Answer:
The inducerd emf is 1.08 V
Solution:
As per the question:
Altitude of the satellite, H = 400 km
Length of the antenna, l = 1.76 m
Magnetic field, B = 
Now,
When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

Here, velocity v is perpendicular to the rod
Thus
e = lvB (1)
For the orbital velocity of the satellite at an altitude, H:

where
G = Gravitational constant
= mass of earth
= radius of earth

Using this value value in eqn (1):

Answer:
diverging light rays of the bulb are collected by the reflector.
Explanation:
Answer:
The separation between the charges was decreased by a factor of 0.2
Explanation:
The Coulomb's force between two charges is given by;

r₂ = 0.2r₁
Therefore, the separation between the charges was decreased by a factor of 0.2.
Explanation:
Work done is the force applied to move a body through a specific or particular direction.
It is also the difference in the amount of energy expended in using an effort.
Work done is given as;
Work done = F x d CosФ
F is the force applied
d is the displacement
Ф is the angle
The unit of work done is in Joules.
Answer:
1) 0.43 meters per second
2) 0.21 meters per second
3) 1.02 
4) 0.66 seconds
Explanation:
part 1
By conservation of energy, the maximum kinetic energy (K) of the block is at equilibrium point where the potential energy is zero. So, at the equilibrium kinetic energy is equal to maximum potential energy (U):


With m the mass, v the speed, k the spring constant and xmax the maximum position respect equilibrium position. Solving for v

part 2
Again by conservation of energy we have kinetic energy equal potential energy:


part 3
Acceleration can be find using Newton's second law:

with F the force, m the mass and a the acceleration, but elastic force is -kx, so:


part 4
The period of an oscillator is the time it takes going from one extreme to the other one, that is going form 4.5 cm to -4.5 cm respect the equilibrium position. That period is:

So between 0 and 4.5 cm we have half a period:
