Answer:
v = 1.28 m/s
Explanation:
Given that,
Maximum compression of the spring, 
Spring constant, k = 800 N/m
Mass of the block, m = 0.2 kg
To find,
The velocity of the block when it first reaches a height of 0.1 m above the ground on the ramp.
Solution,
When the block is bounced back up the ramp, the total energy of the system remains conserved. Let v is the velocity of the block such that,
Initial energy = Final energy

Substituting all the values in above equation,

v = 1.28 m/s
Therefore the velocity of block when it first reaches a height of 0.1 m above the ground on the ramp is 1.28 m/s.
Answer:
It is a measure of the electric force per unit charge on a test charge.
Explanation:
The magnitude of the electric field is defined as the force per charge on the test charge.
Since we define electric field as the force per charge, it will have the units of force divided by the unit of charge. This implies that the SI unit of electric field is given as Newton/Coulomb (N/C).
it depends upon what state they are in like in motion or res
Answer:
16.2 s
Explanation:
Given:
Δx = 525 m
v₀ = 0 m/s
a = 4.00 m/s²
Find: t
Δx = v₀ t + ½ at²
525 m = (0 m/s) t + ½ (4.00 m/s²) t²
t = 16.2 s