Rubber. The other three are metals, and therefore good conductors.
Complete Question:
Metal sphere A has a charge of − Q . −Q. An identical metal sphere B has a charge of + 2 Q . +2Q. The magnitude of the electric force on sphere B due to sphere A is F . F. The magnitude of the electric force on sphere A due to sphere B must be:
A. 2F
B. F/4
C. F/2
D. F
E. 4F
Answer:
D.
Explanation:
If both spheres can be treated as point charges, they must obey the Coulomb's law, that can be written as follows (in magnitude):

As it can be seen, this force is proportional to the product of the charges, so it must be the same for both charges.
As this force obeys also the Newton's 3rd Law, we conclude that the magnitude of the electric force on sphere A due to sphere B, must be equal to the the magnitude of the force on the sphere B due to the sphere A, i.e., just F.
Answer:
35 mph
Explanation:
The key of this problem lies in understanding the way that projectile motion works as we are told to neglect the height of the javelin thrower and wind resistance.
When the javelin is thown, its velocity will have two components: a x component and a y component. The only acceleration that will interact with the javelin after it was thown will be the gravety, which has a -y direction. This means that the x component of the velocity will remain constant, and only the y component will be affected, and can be described with the constant acceleration motion properties.
When an object that moves in constant acceleration motion, the time neccesary for it to desaccelerate from a velocity v to 0, will be the same to accelerate the object from 0 to v. And the distance that the object will travel in both desaceleration and acceleration will be exactly the same.
So, when the javelin its thrown, it willgo up until its velocity in the y component reaches 0. Then it will go down, and it will reach reach the ground in the same amount of time it took to go up and, therefore, with the same velocity.
Answer:
I = 1.38 A
Explanation:
Given that,
Charge, q = 5000 C
Time, t = 1 hour = 3600 s
We need to find the current intensity. The current intensity is equal to the electric charge per unit time. It can be given by :

Substitute all the values in the above formula

So, the current intensity is 1.38 A.