Q=mcΔt
Q= 1kg * 800J/kg°C*4°C
Q= 3200J
At the frequency of 5 MHz, the period of the oscillations is 1/5meg. That's a period of 1/5 microsecond.
There are 5 full cycles in one full microsecond, and there are 2.5 full cycles in a 0.5 us pulse.
You'll have to decide for yourself how damped a pulse of 2.5 cycles is, because the parameters of the definition are corrupted in the question.
Answer:
the speed of the block when it reaches point B is 14 m/s
Explanation:
Given that:
mass of the block slides = 1.5 - kg
height = 10 m
Force constant = 200 N/m
distance of rough surface patch = 20 m
coefficient of kinetic friction = 0.15
In order to determine the speed of the block when it reaches point B.
We consider the equation for the energy conservation in the system which can be represented by:






v = 14 m/s
Thus; the speed of the block when it reaches point B is 14 m/s
The current will only flow if the circuit is closed because if it is open, the connection is severed, therefore cannot produce electricity