The diagram does not represent a real electric field because the field lines__ NEED ANSWERS NOW
1 answer:
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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.