Answer:
Explanation:
(A) True: It is true.
In junction law, the current entering at a junction is equal to teh current leaving at the junction.
(B) False: It is false.
The kirchhoff's junction law is based on the conservation of charge.
(C) True: It is true.
Energy is used in the circuit.
(D) True: It is true.
It is based on the conservation of charge.
Answer:
1408.685 KN/C
Explanation:
Given:
R = 0.45 m
σ = 175 μC/m²
P is located a distance a = 0.75 m
k = 8.99*10^9
- The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:

part a)
Electric Field strength at point P: a = 0.75 m

part b)
Since, R >> a, we can approximate a / R = 0 ,
Hence, E simplified relation becomes:

E = σ / 2*e_o
part c)
Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:
Electric Field strength due to point charge is:
E = k*δ*pi*R^2 / a^2
Since, R << a, Surface area = δ*pi
Hence,
E = (k*δ*pi/a^2)
Answer:
The power will remain the same for a particular load as we are not changing the load. so if we increase the voltage, the current will decrease to make the net power consumed by the load same as before. If we increase the current, the voltage will decrease for making the power same. The power will only change when we changes the load.
Explanation:
Answer:
The height reached by the material on Earth is 91 km.
Explanation:
Given that,
Mass 
Radius = 1821 km
Height 
Suppose we need to find that how high would this material go on earth if it were ejected with the same speed as on Io?
We need to calculate the acceleration due to gravity on Io
Using formula of gravity

Put the value into the formula


Let v be the speed at which the material is ejected.
We need to calculate the height
Using the formula of height

Using ratio of height of earth and height of Io


Put the value into the formula





Hence, The height reached by the material on Earth is 91 km.