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
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.
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.
To find the energy lost due to air friction, we first calculate the initial kinetic energy and the potential energy at the maximum height. 1. Initial kinetic energy (K.E.) when the body is thrown: K.E. = (1/2) * m * v² K.E. = (1/2) * 1 kg * (20 m/s)² K.E. = (1/2) * 1 * 400 K.E. = 200 J. 2. PotentialRead more
To find the energy lost due to air friction, we first calculate the initial kinetic energy and the potential energy at the maximum height.
1. Initial kinetic energy (K.E.) when the body is thrown:
K.E. = (1/2) * m * v²
K.E. = (1/2) * 1 kg * (20 m/s)²
K.E. = (1/2) * 1 * 400
K.E. = 200 J.
2. Potential energy (P.E.) at the maximum height (h = 18 m):
P.E. = m * g * h
P.E. = 1 kg * 10 m/s² * 18 m
P.E. = 180 J.
3. The energy lost to air friction is given by the difference between the initial kinetic energy and the potential energy at the maximum height:
Energy lost = K.E at the start – P.E.
Energy lost = 200 J – 180 J
Energy lost = 20 J.
The coefficient of restitution, denoted by 'e', describes how elastic the collision between two bodies is. When the two bodies collide and go back into their original shape without any changes in momentum or kinetic energy after a perfectly elastic collision, it follows that, by definition of e, eRead more
The coefficient of restitution, denoted by ‘e’, describes how elastic the collision between two bodies is. When the two bodies collide and go back into their original shape without any changes in momentum or kinetic energy after a perfectly elastic collision, it follows that, by definition of e,
e=relative velocity after collision/ relative velocity before collision
To find the velocities of two identical balls A and B after an elastic collision, we can use the conservation of momentum and the fact that kinetic energy is also conserved. 1. Let the initial velocities be: Velocity of A before collision (uA) = +0.5 m/s Velocity of B before collision (uB) = -0.3 m/Read more
To find the velocities of two identical balls A and B after an elastic collision, we can use the conservation of momentum and the fact that kinetic energy is also conserved.
1. Let the initial velocities be:
Velocity of A before collision (uA) = +0.5 m/s
Velocity of B before collision (uB) = -0.3 m/s
2. For two equal masses in an elastic collision, the final velocities can be determined using the following formulas:
vA = (uA + uB) / 2 + (uA – uB) / 2
vB = (uA + uB) / 2 – (uA – uB) / 2
– First, we find the average velocity:
Average velocity (u_avg) = (uA + uB) / 2 = (0.5 – 0.3) / 2 = 0.1 / 2 = 0.05 m/s.
Now, we compute the change in velocities:
Change in velocity = uA – uB = 0.5 – (-0.3) = 0.5 + 0.3 = 0.8 m/s.
3. Using the equations above:
– Final velocity of A (vA):
vA = 0.05 + 0.4 = 0.45 m/s to be corrected depending on the direction.
– Final velocity of B (vB):
vB = 0.05 – 0.4 = -0.35 m/s to be corrected depending on the direction.
4. In a head-on collision, however, for two identical masses, their velocities will swap:
– After collision, ball A takes B’s initial velocity, and ball B takes A’s initial velocity.
– Thus, the final velocities will be:
– vA = -0.3 m/s (B’s initial velocity)
– vB = +0.5 m/s (A’s initial velocity)
Using the principle of conservation of momentum, we can calculate the speed of the combined particle after the collision. 1. Let the initial momentum of the system be: - The momentum of the first particle having mass m and velocity v is m * v. - The momentum of the second particle having mass 2m andRead more
Using the principle of conservation of momentum, we can calculate the speed of the combined particle after the collision.
1. Let the initial momentum of the system be:
– The momentum of the first particle having mass m and velocity v is m * v.
– The momentum of the second particle having mass 2m and at rest is 2m * 0 = 0.
Total initial momentum = mv + 0 = mv.
2. As the collision takes place, the two particles merge together and a total mass as follows:
Total mass = m + 2m = 3m.
3. Supposing the speed of the merged particle after the collision as V. According to the law of conservation of momentum
Initial momentum = Final momentum
mv = (3m) * V.
We can use the work done formula to find the angle between the force and the direction of motion: Work done (W) = Force (F) × Distance (d) × cos(θ), where θ is the angle between the force and the direction of motion. 1. Given values: - Work done (W) = 25 J, - Force (F) = 5 N, - Distance (d) = 10 m.Read more
We can use the work done formula to find the angle between the force and the direction of motion:
Work done (W) = Force (F) × Distance (d) × cos(θ),
where θ is the angle between the force and the direction of motion.
1. Given values:
– Work done (W) = 25 J,
– Force (F) = 5 N,
– Distance (d) = 10 m.
2. Substituting the given values into the formula:
25 J = 5 N × 10 m × cos(θ).
3. Simplifying the equation:
25 = 50 × cos(θ).
4. Dividing both sides by 50:
cos(θ) = 25 / 50 = 0.5.
To find the percentage decrease in kinetic energy (K.E.) when momentum decreases by 20%, we have the following relations: 1. Momentum (p) is defined as: p = m * v where m is the mass and v is the velocity. 2. Kinetic energy (K.E.) is defined as: K.E. = (1/2) * m * v². 3. If momentum decreases by 20%Read more
To find the percentage decrease in kinetic energy (K.E.) when momentum decreases by 20%, we have the following relations:
1. Momentum (p) is defined as:
p = m * v
where m is the mass and v is the velocity.
2. Kinetic energy (K.E.) is defined as:
K.E. = (1/2) * m * v².
3. If momentum decreases by 20%, then the new momentum is given by p’.
p’ = p – 0.2p = 0.8p.
4. As momentum is directly proportional to mass and velocity, we can express the new velocity, v’, in terms of the original velocity, v, as:
p’ = m * v’ = 0.8(m * v).
5. Divide both sides by m
v’ = 0.8v.
6. We can now compute the new kinetic energy, K.E.’:
K.E.’ = (1/2) * m * (v’)²
= (1/2) * m * (0.8v)²
= (1/2) * m * 0.64v²
= 0.64 * (1/2) * m * v²
= 0.64 * K.E.
7. The percentage decrease in kinetic energy is:
Percentage decrease = [(K.E. – K.E.’) / K.E.] * 100
= [(K.E. – 0.64K.E.) / K.E.] * 100
= (0.36K.E. / K.E.) * 100 = 36%.
For finding the value of α as in the extraction of a small disc from the larger disc we take the arrangement, we assume we have large disc with a radius of size 2R and small disc of radius R. After we take out small disc, now we want to know the location of new centre of mass due to this. The largerRead more
For finding the value of α as in the extraction of a small disc from the larger disc we take the arrangement, we assume we have large disc with a radius of size 2R and small disc of radius R. After we take out small disc, now we want to know the location of new centre of mass due to this.
The larger disc has a larger area and mass, so it is initially centered at the origin. The smaller disc that is removed is also centered at the same point. The area and mass of the larger disc are proportionally greater than those of the smaller disc. This difference in mass distribution affects the position of the new center of mass after the smaller disc is removed.
As we analyze the situation, we can infer that the removal of the smaller disc will shift the center of mass toward the larger disc. Applying the principles of mass distribution and balance, we find that the center of mass of the remaining shape is located at a distance of αR from the center of the larger disc. After calculations, we conclude that the value of α is one-third which indicates the relative position of the center of mass in the modified structure.
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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A body of mass 1 kg is thrown upwards with a velocity of 20 m/s. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? [Take g = 10 m/s²]
To find the energy lost due to air friction, we first calculate the initial kinetic energy and the potential energy at the maximum height. 1. Initial kinetic energy (K.E.) when the body is thrown: K.E. = (1/2) * m * v² K.E. = (1/2) * 1 kg * (20 m/s)² K.E. = (1/2) * 1 * 400 K.E. = 200 J. 2. PotentialRead more
To find the energy lost due to air friction, we first calculate the initial kinetic energy and the potential energy at the maximum height.
1. Initial kinetic energy (K.E.) when the body is thrown:
K.E. = (1/2) * m * v²
K.E. = (1/2) * 1 kg * (20 m/s)²
K.E. = (1/2) * 1 * 400
K.E. = 200 J.
2. Potential energy (P.E.) at the maximum height (h = 18 m):
P.E. = m * g * h
P.E. = 1 kg * 10 m/s² * 18 m
P.E. = 180 J.
3. The energy lost to air friction is given by the difference between the initial kinetic energy and the potential energy at the maximum height:
See lessEnergy lost = K.E at the start – P.E.
Energy lost = 200 J – 180 J
Energy lost = 20 J.
The coefficient of restitution e for a perfectly elastic collision is
The coefficient of restitution, denoted by 'e', describes how elastic the collision between two bodies is. When the two bodies collide and go back into their original shape without any changes in momentum or kinetic energy after a perfectly elastic collision, it follows that, by definition of e, eRead more
The coefficient of restitution, denoted by ‘e’, describes how elastic the collision between two bodies is. When the two bodies collide and go back into their original shape without any changes in momentum or kinetic energy after a perfectly elastic collision, it follows that, by definition of e,
e=relative velocity after collision/ relative velocity before collision
For the perfectly elastic collision, e equals 1.
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Two identical balls A and B collide head on elastically. If velocities of A and B, before the collision are +0.5 m/s and -0.3 m/s respectively, then their velocities, after the collision, are respectively
To find the velocities of two identical balls A and B after an elastic collision, we can use the conservation of momentum and the fact that kinetic energy is also conserved. 1. Let the initial velocities be: Velocity of A before collision (uA) = +0.5 m/s Velocity of B before collision (uB) = -0.3 m/Read more
To find the velocities of two identical balls A and B after an elastic collision, we can use the conservation of momentum and the fact that kinetic energy is also conserved.
1. Let the initial velocities be:
Velocity of A before collision (uA) = +0.5 m/s
Velocity of B before collision (uB) = -0.3 m/s
2. For two equal masses in an elastic collision, the final velocities can be determined using the following formulas:
vA = (uA + uB) / 2 + (uA – uB) / 2
vB = (uA + uB) / 2 – (uA – uB) / 2
– First, we find the average velocity:
Average velocity (u_avg) = (uA + uB) / 2 = (0.5 – 0.3) / 2 = 0.1 / 2 = 0.05 m/s.
Now, we compute the change in velocities:
Change in velocity = uA – uB = 0.5 – (-0.3) = 0.5 + 0.3 = 0.8 m/s.
3. Using the equations above:
– Final velocity of A (vA):
vA = 0.05 + 0.4 = 0.45 m/s to be corrected depending on the direction.
– Final velocity of B (vB):
vB = 0.05 – 0.4 = -0.35 m/s to be corrected depending on the direction.
4. In a head-on collision, however, for two identical masses, their velocities will swap:
– After collision, ball A takes B’s initial velocity, and ball B takes A’s initial velocity.
– Thus, the final velocities will be:
– vA = -0.3 m/s (B’s initial velocity)
– vB = +0.5 m/s (A’s initial velocity)
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A particle of mass m moving with velocity v collides with a stationary particle of mass 2 m. After collision, the speed of the combined particle is
Using the principle of conservation of momentum, we can calculate the speed of the combined particle after the collision. 1. Let the initial momentum of the system be: - The momentum of the first particle having mass m and velocity v is m * v. - The momentum of the second particle having mass 2m andRead more
Using the principle of conservation of momentum, we can calculate the speed of the combined particle after the collision.
1. Let the initial momentum of the system be:
– The momentum of the first particle having mass m and velocity v is m * v.
– The momentum of the second particle having mass 2m and at rest is 2m * 0 = 0.
Total initial momentum = mv + 0 = mv.
2. As the collision takes place, the two particles merge together and a total mass as follows:
Total mass = m + 2m = 3m.
3. Supposing the speed of the merged particle after the collision as V. According to the law of conservation of momentum
Initial momentum = Final momentum
mv = (3m) * V.
4. Solving for V:
V = mv / (3m) = v / 3.
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A body moves a distance 0f 10 m along a straight line under the action of a 5 N force. If the work done is 25 J, then angle between the force and direction of motion of the body is
We can use the work done formula to find the angle between the force and the direction of motion: Work done (W) = Force (F) × Distance (d) × cos(θ), where θ is the angle between the force and the direction of motion. 1. Given values: - Work done (W) = 25 J, - Force (F) = 5 N, - Distance (d) = 10 m.Read more
We can use the work done formula to find the angle between the force and the direction of motion:
Work done (W) = Force (F) × Distance (d) × cos(θ),
where θ is the angle between the force and the direction of motion.
1. Given values:
– Work done (W) = 25 J,
– Force (F) = 5 N,
– Distance (d) = 10 m.
2. Substituting the given values into the formula:
25 J = 5 N × 10 m × cos(θ).
3. Simplifying the equation:
25 = 50 × cos(θ).
4. Dividing both sides by 50:
cos(θ) = 25 / 50 = 0.5.
5. Now, finding the angle θ:
θ = cos⁻¹(0.5) = 60°.
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If momentum decreases by 20 %, K.E. will decrease by
To find the percentage decrease in kinetic energy (K.E.) when momentum decreases by 20%, we have the following relations: 1. Momentum (p) is defined as: p = m * v where m is the mass and v is the velocity. 2. Kinetic energy (K.E.) is defined as: K.E. = (1/2) * m * v². 3. If momentum decreases by 20%Read more
To find the percentage decrease in kinetic energy (K.E.) when momentum decreases by 20%, we have the following relations:
1. Momentum (p) is defined as:
p = m * v
where m is the mass and v is the velocity.
2. Kinetic energy (K.E.) is defined as:
K.E. = (1/2) * m * v².
3. If momentum decreases by 20%, then the new momentum is given by p’.
p’ = p – 0.2p = 0.8p.
4. As momentum is directly proportional to mass and velocity, we can express the new velocity, v’, in terms of the original velocity, v, as:
p’ = m * v’ = 0.8(m * v).
5. Divide both sides by m
v’ = 0.8v.
6. We can now compute the new kinetic energy, K.E.’:
K.E.’ = (1/2) * m * (v’)²
= (1/2) * m * (0.8v)²
= (1/2) * m * 0.64v²
= 0.64 * (1/2) * m * v²
= 0.64 * K.E.
7. The percentage decrease in kinetic energy is:
Percentage decrease = [(K.E. – K.E.’) / K.E.] * 100
= [(K.E. – 0.64K.E.) / K.E.] * 100
= (0.36K.E. / K.E.) * 100 = 36%.
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A circular disc of radius R is removed from a bigger circular disc of radius 2R, such that the circumferences of the disc coincide. The centre of mass of the new disc is αR from the centre of the bigger disc. The value of α is
For finding the value of α as in the extraction of a small disc from the larger disc we take the arrangement, we assume we have large disc with a radius of size 2R and small disc of radius R. After we take out small disc, now we want to know the location of new centre of mass due to this. The largerRead more
For finding the value of α as in the extraction of a small disc from the larger disc we take the arrangement, we assume we have large disc with a radius of size 2R and small disc of radius R. After we take out small disc, now we want to know the location of new centre of mass due to this.
The larger disc has a larger area and mass, so it is initially centered at the origin. The smaller disc that is removed is also centered at the same point. The area and mass of the larger disc are proportionally greater than those of the smaller disc. This difference in mass distribution affects the position of the new center of mass after the smaller disc is removed.
As we analyze the situation, we can infer that the removal of the smaller disc will shift the center of mass toward the larger disc. Applying the principles of mass distribution and balance, we find that the center of mass of the remaining shape is located at a distance of αR from the center of the larger disc. After calculations, we conclude that the value of α is one-third which indicates the relative position of the center of mass in the modified structure.
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