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For a given photosensitive material and with a source of constant frequency of incident radiation, how does the photocurrent vary with the intensity of incident light?
For a photosensitive material with constant frequency of incident radiation, the photocurrent increases linearly with the intensity of incident light. This is because a higher intensity of light emits more photons, leading to the ejection of more photoelectrons, thus increasing the photocurrent. ForRead more
For a photosensitive material with constant frequency of incident radiation, the photocurrent increases linearly with the intensity of incident light. This is because a higher intensity of light emits more photons, leading to the ejection of more photoelectrons, thus increasing the photocurrent.
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Why is photoelectric emission not possible at all frequencies?
Photoelectric emission occurs only if the frequency of incident light exceeds the material's threshold frequency. Below this frequency, photons lack sufficient energy to overcome the work function, the minimum energy required to eject electrons from the material's surface, making emission impossibleRead more
Photoelectric emission occurs only if the frequency of incident light exceeds the material’s threshold frequency. Below this frequency, photons lack sufficient energy to overcome the work function, the minimum energy required to eject electrons from the material’s surface, making emission impossible.
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An engine pumps water through a hose pipe. Water passes through the pipe and leaves it with a velocity of 2 m/s. The mass per unit length of water in the pipe is 100 kg/m. What is the power of the engine?
To calculate the power of the engine, we can use the formula for power related to the flow of water through the hose: Power (P) = (mass flow rate) × (velocity)² / 2 Step 1: Calculate the mass flow rate Mass per unit length of water = 100 kg/m Velocity of water = 2 m/s The mass flow rate (ṁ) can be cRead more
To calculate the power of the engine, we can use the formula for power related to the flow of water through the hose:
Power (P) = (mass flow rate) × (velocity)² / 2
Step 1: Calculate the mass flow rate
Mass per unit length of water = 100 kg/m
Velocity of water = 2 m/s
The mass flow rate (ṁ) can be calculated as:
ṁ = (mass per unit length) × (velocity)
ṁ = 100 kg/m × 2 m/s
ṁ = 200 kg/s
Step 2: Calculate the power
Now using the power formula:
P = (ṁ × v²) / 2
P = (200 kg/s × (2 m/s)²) / 2
P = (200 kg/s × 4 m²/s²) / 2
P = (800 kg·m²/s³) / 2
P = 400 W
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A body projected vertically from the earth reaches a height equal to earth’s radius before returning to the earth. The power exerted by the gravitational force is greatest
When a body is projected vertically from the Earth and reaches a height equal to the Earth's radius before returning, we can analyze the power exerted by the gravitational force throughout its motion. The gravitational force acts as a constant downward force, affecting the body's velocity as it riseRead more
When a body is projected vertically from the Earth and reaches a height equal to the Earth’s radius before returning, we can analyze the power exerted by the gravitational force throughout its motion. The gravitational force acts as a constant downward force, affecting the body’s velocity as it rises and falls.
Initially, at the moment just after the body is projected, it has its maximum velocity, resulting in a certain amount of kinetic energy. As the body ascends, gravity does negative work on it, causing its velocity to decrease until it reaches its maximum height, where the velocity becomes zero. At this point, the power exerted by the gravitational force is also zero, as power is the product of force and velocity.
As the body starts descending, it accelerates due to gravity, and its velocity increases. Just before the body hits the Earth, it reaches its maximum velocity once again. The gravitational force, although constant, does more work as the velocity of the body increases. Therefore, the power exerted by the gravitational force is greatest at the instant just before the body strikes the Earth, as both the gravitational force and the velocity are at their maximum values at that point.
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A small mass attached to string rotates on a frictionless table top as shown. If the tension in the string is increased by pulling the string causing the radius of the circular motion to decrease by a factor of 2, the kinetic energy of the mass will
To find how the kinetic energy of a mass changes when the radius of its circular motion is decreased by a factor of 2 while the tension in the string is increased, we can use the following relationships: 1. The centripetal force required to keep the mass moving in a circular path is provided by theRead more
To find how the kinetic energy of a mass changes when the radius of its circular motion is decreased by a factor of 2 while the tension in the string is increased, we can use the following relationships:
1. The centripetal force required to keep the mass moving in a circular path is provided by the tension in the string:
T = (m * v²) / r
where T is the tension, m is the mass, v is the velocity, and r is the radius.
2. The kinetic energy (K.E.) of the mass is given by:
K.E. = (1/2) * m * v².
3. If the radius is reduced to half of the previous value (r’ = r/2) and assuming tension is increased to a level at which circular motion will be maintained, we have
T’ = (m * v’²) / (r/2)
T’ = (2 * m * v’²) / r.
4. Keeping T’ constant and solving for v’, we see that v’ must be greater to be in equilibrium.
5. Putting it all back together to find the kinetic energy:
K.E.’ = (1/2) * m * (v’)².
Since v’ = √2 * v, the kinetic energy is multiplied by a factor of 4.
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