A few physics questions i dont know..?

Question

If the kinetic energy of a 10 kilogram object is 2000 Joules, its velocity is

10 m/s.

20 m/s.

100 m/s.

400 m/s.

An object is dropped from the top of a building through the air (friction is present) to the ground below. How does its kinetic energy (KE) just before striking the ground compare to its gravitational potential energy at the top of the building?

Kinetic Energy is equal to Potential Energy.

Kinetic Energy is greater than Potential Energy.

Kinetic Energy is less than Potential Energy.

It is impossible to tell.

A block of copper with mass 0.500 kg is heated to 400. degrees Celsius from 290. degrees Celsius. The specific heat capacity of copper is 387 J/ (kg*Co). The heat absorbed by the copper block is

7.74 E 5 J.

7.74 E 4 J.

5.61 E 4 J.

2.13 E 5 J.

2.13 E 4 J.

What is the work required to raise a 10. kg object from the surface of the Earth to a height of 2.0 m?

5.0 J

20. J

2.0 E2 J

2.0 E3 J

A crane raises a 150 N weight to a height of 2.0 m in in 5.0 s. The crane does work at a rate of

10. W

30. W

60. W

380 W

750 W

Laura is a star runner of the school's track team. During a trial, her speed was measured to be 9.80 m/s. If Laura’s mass is 45.2 kg, her kinetic energy during the trial was

2.17 E 3 J.

2.21 E 3 J.

2.21 E 2 J.

3.21 E 3 J.

4.43 E 2 J.

An unknown solid with a mass of 2.00 kilograms remains in the solid state while it absorbs 32.0 kilojoules of heat. Its temperature rises 4.00 degrees Celsius. What is the specific heat of the unknown solid?

0.250 kJ/kg*C

4.00 kJ/kg*C

8.00 kJ/kg*C

128 kJ/kg*C

256 kJ/kg*C

0.5000 kg of water at 35.00 degrees Celsius is cooled, with the removal of 6.300 E4 J of heat. What is the final temperature of the water? Specific heat capacity of water is 4186 J/(kg Co).

Answer 1

1) K = 1/2 mv^2

Or, 2K = mv^2

Or, v^2 = 2K/m

Or, v = sqrt(2K/m)

v = sqrt(2*2000/10) m/s = sqrt(400) m/s = 20 m/s

Ans: 20 m/s

2) Ans: Kinetic Energy is less than Potential Energy.

The remaining energy is used in doing work against friction.

3) Heat absorbed = mass * specific heat capacity * change in temperature

= 0.5 * 387 * (400 - 290) J

= 0.5 * 387 * 110 J = 21285 J = 2.13 E 4 J (approximately)

4) Work is to be done against gravity. Work done = change in gravitational potential energy = mgh = 10 * 9.8 * 2 J

= 196 J

= 1.96 E2 J

If you want to take approximate value, then 2.0 E2 J is correct.

5) Work done = weight * height = 150 * 2.0 = 300 J

Rate of work = work/time = 300/5.0 W = 60 W

Ans: 60 W

6) Kinetic energy = 1/2 mv^2

= 1/2 * 45.2 * 9.80^2 J

= 2170 J = 2.17 E3 J

7) Mass m = 2 kg

Heat H = 32 kilojoule

Change in temperature deltaT = 4 degree C

Specific heat capacity s = ?

H = m*s*deltaT

Or, s = H/(m * deltaT)

= 32/(2 * 4) kJ/kg*C

= 32/8 kJ/kg*C

= 4 kJ/kg*C

Ans: 4.00 kJ/kg*C

8) Heat removed H = 6.3 * 10^4 J = 63000 J

Specific heat capacity s = 4186 J/(kg Co)

Mass m = 0.5 kg

Let deltaT = change in temperature

H = m*s*deltaT

Or, deltaT = H/(m*s) = 63000/(0.5 * 4186) = 30.1 degree C

So, temperature drops by 30.1 deg C

Initial temperature is 35 deg C

Therefore, final temperature = 35 - 30.1 deg C = 4.9 deg C

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Note: Please do not mind but your questions show that you have doubts in fundamentals. Therefore, please try to learn theory first. If you want, you can contact me for clarifications in theory.