Answer:
a) 60 V
b) 125 V
c) 125 V
Explanation:
<u>Given</u>
We are given the total electric charge q = 6.75 nC = 6.75x 10^-9 C distributed uniformly over the surface of a metal sphere with a radius of R = 20.0 cm = 0.020 m.
<u>Required </u>
We are asked to calculate the potential at the distances
(a) r = 10.0 cm
(b) r = 20.0 cm
(c) r = 40.0 cm
<u>Solution</u>
(a) Here, the distance r > R so, we can get the potential outside the sphere (r > R) where the potential is given by
V = q/4
∈_o (1)
r is the distance where the potential is measured and the term 1/4∈_o equals 9.0 x 10^9 Nm^2/C^2. Now we can plug our values for q and r into equation (1) to get the potential V where r = 0.10 m
V= 1*q/4∈_o*r
=60 V
(b) Here the distance r is the same for the radius R, so we can get the potential inside the sphere (r = R) where the potential is given by
V = 1*q/4∈_o*R (2)
Now we can plug our values for q and R into equation (2) to get the potential V where R = 0.20 m
V = 1*q/4∈_o*R
= 125 V
(c) Inside the sphere the electric field is zero therefore, no work is done on a test charge that moves from any point to any other point inside the sphere. Thus the potential is the same at every point inside the sphere and is equal to the potential on the surface. and it will be the same as in part (b)
V= 125 V
Answer:
13 blocks
Explanation:
The total distance the student travels is 13 blocks.
Distance is the length of path covered during the motion of a body.
To find distance:
Total distance = Number of blocks to the west + number of blocks to the north + number of blocks to the east
Total distance = 3blocks + 4blocks + 6blocks = 13blocks
Answer:
Depth of the pool, h = 4.004 cm
Explanation:
Pressure at the bottom, P = 39240 N/m²
The density of water, d = 1000 kg/m³
The pressure at the bottom is given by :
P = dgh
We need to find the depth of pool. Let h is the depth of the pool. So,
h = 4.004 m
So, the pool is 4.004 meters pool. Hence, this is the required solution.
Answer:
Newton's Cradle experiment perfectly demonstrates the law of conservation of momentum which states that in a closed system, momentum before the collision is equal to momentum after the collision of the system.
As the first ball swings in the air, it gains momentum. When it strikes the second ball, it loses momentum and second ball gains equal amount of momentum. The second ball transfers the momentum to third, then fourth and till the last. The last ball when gains the same momentum swings up in the air. This continues. This experiment is done in drag free condition. This means there is no loss of momentum or opposing forces present.
