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How much water a pump of 2 kW can raise in one minute to a height of 10 m? (take g = 10m /s²)
To determine how much water a 2 kW pump can raise in one minute to a height of 10 m, we can use the formula for power: P = W / t Step 1: Calculate Work Done The work done to raise water to a height h is given by: W = mgh Where: - m is the mass of the water in kilograms (kg) - g is the acceleration dRead more
To determine how much water a 2 kW pump can raise in one minute to a height of 10 m, we can use the formula for power:
P = W / t
Step 1: Calculate Work Done
The work done to raise water to a height h is given by:
W = mgh
Where:
– m is the mass of the water in kilograms (kg)
– g is the acceleration due to gravity (10 m/s²)
– h is the height in meters (10 m)
Step 2: Convert Power to Work Done in One Minute
Given:
– Power, P = 2 kW = 2000 W
– Time, t = 1 minute = 60 s
Now calculate the work done:
W = P × t
W = 2000 W × 60 s
W = 120000 J
Step 3: Calculate the Mass of Water
Now we can use the work done to calculate the mass of water:
W = mgh => m = W / (gh)
Substitute known values:
m = 120000 J / (10 m/s² × 10 m)
m = 120000 / 100
m = 1200 kg
Step 4: Convert Mass to Volume
Knowing that the density of water is around 1000 kg/m³, the volume V of water lifted will be:
V = m / density
V = 1200 kg / (1000 kg/m³)
V = 1.2 m³
Converting the above value to cubic meters into liters:
Since,
1 m³ = 1000 liters,
V = 1.2 m³ × 1000 liters/m³
V = 1200 liters
Final Answer:
In one minute, the pump is able to pump 1200 liters of water up a height of 10 m.
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The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm, the potential energy stored in it is
To determine the potential energy in a stretched spring, we have the formula for elastic potential energy: U = (1/2) k x² Where: - U is the potential energy - k is the spring constant - x is the extension (or compression) of the spring from its equilibrium position Given: - When the spring is stretcRead more
To determine the potential energy in a stretched spring, we have the formula for elastic potential energy:
U = (1/2) k x²
Where:
– U is the potential energy
– k is the spring constant
– x is the extension (or compression) of the spring from its equilibrium position
Given:
– When the spring is stretched by x₁ = 2 cm:
U = (1/2) k (2 cm)² = (1/2) k (0.02 m)²
– When the spring is stretched by x₂ = 8 cm:
U’ = (1/2) k (8 cm)² = (1/2) k (0.08 m)²
Step 1: Calculate the ratio of potential energies
We are required to find the ratio of the potential energies when stretched by 8 cm compared to when stretched by 2 cm:
(U’) / (U) = [(1/2) k (0.08 m)²] / [(1/2) k (0.02 m)²] = [(0.08)²] / [(0.02)²]
Step 2: Simplify the ratio
Compute the squares:
(U’) / (U) = [(0.08)²] / [(0.02)²] = [0.0064] / [0.0004] = 16
Conclusion
The potential energy in the spring stretched by 8 cm is given by:
U’ = 16 U
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A moving body of mass m and velocity 3 km/h collides with a rest body of mass 2 m and stricks to it. Now the combined mass starts to move. What will be the combined velocity?
To find the resultant velocity after the collision, we can use the principle of conservation of momentum. Momentum before collision is equal to the momentum after collision. Let: - Mass of moving body = m - Velocity of moving body = v₁ = 3 km/h - Mass of resting body = 2m - Velocity of resting bodyRead more
To find the resultant velocity after the collision, we can use the principle of conservation of momentum. Momentum before collision is equal to the momentum after collision.
Let:
– Mass of moving body = m
– Velocity of moving body = v₁ = 3 km/h
– Mass of resting body = 2m
– Velocity of resting body = v₂ = 0 km/h
– Combined mass after the collision = m + 2m = 3m
– Combined velocity after the collision = vբ
Total momentum before collision is:
Momentum Initial = mv₁ + (2m)v₂ = m(3) + (2m)(0) = 3m
Total momentum after collision:
Final Momentum = (3m)vբ
Put the initial momentum equal to final momentum:
3m = 3mvբ
Now solve for բv:
բv = 3m / 3m = 1 km/h
Thus, the resultant velocity after collision is 1 km/hour.
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For a hollow cylinder and a solid cylinder rolling without slipping on an inclined plane, then which of these reaches earlier
The acceleration of the two rolling cylinders on an inclined plane without slipping is dominated by the moment of inertia. For the same mass and radius, the moment of inertia for the hollow cylinder is greater than that of the solid cylinder. Thus, the fraction of potential energy converted to transRead more
The acceleration of the two rolling cylinders on an inclined plane without slipping is dominated by the moment of inertia. For the same mass and radius, the moment of inertia for the hollow cylinder is greater than that of the solid cylinder. Thus, the fraction of potential energy converted to translational kinetic energy is greater for the solid cylinder, resulting in greater acceleration.
The hollow cylinder, however, has a greater moment of inertia and, hence, more energy in rotational rather than translational motion. This results in a smaller acceleration of the hollow cylinder along the slope.
The acceleration is greater in the solid cylinder; therefore, the solid cylinder will reach the bottom of the slope before the hollow cylinder does. This happens for all objects with different moments of inertia which roll down any incline.
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If force acts on a body, whose line of action does not pass through its centre of gravity, then the body will experience
When a force acts on a body, and its line of action does not pass through the center of gravity, the body experiences two effects: linear acceleration and angular acceleration. Linear acceleration occurs as the force pushes or pulls the body in a specific direction. This is due to the body's responsRead more
When a force acts on a body, and its line of action does not pass through the center of gravity, the body experiences two effects: linear acceleration and angular acceleration. Linear acceleration occurs as the force pushes or pulls the body in a specific direction. This is due to the body’s response to the applied force, which causes a change in its state of motion.
At the same time, since the line of action of this force does not pass through the center of gravity, a torque is developed. Torque is defined as the rotational effect developed due to the application of the force at a distance from the center of gravity. Thus, this torque causes angular acceleration, and hence, the body rotates around an axis. The combined effects of the linear acceleration and angular acceleration influence the motion of the body.
For example, when you push a door at its edge, it rotates around its hinges because of the torque generated, and the force exerts a linear effect as well. Similarly, if you apply force to a spinning top at any point other than its center, it causes rotation and a change in position as well. So, whenever the line of action of the force misses the center of gravity, both types of acceleration occur.
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