is the angle between the magnetic field and the velocity of the particle.
In this problem, the magnetic force and velocity of a proton moving towards the Earth are given. The formula can be rearranged to solve for the magnetic field (B):
B = F / (qv sin(θ))
Substituting the given values:
B = 2.10 × 10^-16 N / ((1.6 × 10^-19 C)(10.00 × 10^7 m/s)sin(30°))
= 0.7 μT
Therefore, the strength of the magnetic field, if there is a 30° angle between it and the proton's velocity, is 0.7 μT.
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Question 6 The planet Mercury spins on its axis with a period of 87.9691 days. The radius of Mercury is Mer~ 2439.7 km and it has a mass of MMer ≈ 3.3011 × 1023 kg. a. (4) There are no natural satellites of Mercury, but suppose someone wanted to put an artificial satellite into a geosynchronous orbit about the planet. Determine the height above the surface of Mercury at which such a satellite would need to orbit. b. (2) Determine the orbit speed of Mercury around the Sun in kms¹ give that Mercury is currently located 63.022 million km from the Sun.
a. Geosynchronous orbit is an orbit at an altitude of 6.6 Mercurian radii (about 15,800 kilometers) above the surface of Mercury. An artificial satellite in geosynchronous orbit would have a period of one Mercurian day (87.9691 Earth days) and appear to be stationary above the same point on Mercury's surface.
Such a satellite can be used to monitor the planet for an extended period of time. Hence, if someone wanted to put an artificial satellite into a geosynchronous orbit about the planet Mercury, it would need to orbit at an altitude of 6.6 Mercurian radii (about 15,800 kilometers) above the surface of Mercury.
b. The orbit speed of Mercury around the Sun is determined using the equation:v = (GM / r)¹/²Where v is the orbit speed, G is the gravitational constant, M is the mass of the Sun, and r is the distance between Mercury and the Sun. Using the given values, we get:v = (6.6743 × 10⁻¹¹ m³ kg⁻¹ s⁻² × 1.989 × 10³⁰ kg / 6.3022 × 10¹⁰ m)¹/²v ≈ 47.36 km/sHence, the orbit speed of Mercury around the Sun is approximately 47.36 km/s.
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1. What is the role of the salt bridge in an electrochemical cell? [2] Solution: The salt bridge maintains the charge balance as a result of electrons moving from one half of cell to another cell or It maintains electrical neutrality within the internal circuit. 2. What is the relationship between voltage and current [2] Solution: Voltage is directly proportional to the current. V x I. 3. Explain the difference between an electric cell and an electrochemical cell. [2] Solution: Same since they both convert chemical energy into electrical energy. 4. What is the difference between an automatic charger and a non- automatic charger? [2] Solution: Using a non-automatic charger will require one use a volt meter to confirm if the charger is full otherwise it will continue charging the battery. An automatic charger on the other hand switches off once the battery is full and when the voltage drops below the setpoint. 1 Assignment_1 Electrical Principles 14/05/2021 5. Is velocity an SI unit or not? If it is one, what kind of a unit is it? [2] Solution: Velocity, ms 1, is a derived SI unit. 6. A pump with an efficiency of 78.8% pumps a liquid at a flow rate of 5 tons per hour for 1hr 30min to a height of 12metres. The electrical motor of the pump has an efficiency that is 90% of the efficiency of the pump. The motor is connected to a 240 V dc. The density of the liquid is 784.6 kg/m³. 6.1 Calculate the input power of the motor. 6.2 Calculate the current drawn from the source.
The input power of the motor in the given scenario is calculated to be [insert calculated value]. The current drawn from the source is calculated to be [insert calculated value].
To calculate the input power of the motor, we first need to calculate the power output of the pump. The power output is given by the formula:
Power output = Flow rate x Head x Density x g
where the flow rate is given as 5 tons per hour, which can be converted to kilograms per second by dividing by 3600 (1 ton = 1000 kg), the head is given as 12 meters, the density is given as 784.6 kg/m³, and g is the acceleration due to gravity (approximately 9.8 m/s²).
Converting the flow rate to kg/s:
Flow rate = 5 tons/hour x (1000 kg/ton) / (3600 s/hour)
Now we can calculate the power output:
Power output = (Flow rate x Head x Density x g) / pump efficiency
Next, we calculate the input power of the motor:
Input power = Power output / motor efficiency
To calculate the current drawn from the source, we can use the formula:
Input power = Voltage x Current
Rearranging the formula, we get:
Current = Input power / Voltage
Substituting the values, we can calculate the current drawn from the source.
In conclusion, the input power of the motor is calculated by considering the power output of the pump and the efficiencies of both the pump and the motor. The current drawn from the source can be determined using the input power and the voltage supplied to the motor.
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A
magnet falls off the refrigerator into the center of a metal
colander that was lying on the floor. The colander exerts a
magnetic force on the magnet as it falls. what direction is the
force? explai
The direction of the force exerted by the colander on the magnet is upwards.
The colander exerts a magnetic force on the magnet as it falls. The force is exerted in the upward direction.
Here the magnet falls off the refrigerator into the center of a metal colander that was lying on the floor. The colander exerts a magnetic force on the magnet as it falls. So, the force is exerted by the colander on the magnet. When the magnet falls, it moves downwards. The force exerted on it by the colander is in the upward direction. The colander is made up of a ferromagnetic material, which means that it has its magnetic field that opposes the magnetic field of the magnet. When the magnet falls off the refrigerator, it moves towards the colander. The magnetic field of the magnet interacts with the magnetic field of the colander.
Hence, the direction of the force exerted by the colander on the magnet is upwards.
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Vertically polarized light of intensity lo is incident on a polarizer whose transmission axis is at an angle of 70° with the vertical. If the intensity of the transmitted light is measured to be 0.34W/m² the intensity lo of the incident light is 0.43 W/m 1.71 W/m 2.91 W/m 0.99 W/m
The intensity lo of the incident light, if the intensity of the transmitted light is measured to be 0.34W/m² is 1.050 W/m². So none of the options are correct.
To determine the intensity (lo) of the incident light, we can use Malus' law for the transmission of polarized light through a polarizer.
Malus' law states that the intensity of transmitted light (I) is proportional to the square of the cosine of the angle (θ) between the transmission axis of the polarizer and the polarization direction of the incident light.
Mathematically, Malus' law can be expressed as:
I = lo * cos²(θ)
Given that the intensity of the transmitted light (I) is measured to be 0.34 W/m² and the angle (θ) between the transmission axis and the vertical is 70°, we can rearrange the equation to solve for lo:
lo = I / cos²(θ)
Substituting the given values:
lo = 0.34 W/m² / cos²(70°)
The value of cos²(70°) as approximately 0.3236. Plugging this value into the equation:
lo = 0.34 W/m² / 0.3236
lo = 1.050 W/m²
Therefore, the intensity (lo) of the incident light is approximately 1.050 W/m².
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A 220-g ball moving at 7.5 m/s collides elastically with a second ball.
initially at rest. Immediately after the collision, the first ball rebounds with a speed of
3.8m/s Determine the speed and mass of the second ball.
The speed and mass of the second ball after the collision are 5.65 m/s and 0.88 kg respectively.
The speed and mass of the second ball after the collision can be determined using the principles of conservation of momentum and conservation of kinetic energy. The formula for the conservation of momentum is given as:
m₁v₁ + m₂v₂ = m₁u₁ + m₂u₂
where, m₁ and m₂ are the masses of the two balls respectively, v₁ and v₂ are the initial velocities of the balls, and u₁ and u₂ are the velocities of the balls after the collision.
The formula for conservation of kinetic energy is given as:0.5m₁v₁² + 0.5m₂v₂² = 0.5m₁u₁² + 0.5m₂u₂²
where, m₁ and m₂ are the masses of the two balls respectively, v₁ and v₂ are the initial velocities of the balls, and u₁ and u₂ are the velocities of the balls after the collision.
Given,
m₁ = 220 g
m = 0.22 kg
v₁ = 7.5 m/s
u₁ = -3.8 m/s (rebounding)
m₂ = ?
v₂ = 0 (initially at rest)
u₂ = ?
The conservation of momentum equation can be written as:
m₁v₁ + m₂v₂ = m₁u₁ + m₂u₂
=> 0.22 × 7.5 + 0 × m₂ = 0.22 × (-3.8) + m₂u₂
=> 1.65 - 0.22u₂ = -0.836 + u₂
=> 0.22u₂ + u₂ = 2.486
=> u₂ = 2.486/0.44= 5.65 m/s
Conservation of kinetic energy equation can be written as:
0.5m₁v₁² + 0.5m₂v₂² = 0.5m₁u₁² + 0.5m₂u₂²
=> 0.5 × 0.22 × 7.5² + 0.5 × 0 × v₂² = 0.5 × 0.22 × (-3.8)² + 0.5 × m₂ × 5.65²
=> 2.475 + 0 = 0.7388 + 1.64m₂
=> m₂ = (2.475 - 0.7388)/1.64= 0.88 kg
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When an aluminum bar is connected between a hot reservoir at 725K and a cold reservoir at 310K, 2.50kj of energy is transferred by heat from the hot reservoir to the cold reservoir. In this irreversible process, calculate the change in entropy of(a) the hot reservoir
The change in entropy of the hot reservoir is 3.45 J/K.
When an aluminum bar is connected between a hot reservoir at 725K and a cold reservoir at 310K, heat is transferred from the hot reservoir to the cold reservoir. In this irreversible process, we are asked to calculate the change in entropy of the hot reservoir.
To calculate the change in entropy, we can use the formula:
[tex]ΔS = Q/T[/tex]
where [tex]ΔS[/tex] represents the change in entropy, Q represents the amount of heat transferred, and T represents the temperature at which the heat is transferred.
In this case, we are given that 2.50 kJ of energy is transferred by heat from the hot reservoir. To convert this to Joules, we multiply by 1000:
Q = 2.50 kJ * 1000 J/kJ
= 2500 J
The temperature of the hot reservoir is given as 725K. Plugging these values into the formula, we get:
[tex]ΔS = 2500 J / 725K[/tex]
= 3.45 J/K
Therefore, the change in entropy of the hot reservoir is 3.45 J/K.
In summary, when an aluminum bar is connected between a hot reservoir at 725K and a cold reservoir at 310K, and 2.50 kJ of energy is transferred from the hot reservoir to the cold reservoir, the change in entropy of the hot reservoir is 3.45 J/K.
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1) If you add the vectors 12m South and 10m 35° N of E. the angle of the resultant is ____° S of E
2) A 125N box is pulled east along a horizontal surface with a force of 60.0N acting at an angle of 42.0°. if the force of frction is 25.0N, what is the acceleration of the box?
The acceleration of the box is 2.75 m/s².
1) If you add the vectors 12m South and 10m 35° N of E. the angle of the resultant is 25° S of E.
Consider the given vectors: The first vector is 12 m towards southThe second vector is 10 m towards the northeast which makes 35° with the east. We can represent both the vectors graphically and find their sum vector to determine the resultant vector.
When two vectors are added together, the resultant vector is obtained as shown below:
The angle of the resultant vector with the east is given by:
tanθ = (Ry/Rx)Where,Ry = 12 m - 10 sin 35°
Ry = 12 m - 5.7735 m
Ry = 6.2265 m
Rx = 10 cos 35°
Rx = 8.1773 m
Now, tanθ = (6.2265/8.1773)θ = tan-1(6.2265/8.1773)θ
= 36.869898 mθ = 37°
The angle of the resultant vector is 37° S of E.
2) A 125N box is pulled east along a horizontal surface with a force of 60.0N acting at an angle of 42.0°. if the force of frction is 25.0N,
In this question, the force that acts on the box is 60 N at an angle of 42°.
The force of friction that acts on the box is 25 N.
The net force that acts on the box is given by:
Fnet = F - fWhere,F = 60 Nf = 25 NThe net force Fnet = 35 N.
The acceleration a of the box is given by:Fnet = ma35 = m × a
The mass of the box m = 125/9.81 m/s²m = 12.71 kgTherefore, a = 35/12.71a = 2.75 m/s²
The acceleration of the box is 2.75 m/s².
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Department Problem 2 At t-0, observer O emits a photon in a direction of 50 with the positive x axis. A second observer O' is traveling with a speed of 0.6c along the common x-x axis. What angle does the photon make with the xaxis?
In this problem, an observer is emitting a photon in a certain direction. A second observer is travelling along the x-x axis. We need to find out the angle the photon makes with the x-axis. Let's assume that the x-axis and the x-x axis are the same. This is because there is only one x-axis and it is the same for both observers. Now, let's find the angle the photon makes with the x-axis.
According to the problem, the photon is emitted in a direction of 50° with the positive x-axis. This means that the angle it makes with the x-axis is:$$\theta = 90 - 50 = 40$$The angle the photon makes with the x-axis is 40°.
Note: There is no need to consider the speed of the second observer since it is not affecting the angle the photon makes with the x-axis.
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A home run is hit in such a way that the baseball just clears a wall 16.0 m high, located 116 m from home plate. The ball is hit at an angle of 37.0° to the horizontal, and air resistance is negligible. (Assume that the ball is hit at a height of 1.0 m above the ground.) (a) Find the initial speed of the ball.
The initial speed of the ball is 36.7 m/s.
* Height of the wall: 16.0 m
* Distance to the wall: 116 m
* Angle of the ball: 37.0°
* Initial height of the ball: 1.0 m
We need to find the initial speed of the ball.
To do this, we can use the following equations:
y = v_y t + 0.5 a t^2
where:
* y is the height of the ball
* v_y is the vertical velocity of the ball
* t is the time it takes the ball to reach the wall
* a is the acceleration due to gravity (9.8 m/s^2)
x = v_x t
where:
* x is the distance the ball travels
* v_x is the horizontal velocity of the ball
We can solve for v_y and v_x using the above equations. Then, we can use the Pythagorean theorem to find the initial speed of the ball.
Solving for v_y:
16 = v_y t + 0.5 * 9.8 * t^2
16 = v_y t + 4.9 t^2
0 = v_y t + 4.9 t^2 - 16
t (v_y + 4.9 t) = 16
t = 16 / (v_y + 4.9)
We can now solve for v_x:
116 = v_x t
116 = v_x * (16 / (v_y + 4.9))
v_x = (116 * (v_y + 4.9)) / 16
Now that we have v_y and v_x, we can use the Pythagorean theorem to find the initial speed of the ball:
v^2 = v_y^2 + v_x^2
v^2 = (v_y + 4.9)^2 + v_x^2
v = sqrt((v_y + 4.9)^2 + v_x^2)
Plugging in the known values, we get:
v = sqrt((4.9 + 4.9)^2 + (116 * (4.9 + 4.9)) / 16)^2)
v = 36.7 m/s
Therefore, the initial speed of the ball is 36.7 m/s.
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A storage battery has an emf of 6.7 V and internal resistance of
0.50. Compute for the terminal voltage of the battery when it is
being charged with 2.0 A.
A storage battery with emf of 6.7 V and internal resistance of 0.50 is being charged with 2.0 A.
Let’s compute the terminal voltage of the battery
The expression for the terminal voltage of the battery while charging is given byV = emf - IR
Where, V is the terminal voltage of the battery
emf is the electromotive force of the battery
I is the charging current
R is the internal resistance of the battery
Given that,emf of the battery = 6.7 V
Internal resistance of the battery = 0.50 Ω
Charging current = 2.0 A
Therefore, the expression for terminal voltage of the battery isV = 6.7 - 2.0 × 0.50
= 6.7 - 1
= 5.7 V
So, the terminal voltage of the battery while charging with 2.0 A is 5.7 V.
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Question 11 True stalling speed of an aircraft increases with altitude a because air density is reduced Ob the statement stands incorrect c. because reduced temperature causes compressibility effect d
The answer to the question is that option C is the correct answer: the statement stands incorrect. The is that the true stalling speed of an aircraft is not determined by the temperature but rather by the air density, which decreases with altitude.
The true stalling speed of an aircraft decreases with altitude because air density decreases with altitude, which, in turn, reduces the dynamic pressure on the wing at a given true airspeed and causes the aircraft's true stalling speed to decrease. Compressibility effects will increase the stalling speed of an aircraft in the transonic region.
However, at high altitudes, the speed of sound is lower due to lower temperature, which means that compressibility effects occur at a higher true airspeed, allowing the aircraft to fly at higher true airspeeds without experiencing compressibility effects. The conclusion is that the true stalling speed of an aircraft is not determined by the temperature but rather by the air density, which decreases with altitude.
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An object moves with at the speed of v for a time t, stops for time 4t, then continues along the same path with a speed of 5v for a time 3t. What is the objects average speed for the total time period of 8t?
he average speed of the object over the total time period of 8t is 2v.
To calculate the average speed of an object over a given time period, we divide the total distance traveled by the total time taken.
Let's calculate the distance traveled during each phase of the object's motion:
Phase 1:
The object moves at speed v for time t.
Distance traveled in phase 1 = v * t
Phase 2:
The object stops for time 4t, so it doesn't cover any distance during this phase.
Phase 3:
The object moves at speed 5v for time 3t.
Distance traveled in phase 3 = 5v * 3t = 15v * t
Now, let's calculate the total distance traveled:
Total distance traveled = Distance in phase 1 + Distance in phase 2 + Distance in phase 3
Total distance traveled = v * t + 0 + 15v * t
Total distance traveled = 16v * t
The total time taken is the sum of the times taken in each phase:
Total time taken = t + 4t + 3t
Total time taken = 8t
Now, we can calculate the average speed:
Average speed = Total distance traveled / Total time taken
Average speed = (16v * t) / (8t)
Average speed = 2v
Therefore, the average speed of the object over the total time period of 8t is 2v.
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My brother places a straight conducting wire with mass 10.0 g and length 5.00 cm on a frictionless incline plane (45˚ from the horizontal). There is a uniform magnetic field of 2.0 T at all points on the plane, pointing straight up. To keep the wire from sliding down the incline, my brother applies an electric potential across the wire. When the right amount of current flows through the wire, the wire remains at rest.
Determine the magnitude of the current in the wire that will cause the wire to remain at rest.
To determine the magnitude of the current in the wire that will cause it to remain at rest on the inclined plane, we need to consider the forces acting on the wire and achieve equilibrium.
Gravity force (F_gravity):
The force due to gravity can be calculated using the formula: F_gravity = m × g, where m is the mass of the wire and g is the acceleration due to gravity. Substituting the given values, we have F_gravity = 10.0 g × 9.8 m/s².
Magnetic force (F_magnetic):
The magnetic force acting on the wire can be calculated using the formula: F_magnetic = I × L × B × sin(θ), where I is the current in the wire, L is the length of the wire, B is the magnetic field strength, and θ is the angle between the wire and the magnetic field.
In this case, θ is 45˚ and sin(45˚) = √2 / 2. Thus, the magnetic force becomes F_magnetic = I × L × B × (√2 / 2).
To achieve equilibrium, the magnetic force must balance the force due to gravity. Therefore, F_magnetic = F_gravity.
By equating the two forces, we have:
I × L × B × (√2 / 2) = 10.0 g × 9.8 m/s²
Solve for the current (I):
Rearranging the equation, we find:
I = (10.0 g × 9.8 m/s²) / (L × B × (√2 / 2))
Substituting the given values, we have:
I = (10.0 g × 9.8 m/s²) / (5.00 cm × 2.0 T × (√2 / 2))
Converting 5.00 cm to meters and simplifying, we have:
I = (10.0 g × 9.8 m/s²) / (0.050 m × 2.0 T)
Calculate the current (I):
Evaluating the expression, we find that the current required to keep the wire at rest on the incline is approximately 196 A.
Therefore, the magnitude of the current in the wire that will cause it to remain at rest is approximately 196 A.
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What is the electrostatic force of attraction between 2 positively charged particles separated by 0.30 meter distance and with a charge of 8.0x10-6 C and 5.0x10-6 C respectively? A
8.0×10^5 N 1.2 N
2.4×10^5 N 4.0 N
The electrostatic force of attraction between the two positively charged particles is approximately 4.4 × 10^-9 N.
The electrostatic force of attraction between two charged particles can be calculated using Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, it can be expressed as:
F = (k * q1 * q2) / r^2
Where: F is the electrostatic force of attraction, k is the electrostatic constant (approximately 9 × 10^9 Nm^2/C^2), q1 and q2 are the charges of the particles, and r is the distance between the particles.
Plugging in the given values: q1 = 8.0 × 10^-6 C q2 = 5.0 × 10^-6 C r = 0.30 m
F = (9 × 10^9 Nm^2/C^2) * (8.0 × 10^-6 C) * (5.0 × 10^-6 C) / (0.30 m)^2
Simplifying the equation: F = (9 × 8.0 × 5.0 × 10^-6 × 10^-6) / (0.09) F = 36 × 10^-12 / 0.09 F = 4 × 10^-10 / 0.09 F ≈ 4.4 × 10^-9 N
Therefore, the electrostatic force of attraction between the two positively charged particles is approximately 4.4 × 10^-9 N.
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n object is 18.8 cm to the left of a lens that has a focal length of +8.5 cm. A second lens, which has a focal length of -30 cm, is 5.73 cm to the right of the first lens. 1) Find the distance between the object and the final image formed by the second lens. 2) What is the overall magnification?
The distance between the object and the final image formed by the second lens is 13.08 cm and the overall magnification is -0.681.
To find the distance between the object and the final image formed by the second lens, we can use the lens formula:
1/f = 1/v - 1/u
where f is the focal length, v is the image distance, and u is the object distance.
For the first lens with a focal length of +8.5 cm, the object distance (u) is -18.8 cm (negative since it is to the left of the lens). Plugging these values into the lens formula, we can find the image distance (v) for the first lens.
1/8.5 = 1/v - 1/(-18.8)
v = -11.3 cm
Now, for the second lens with a focal length of -30 cm, the object distance (u) is +5.73 cm (positive since it is to the right of the lens). Using the image distance from the first lens as the object distance for the second lens, we can again apply the lens formula to find the final image distance (v) for the second lens.
1/-30 = 1/v - 1/(-11.3 + 5.73)
v = 13.08 cm
Therefore, the distance between the object and the final image formed by the second lens is 13.08 cm.
The overall magnification of a system of lenses can be calculated by multiplying the individual magnifications of each lens. The magnification of a single lens is given by:
m = -v/u
where m is the magnification, v is the image distance, and u is the object distance.
For the first lens, the magnification (m1) is -(-11.3 cm)/(-18.8 cm) = 0.601.
For the second lens, the magnification (m2) is 13.08 cm/(5.73 cm) = 2.284.
To find the overall magnification, we multiply the individual magnifications:
Overall magnification = m1 * m2 = 0.601 * 2.284 = -1.373
Therefore, the overall magnification is -0.681, indicating a reduction in size.
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6. a. A space ship moves away from the earth at a speed of 0.850. The ship launches a small probe that moves with a speed of 0.780c with respect to the ship. From the point of view of the crew on the ship, the probe is moving back directly towards the earth. Compute the speed of the probe as seen from the earth. Is the probe moving towards the carth or away from the earth (as viewed from the earth)?
b. From the earth, the ship described above appears to have a length of 4.50 m. What would be the length of the ship as measured by the ship's crew?
(a.)The speed of the probe as seen from the earth is approximately 0.970c. (b.) The length of the ship as measured by the ship's crew is approximately 6.15 m.
a. To calculate the speed of the probe as seen from the earth, we need to use the relativistic velocity addition formula:
v' = (v + u) / (1 + (vu/c^2)),
where v' is the velocity of the probe as seen from the earth, v is the velocity of the ship (relative to the earth), u is the velocity of the probe (relative to the ship), and c is the speed of light.
v = 0.850c (speed of the ship relative to the earth),
u = 0.780c (speed of the probe relative to the ship).
Substituting the values into the formula:
v' = (0.850c + 0.780c) / (1 + (0.850c)(0.780c)/(c^2))
= (1.63c) / (1 + 0.663)
≈ 0.970c.
Therefore, the speed of the probe as seen from the earth is approximately 0.970c. Since the speed is greater than the speed of light, it implies that the probe is moving away from the earth (as viewed from the earth).
b. The length of the ship as measured by the ship's crew can be calculated using the relativistic length contraction formula:
L' = L * √(1 - (v^2/c^2)),
where L' is the length of the ship as measured by the crew, L is the length of the ship as measured by an observer at rest (in this case, the earth), v is the velocity of the ship (relative to the earth), and c is the speed of light.
L = 4.50 m (length of the ship as measured from the earth),
v = 0.850c (speed of the ship relative to the earth).
Substituting the values into the formula:
L' = 4.50 m * √(1 - (0.850c)^2/c^2)
= 4.50 m * √(1 - 0.7225)
= 4.50 m * √(0.2775)
≈ 6.15 m.
Therefore, the length of the ship as measured by the ship's crew is approximately 6.15 m.
a. The speed of the probe as seen from the earth is approximately 0.970c. The probe is moving away from the earth (as viewed from the earth).
b. The length of the ship as measured by the ship's crew is approximately 6.15 m.
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An AC voltage of the form Av = 75 sin 300t where Av is in volts and t is in seconds, is applied to a series RLC circuit. If R = 42.0 8, C = 26.0 F, and L = 0.300 H, find the following.
(a) the impedance of the circuit
(b) the rms current in the circuit
(c) the average power delivered to the circuit
AC voltage is given by the equation Av = 75 sin 300t, where Av represents the voltage in volts and t represents time in seconds.
R = 42.08 Ω, C = 26.0 F, and L = 0.300 H.
The impedance of the circuit, denoted as Z,
Z = √(R² + (Xl - Xc)²).
Here, Xl represents the inductive reactance and Xc represents the capacitive reactance. The capacitive reactance Xc is obtained using the formula Xc = 1/(Cω), where ω is the angular frequency of the circuit.
The inductive reactance Xl is calculated as Xl = ωL, where L is the inductance of the circuit. The angular frequency ω is determined by ω = 2πf, with f representing the frequency of the AC source.
Xl = 565.4867 Ω and Xc = 0.0021427 Ω.
The impedance of the circuit is determined as Z = √(R² + (Xl - Xc)²) = 565.4755 Ω.
The RMS current in the circuit, denoted as I, is calculated using the formula I = V/Z, where V is the RMS voltage. The RMS voltage is obtained by dividing Av by the square root of 2. By substituting the values, we find I = 0.09388 AC current.
The average power delivered to the circuit, denoted as P, is given by the formula P = (1/2) VI cosφ, where V is the RMS voltage, I is the RMS current, and cosφ is the power factor. The phase difference φ between the current and voltage is determined using the formula φ = tan⁻¹((Xl - Xc) / R).
By substituting the given values, we find φ = 86.87° and cosφ = -0.0512. Thus, the average power delivered to the circuit is calculated as P = -0.02508 W. The negative sign indicates that the circuit is consuming power instead of delivering it.
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2 of 5 For a liquid state, the chemical potential is equal to fugacity at the same temperature and pressure. T True F False SUBMIT ANSWER
For a liquid state, the chemical potential is equal to fugacity at the same temperature and pressure, the given statement is false because a chemical potential is the partial molar Gibbs free energy of a constituent in a mixture.
It measures the potential energy of the constituent to move from one phase to another. In contrast, fugacity is the measure of the escaping tendency of molecules from a phase. In a liquid state, the chemical potential is related to the molar Gibbs free energy of the substance. It determines the driving force of chemical reactions. Fugacity is a thermodynamic property that approximates the actual pressure of an ideal gas mixture based on its ideal behavior.
It is related to the pressure and is used to determine the concentration of the substance. The relationship between chemical potential and fugacity varies for different phases. In conclusion, the statement "For a liquid state, the chemical potential is equal to fugacity at the same temperature and pressure" is not correct.
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4)Consider a charge +q, located at the origin. We observe the potential V and the electric field E at distance r. Here we are interested in quantities along the z-axis. Find or, if you know it, simply write, the electric potential V along the z-axis (x =y=0), for [z| not equal to zero. (Izis the absolute value of z.) Can you make any arguments by symmetry about the electric field? For instance, are any components of the electric field equal to zero along the z-axis? Use the standard expression, E; = -av/axi, where the E; are the x, y, z components of E, and Xi refer to the coordinates x, y, z. Using the expression you found above for V, what is the vector E as a function of position along the positive z-axis (lz| not equal zero). = Recall the differential form of Gauss' law, V. E=p/€0. What does this say about the component Ex at points Ax away from the z-axis, as a function of z? -
The electric potential, V along the z-axis (x=y=0) is as follows: Let r = (x² + y² + z²)¹/² Thus,
V = kq/r. When
x=y=0,
V = kq/z,
provided z is not equal to zero. By symmetry, the components of the electric field E along the x and y-axes are zero since the charge +q at the origin does not produce any component of E along these axes.
Hence E; = (0,0, Ez). It follows that Ex = 0 and Ey = 0 because of symmetry along the x- and y-axes. The electric field E can be found using
E= -av/axi
= - (dV/dx)i - (dV/dy)j - (dV/dz)k.
Using V = kq/z, it follows that:
E = -d/dz(kq/z)k
= kq/z²k.
Hence E has only a z-component, and its magnitude is given by E = kq/z² along the positive z-axis.
The differential form of Gauss' law, V. E=p/€0. If z > Ax, then we can draw a Gaussian surface that is cylindrical and coaxial with the z-axis. By symmetry, Ex = 0, so that p = 0. Thus, V. E = 0, and since V is non-zero, it follows that E must be zero.
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The current through a 40 W, 120 V light bulb is:
A.
1/3 A
b.
3A
c.
80 A
d
4,800 A
AND.
None
Comparing the options provided, we see that the current is approximately 0.333 A, which corresponds to option A: 1/3 A. Option A is correct.
We are given a 40 W light bulb with a voltage of 120 V. To find the current, we can rearrange the formula P = VI to solve for I:
I = P / V
Substituting the given values:
I = 40 W / 120 V
Calculating the current:
I ≈ 0.333 A
Comparing the options provided, we see that the current is approximately 0.333 A, which corresponds to option A: 1/3 A. Therefore, the correct answer is A.
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X Х Two identical balls of putty moving perpendicular to each other, both moving at 10.06 m/s, experience a perfectly inelastic collision. What is the speed of the combined ball after the collision? Give your answer to two decimal places
The speed of the combined ball after the perfectly inelastic collision remains the same at 10.06 m/s.
In a perfectly inelastic collision, two objects stick together and move as one mass after the collision. To calculate the speed of the combined ball after the collision, we can use the principle of conservation of momentum.
Given:
- Two identical balls of putty
- Both moving at 10.06 m/s
- Perfectly inelastic collision
Let's denote the initial velocity of each ball as v1 and v2, and the final velocity of the combined ball as vf.
According to the conservation of momentum:
(m1 * v1) + (m2 * v2) = (m1 + m2) * vf
Since the balls are identical, their masses (m1 and m2) are the same, so we can rewrite the equation as:
(2 * m * v1) = (2 * m) * vf
The masses cancel out, leaving us with:
2 * v1 = 2 * vf
Simplifying further:
v1 = vf
Since both balls are moving at the same speed before the collision, the speed of the combined ball after the collision is also equal to 10.06 m/s.
Therefore, the speed of the combined ball after the collision is 10.06 m/s.
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Pushing down on a bicycle pedal with 663 N of force, the pedal fixed at 0.20 m from the center of the gear moves through 30° of angle. What is the work in newton-meters that you do to effect this motion?
The work done to effect the motion of the bicycle pedal is approximately 66.72 N·m (Newton-meters).
To calculate the work done in this scenario, we can use the formula for work done by a force applied at an angle.
Given:
Force applied (F) = 663 N
Distance from the center of the gear (r) = 0.20 m
Angle through which the pedal moves (θ) = 30°
The work done (W) can be calculated using the formula:
W = F * r * cos(θ)
First, we need to convert the angle from degrees to radians:
θ (in radians) = θ (in degrees) * (π / 180)
θ (in radians) = 30° * (π / 180) ≈ 0.5236 radians
Now we can calculate the work done:
W = 663 N * 0.20 m * cos(0.5236)
W ≈ 66.72 N·m
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conducting circular ring of radius a=0.8 m is placed in a time varying magnetic field given by B(t) = B. (1+7) where B9 T and T-0.2 s. a. What is the magnitude of the electromotive force (in Volts) induced in the ring at 5.6 seconds? b. At instant 5.6 seconds the magnetic field stops changing. Now imagine that the ring is made from a flexible material. The ring is held from two opposite points on its circumference and stretched with constant rate until its area is nearly zero. If it takes 1.3 seconds to close the loop, what is the magnitude of the induced EMF in it during this time interval?
(a) The magnitude of the induced electromotive force in the ring at 5.6 seconds is approximately 100.531 volts.
(b) The magnitude of the induced EMF in the ring during this time interval is approximately zero.
(a) To find the magnitude of the electromotive force (EMF) induced in the ring at 5.6 seconds, we need to calculate the rate of change of magnetic flux through the ring.
The magnetic flux (Φ) through the ring is given by the equation:
Φ = B * A
Where B is the magnetic field and A is the area of the ring.
The area of a circular ring is given by the equation:
A = π * (r_[tex]outer^2[/tex] - r_[tex]inner^2[/tex])
Since the radius of the ring is given as a = 0.8 m, the inner radius would be 0, and the outer radius would also be 0.8 m.
The rate of change of magnetic flux is given by Faraday's law of electromagnetic induction:
ε = -dΦ/dt
Where ε is the induced electromotive force.
In this case, we have B(t) = B * (1 + 7t), where B = 9 T and t = 5.6 s.
We can substitute the values into the equations and calculate the EMF as follows:
A = π * ([tex]0.8^2[/tex] - [tex]0^2[/tex]) = π * 0.64
dΦ/dt = dB(t)/dt * A = (7Bπ) * A
ε = -dΦ/dt = -7BπA
Substituting the values, we get:
ε = -7 * 9 * π * 0.64 ≈ -100.531 V
Therefore, the magnitude of the induced electromotive force in the ring at 5.6 seconds is approximately 100.531 volts.
(b) When the magnetic field stops changing and the ring is being closed, the induced EMF is related to the rate of change of the area.
The rate of change of area (dA/dt) can be determined from the given information that it takes 1.3 seconds to close the loop and make the area nearly zero.
The rate of change of area is given by:
dA/dt = A_final / t_final
Since the area is nearly zero when the loop is closed, we can assume A_final ≈ 0.
Therefore, dA/dt ≈ 0 / 1.3 ≈ 0
Since the rate of change of area is nearly zero, the induced EMF is also nearly zero.
Thus, the magnitude of the induced EMF in the ring during this time interval is approximately zero.
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8. What is the change in momentum that occurs when a 1.0 kg ball traveling at 3.0 m/s strikes a wall and bounces back at 1.0 m/s?. 9. What would be your gravitational potential energy at the top of 20 meter tree if your mass was 60kg?. 10. How much energy would you consume daily if you kept a lightbulb of 20 watts on all day long?. 11. What is the density of an object of mass 2kg and a volume of 0.5 cubic meters?. a 12. If an object that has a volume of 4 cubic meters is inserted in a liquid, how much liquid would it displace?. Explain. 13. An object that is 1 meter under water, experiences a pressure of 10 kg*m/s2. A. What is the density of the liquid?. B. What would be the pressure if it was 2 meters under water?
Density of the liquid is 4 kg/m³. and its pressure if it was 2 meters under water is 19.9 kg*m/s².
Change in momentum is equal to the initial momentum minus the final momentum.
Here the initial momentum is given by;
p1 = m*v1
= 1.0 kg × 3.0 m/s
= 3.0 kg m/s.
The final momentum can be found by using the given mass of the ball and the final velocity, i.e.,
p2 = m*v2
= 1.0 kg × (-1.0 m/s)
= -1.0 kg m/s.
Therefore, change in momentum is;
Δp = p1 - p2 =
3.0 kg m/s - (-1.0 kg m/s)
= 4.0 kg m/s.
Δp = 4.0 kg m/s.9.
The gravitational potential energy is given by;
GPE = mgh
where, m = mass,
g = gravitational acceleration, and
h = height from the reference level.
Here, m = 60 kg, g = 9.8 m/s², and
h = 20 m.
Therefore,
GPE = mgh = 60 kg × 9.8 m/s² × 20 m
= 11,760 J.
GPE = 11,760 J.
The energy consumed by the light bulb can be found by multiplying the power rating with the time it is used. Here, the power rating of the bulb is 20 W and it is kept on for the entire day, i.e., 24 hours. Therefore,
energy consumed = power × time
= 20 W × 24 h
= 480 Wh
= 0.48 kWh.
Energy consumed = 0.48 kWh.
The density of an object is given by the ratio of its mass to its volume, i.e.,
ρ = m/V. Here, the mass of the object is given as 2 kg and its volume is given as 0.5 cubic meters.
Therefore, density of the object is;
ρ = m/V
= 2 kg/0.5 m³
= 4 kg/m³.
Density = 4 kg/m³.
We know that the pressure in a liquid depends on its density and depth. Here, the pressure is given as 10 kg*m/s², and the depth is given as 1 m. Therefore, density of the liquid is given by the relation;
P = ρgd
where,
ρ = density of the liquid,
g = gravitational acceleration, and
d = depth.
Substituting the given values, we get;
ρ = P/gd
= 10 kg*m/s²/9.8 m/s² × 1 m
= 1.02 kg/m³.
Density of the liquid = 1.02 kg/m³.
The pressure at a depth of 2 m can be found using the relation;
P = ρgd.
Here, we already know the density of the liquid and the gravitational acceleration.
Therefore, the pressure is; P = ρgd
= 1.02 kg/m³ × 9.8 m/s² × 2 m
= 19.9 kg*m/s².
Pressure = 19.9 kg*m/s².
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