Thermal energy is converted to radiative energy via molecular collisions and released as photons.
I would say B. Because actual mass would ricochet off the sidewalk.
Answer:
2 hours
Explanation:
The motorcycle travels 40 km per hour.
80km / 40km/h = 2 hours.
The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J
![\texttt{ }](https://tex.z-dn.net/?f=%5Ctexttt%7B%20%7D)
<h3>Further explanation</h3>
Let's recall Elastic Potential Energy formula as follows:
![\boxed{E_p = \frac{1}{2}k x^2}](https://tex.z-dn.net/?f=%5Cboxed%7BE_p%20%3D%20%5Cfrac%7B1%7D%7B2%7Dk%20x%5E2%7D)
where:
<em>Ep = elastic potential energy ( J )</em>
<em>k = spring constant ( N/m )</em>
<em>x = spring extension ( compression ) ( m )</em>
Let us now tackle the problem!
![\texttt{ }](https://tex.z-dn.net/?f=%5Ctexttt%7B%20%7D)
<u>Given:</u>
mass of object = m = 1.25 kg
initial extension = x = 0.0275 m
final extension = x' = 0.0735 - 0.0275 = 0.0460 m
<u>Asked:</u>
kinetic energy = Ek = ?
<u>Solution:</u>
<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>
![F = k x](https://tex.z-dn.net/?f=F%20%3D%20k%20x)
![mg = k x](https://tex.z-dn.net/?f=mg%20%3D%20k%20x)
![k = mg \div x](https://tex.z-dn.net/?f=k%20%3D%20mg%20%5Cdiv%20x)
![k = 1.25(9.8) \div 0.0275](https://tex.z-dn.net/?f=k%20%3D%201.25%289.8%29%20%5Cdiv%200.0275)
![k = 445 \frac{5}{11} \texttt{ N/m}](https://tex.z-dn.net/?f=k%20%3D%20445%20%5Cfrac%7B5%7D%7B11%7D%20%5Ctexttt%7B%20N%2Fm%7D)
![\texttt{ }](https://tex.z-dn.net/?f=%5Ctexttt%7B%20%7D)
<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>
![Ep_1 + Ek_1 = Ep_2 + Ek_2](https://tex.z-dn.net/?f=Ep_1%20%2B%20Ek_1%20%3D%20Ep_2%20%2B%20Ek_2)
![\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dk%20%28x%27%29%5E2%20%2B%20mgh%20%2B%200%20%3D%20%5Cfrac%7B1%7D%7B2%7Dk%20x%5E2%20%2B%20Ek)
![Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2](https://tex.z-dn.net/?f=Ek%20%3D%20%5Cfrac%7B1%7D%7B2%7Dk%20%28x%27%29%5E2%20%2B%20mgh%20-%20%5Cfrac%7B1%7D%7B2%7Dk%20x%5E2)
![Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh](https://tex.z-dn.net/?f=Ek%20%3D%20%5Cfrac%7B1%7D%7B2%7Dk%20%28%20%28x%27%29%5E2%20-%20x%5E2%20%29%20%2B%20mgh)
![Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)](https://tex.z-dn.net/?f=Ek%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28445%20%5Cfrac%7B5%7D%7B11%7D%29%20%28%200.0460%5E2%20-%200.0275%5E2%20%29%20%2B%201.25%289.8%29%280.0735%29)
![\boxed {Ek \approx 1.20 \texttt{ J}}](https://tex.z-dn.net/?f=%5Cboxed%20%7BEk%20%5Capprox%201.20%20%5Ctexttt%7B%20J%7D%7D)
![\texttt{ }](https://tex.z-dn.net/?f=%5Ctexttt%7B%20%7D)
<h3>Learn more</h3>
![\texttt{ }](https://tex.z-dn.net/?f=%5Ctexttt%7B%20%7D)
<h3>Answer details</h3>
Grade: High School
Subject: Physics
Chapter: Elasticity