Answer:
A
Explanation:
When you dropping the swing, it increases the kinetic energy. In the middle, that is when it is at it's fullest potential. As it makes its way up, it pushes again the gravitational force, and soon, the swing will stop at the top for a VERY SHORT AND SMALL period of time. That is when it is at it's very weakest point, making the answer, indeed, A.
The electric flux on a surface of area A in a uniform electric field E is given by
φ = EA cos θ
where θ = the angle between the directions of E and a normal vector to the surface of the area.
See the diagram shown below.
When the electric field is perpendicular to the surface, then θ = 0°, and
φ = EA cos(0°) = (6.20 x 10⁵ N/C)*(3.2 m²) = 1.984 x 10⁶ (N-m²)/C
When the electric field is parallel to the area, then θ = 90°, and
φ = EA cos(90°) = 0
Answer:
(a) 1.984 x 10⁶ (N-m²)/C
(b) 0
Answer: I am pretty sure it is (b) what is the value of mass suspended at the end of the spring.
Explanation:
Answer:
121440 feet is 23 miles.
5280 x 23= 121440
To check your answer just divide 121440 by 5280, which gives you 23.
Hopefully this helps you!
Answer:
Figure A
Explanation:
At first, the inflated balloon is rubbed against the hair.
In this situation, the balloon is charged by friction: because of the friction between the surface of the balllon and the hair, electrons are transferred from the hair to the surface of the balloon.
As a result, when the balloon is detached from the hair, it will have an excess of negative charge (due to the acquired electrons).
Then, the balloon is placed in contact with the non-conducting wall.
The non-conducting wall is initially neutral (equal number of positive and negative charges).
Because the wall is made of a non-conducting material (=isolant), the charges cannot move easily through it. Therefore, even though the charges on the wall feel a force due to the presence of the electrons in the balloon, they will not redistribute along the wall.
Therefore, the charges on the wall will remain equally distributed, as shown in figure A.