The kinetic energy of an object increases as its decreases <span>its potential energy as the sum of energy will remain constant.
In short, Your Answer would be "Decreases"
Hope this helps!</span>
Answer:
Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:
![E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}](https://tex.z-dn.net/?f=E_%7Boutside%7D%20%3D%5Cfrac%7B1%7D%7B4%5Cpi%28e_%7B0%7D%29%7D%5Cfrac%7BQ%7D%7Br%5E%7B2%7D%20%7D%20r%5E%7B%27%7D)
lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:
![E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}](https://tex.z-dn.net/?f=E_%7Binside%7D%20%3D%5Cfrac%7B1%7D%7B4%5Cpi%28e_%7B0%7D%29%7D%5Cfrac%7BQ%7D%7BR%5E%7B2%7D%20%7D%20%5Cfrac%7Br%7D%7BR%7D%20r%5E%7B%27%7D)
To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):
![V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2} } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r](https://tex.z-dn.net/?f=V%28r%3ER%29%3D%5Cint%5Climits%5Er_%5Cinfty%20%7B%5Cfrac%7B1%7D%7B4%5Cpi%28e_%7B0%29%7D%20%7D%5Cfrac%7BQ%7D%7Br%5E2%7D%20%20%7D%20%5C%2C%20dr%3D%5Cfrac%7Bq%7D%7B4%5Cpi%28e_%7B0%29%7D%20%7D%20%5Cfrac%7B1%7D%7Br%7D%20%5C%5CV%28r%3CR%29%3D-%20%5Cint%5Climits%5Er_%5Cinfty%7BE.dl%5C%5C%5C%5C%3D%20-%5Cint%5Climits%5ER_%5Cinfty%7B%5Cfrac%7B1%7D%7B4%5Cpi%28e_%7B0%29%7D%20%7D%5Cfrac%7BQ%7D%7Br%5E2%7D%20%20%7D%20-%5Cint%5Climits%5Er_R%7B%5Cfrac%7B1%7D%7B4%5Cpi%28e_%7B0%29%7D%20%7D%5Cfrac%7BQ%7D%7BR%5E2%7D%5Cfrac%7Br%7D%7BR%7D%20dr%5C%5C)
=![\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2} }{2R^{3} } ]](https://tex.z-dn.net/?f=%5Cfrac%7Bq%7D%7B4%5Cpi%20e_%7B0%7D%20%7D%20%5B%5Cfrac%7B1%7D%7BR%7D%20-%5Cfrac%7Br%5E%7B2%7D-R%5E%7B2%7D%20%20%7D%7B2R%5E%7B3%7D%20%7D%20%5D)
∴NOTE: Graph is attached
Answer:mass of the object,how much force earth exerts on the object,and shape of the object
Explanation:
The thermal energy that is generated due to friction is 344J.
<h3>What is the thermal energy?</h3>
Now we know that the total mechanical energy in the system is constant. The loss in energy is given by the loss in energy.
Thus, the kinetic energy is given as;
KE = 0.5 * mv^2 =0.5 * 15.0-kg * (1.10 m/s)^2 = 9.1 J
PE = mgh = 15.0-kg * 9.8 m/s^2 * 2.40 m = 352.8 J
The thermal energy is; 352.8 J - 9.1 J = 344J
Learn more about thermal energy due to friction:brainly.com/question/7207509
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