Answer:
Energy Lost for group A's car = 0.687 J
Energy Lost for group B's car = 0.55 J
Explanation:
The exact question is as follows :
Given - The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potential energy of the car and its kinetic energy at the bottom of the hill equals the energy lost due to friction.
To find - How much energy is lost due to heat for group A's car ?
How much for Group B's car ?
Solution -
We know that,
GPE = 1 Joule (Potential Energy)
Now,
For Group A -
Energy Lost = GPE - KE
= 1 J - 0.313 J
= 0.687 J
So,
Energy Lost for group A's car = 0.687 J
Now,
For Group B -
Energy Lost = GPE - KE
= 1 J - 0.45 J
= 0.55 J
So,
Energy Lost for group B's car = 0.55 J
Answer:
No sand doesn't stay sand forever.
Explanation:
- We may have a thought that the sand we see on the beach areas are always the same one for eternal, but it is not true.
- Due to different activities like beach nourishment, sand replenishment etc. the sand in the beach areas are changed and replaced.
- If the sand remained there for long time, it also affects the sand eating organisms and plants.
Answer: The specific heat capacity is very low.
Explanation:
The specific heat capacity of a body is defined as the heat energy required by a body to cause a unit change in its temperature. The value is over low that is why it is easier for the desert sand to easily get very hot during the day. Conversely, it is very easy for the desert sand to lose it's heat a cool breeze pass over it in the night making it very cold in the night. This value also defines how long the desert sand can retain heat. Therefore, the desert sand has a low specific heat capacity.
Answer:
A) would be the best explanation because hot air is less dense and would rise to the top of a column of air
The maximum efficiency is 10%
Explanation:
The maximum efficiency of a machine is given by:
where
is the temperature of the hot reservoir
is the temperature of the cold reservoir
For the machine in this problem, we have:
is the temperature of the hot reservoir
is the temperature of the cold reservoir
substituting, we find the max efficiency:
So, the efficiency is 0.10 (10%).
Learn more about temperature:
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