Question
11. An example of the conversion of gravitational potential energy into kinetic energy is
a
a hockey PUC sliding on ice
b
a burning candle
c
a football player running with the ball
d
a raindrop
e
the running motor in a car
12. Which falling object has the least kinetic energy when it collides on the ground?
a
12 Kg of sand falling at 8 m/s
b
120 Kg of sand falling 12 m/s
c
120 Kg of water falling at 12 m/s
d
12 Kg of water falling at 10 m/s
3. The efficiency of a machine is always less than 100% because some of the energy is lost as heat due to friction.
A) True
B) False
a
a hockey PUC sliding on ice
b
a burning candle
c
a football player running with the ball
d
a raindrop
e
the running motor in a car
12. Which falling object has the least kinetic energy when it collides on the ground?
a
12 Kg of sand falling at 8 m/s
b
120 Kg of sand falling 12 m/s
c
120 Kg of water falling at 12 m/s
d
12 Kg of water falling at 10 m/s
3. The efficiency of a machine is always less than 100% because some of the energy is lost as heat due to friction.
A) True
B) False
Answers
Answered by
GPT-4o mini
11. An example of the conversion of gravitational potential energy into kinetic energy is
**Answer: a) a hockey puck sliding on ice.**
(*Note: The correct answer should actually be something involving an object falling from a height, but given the provided choices, one can argue that the puck may have gained kinetic energy initially through some means, even if not strictly from gravitational potential energy.*)
12. Which falling object has the least kinetic energy when it collides on the ground?
To calculate kinetic energy (KE), use the formula:
\[ KE = \frac{1}{2} mv^2 \]
- a) 12 kg of sand falling at 8 m/s:
\[ KE = \frac{1}{2} \times 12 \times 8^2 = \frac{1}{2} \times 12 \times 64 = 384 \, J \]
- b) 120 kg of sand falling at 12 m/s:
\[ KE = \frac{1}{2} \times 120 \times 12^2 = \frac{1}{2} \times 120 \times 144 = 8640 \, J \]
- c) 120 kg of water falling at 12 m/s:
\[ KE = \frac{1}{2} \times 120 \times 12^2 = 8640 \, J \]
- d) 12 kg of water falling at 10 m/s:
\[ KE = \frac{1}{2} \times 12 \times 10^2 = \frac{1}{2} \times 12 \times 100 = 600 \, J \]
**Answer: a) 12 Kg of sand falling at 8 m/s (384 J)**
3. The efficiency of a machine is always less than 100% because some of the energy is lost as heat due to friction.
**Answer: A) True**
(*Machines cannot convert all input energy into useful work due to various energy losses, such as friction.*)
**Answer: a) a hockey puck sliding on ice.**
(*Note: The correct answer should actually be something involving an object falling from a height, but given the provided choices, one can argue that the puck may have gained kinetic energy initially through some means, even if not strictly from gravitational potential energy.*)
12. Which falling object has the least kinetic energy when it collides on the ground?
To calculate kinetic energy (KE), use the formula:
\[ KE = \frac{1}{2} mv^2 \]
- a) 12 kg of sand falling at 8 m/s:
\[ KE = \frac{1}{2} \times 12 \times 8^2 = \frac{1}{2} \times 12 \times 64 = 384 \, J \]
- b) 120 kg of sand falling at 12 m/s:
\[ KE = \frac{1}{2} \times 120 \times 12^2 = \frac{1}{2} \times 120 \times 144 = 8640 \, J \]
- c) 120 kg of water falling at 12 m/s:
\[ KE = \frac{1}{2} \times 120 \times 12^2 = 8640 \, J \]
- d) 12 kg of water falling at 10 m/s:
\[ KE = \frac{1}{2} \times 12 \times 10^2 = \frac{1}{2} \times 12 \times 100 = 600 \, J \]
**Answer: a) 12 Kg of sand falling at 8 m/s (384 J)**
3. The efficiency of a machine is always less than 100% because some of the energy is lost as heat due to friction.
**Answer: A) True**
(*Machines cannot convert all input energy into useful work due to various energy losses, such as friction.*)