Answer:
a) 50μC
b) 37.45 m/s
Explanation:
a) If the spheres are connected the charge in both spheres tends to be equal. This because is the situation of minimum energy.
Thus, you have:
Hence, each sphere has a charge of 50μC.
b) You use the fact that the total work done by the electric force is equal to the change in the kinetic energy of the sphere. Then, you use the following equations:
![\Delta W=\Delta K\\\\\int_{0.4}^\infty Fdr=\frac{1}{2}m[v^2-v_o^2]\\\\F=k\frac{Q^2}{r^2}\\\\v_o=0m/s\\\\m=0.08kg\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=kQ^2[-\frac{1}{r}]_{0.4}^{\infty}=\frac{kQ^2}{0.4m}=\frac{(8.98*10^9Nm^2/C^2)(50*10^{-6}C)^2}{0.4m}\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=56.125J](https://tex.z-dn.net/?f=%5CDelta%20W%3D%5CDelta%20K%5C%5C%5C%5C%5Cint_%7B0.4%7D%5E%5Cinfty%20Fdr%3D%5Cfrac%7B1%7D%7B2%7Dm%5Bv%5E2-v_o%5E2%5D%5C%5C%5C%5CF%3Dk%5Cfrac%7BQ%5E2%7D%7Br%5E2%7D%5C%5C%5C%5Cv_o%3D0m%2Fs%5C%5C%5C%5Cm%3D0.08kg%5C%5C%5C%5CkQ%5E2%5Cint_%7B0.4%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7Bdr%7D%7Br%5E2%7D%3DkQ%5E2%5B-%5Cfrac%7B1%7D%7Br%7D%5D_%7B0.4%7D%5E%7B%5Cinfty%7D%3D%5Cfrac%7BkQ%5E2%7D%7B0.4m%7D%3D%5Cfrac%7B%288.98%2A10%5E9Nm%5E2%2FC%5E2%29%2850%2A10%5E%7B-6%7DC%29%5E2%7D%7B0.4m%7D%5C%5C%5C%5CkQ%5E2%5Cint_%7B0.4%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7Bdr%7D%7Br%5E2%7D%3D56.125J)
where you have used the Coulomb constant = 8.98*10^9 Nm^2/C^2
Next, you equal the total work to the change in K:
hence, the speed of the spheres is 37.45 m/s
Answer:
The kinetic energy of the phone would increase. The gravitational potential energy of the phone would decrease.
Explanation:
The kinetic energy 
of an object is proportional to the square of the speed of that object. If air resistance is negligible, the phone would accelerate under gravitational pull and speed up. Hence, the kinetic energy of the phone would increase.
The gravitational field near the surface of the earth is approximately constant. Hence, the gravitational potential energy 
of the phone would be proportional to its height. As the phone approaches the ground, the height of the phone becomes lower and the gravitational potential energy of the phone would decrease.
Answer:
The correct option is D
Explanation:
This question can be better understood when discussed using the Newton's first law of motion which states that an object would continue to move with a uniform speed (in a straight line) unless acted upon by an external force. What happens here (in the question) is that the bike rider would have continued moving at a constant speed (to the right) if not for the opposing force of the wind that acted against her (to the left). <u>This wind/force would cause her speed to reduce or slow down (as posited by the law)</u>.
Answer:
4.17 m/s²
Explanation:
We are told the reaction time is 0.2 s. Now, during this reaction time the car is going to travel an additional distance of
: x = u × t = 40 × 0.2 = 8 m
where u is the initial velocity of the car which is 40.0 m/s.
We are told that he had 200 m to stop before applying brakes. Thus, after applying brakes, he now has a distance to cover of; s = 200 - 8 = 192 m
Since vehicle is coming to rest acceleration would be negative, thus using Newton's equation of motion, we have;
v
² = u² - 2as
v = 0 m/s since it's coming to rest
u = 40 m/s
s = 192 m
Thus;
0² = 40² - 2(a)(192)
0² = 1600 - 384a
a = 1600/384
a = 4.17 m/s²
