As an object rolls downhill, some of the energy is

2 answers:

7 0

When the object is at the top of the hill it has the most potential energy. If it is sitting still, it has no kinetic energy. As the object begins to roll down the hill, it loses potential energy, but gains kinetic energy. The potential energy of the position of the object at the top of the hill is getting converted into kinetic energy. Hope this helped. :)

mr_godi [17]3 years ago

6 0

Well this is kind of a trick question. When the object is still before it rolls downhill it is potential. But as the object is rolling down the hill, it is kinetic energy. So, I believe your answer would be kinetic since it doesn't specify everything that's happening.

You might be interested in

15 points. give me the method.

AveGali [126]

Answer:

$\boxed{{160 \: m(s)}^{ - 1} }$

Explanation:

$if \:the \: frequencies \: are \to \\ f_{1} = 640Hz \\ and \\f_{2} = 480Hz \: \\ but \: \boxed{v = f \gamma }: f = \frac{v}{ \gamma } \\ if \: \gamma_{1} - \gamma _{2} = 1 = \gamma \\ f_{1} - f_{2} = 640 - 480 = \boxed{ 160Hz} = f \\ v = f \gamma = 160 \times 1 = \boxed{{160 \: m(s)}^{ - 1} }$

5 0

2 years ago

Which one of the following statements concerning the Stefan-Boltzmann equation is correct? The equation can be used to calculate

Helen [10]

"The equation can be used to calculate the power absorbed by any surface" statement concerning the Stefan-Boltzmann equation is correct.

Answer: Option A

<u>Explanation:</u>

According to Stefan Boltzmann equation, the power radiated by black body radiation source is directly proportionate to the fourth power of temperature of the source. So the radiation transferred is absorbed by another surface and that absorbed power will also be equal to the fourth power of the temperature. So the equation describes the relation of net radiation loss with the change in temperature from hotter temperature to cooler temperature surface.

$P=e \sigma A\left(T^{4}-T_{c}^{4}\right)$