Answer: The work done is W = 3528J.
Explanation: Work (W) is an amount of energy used to do determined movement. To calculate it, we use the formula, as the movement is horizontal:
W = F.d, where
F is force on the object
d is distance it went through.
The force acting on the crate is due to friction.
= μ .
μ is the kinetic friction
is the normal force
There is no up and down movement, so , with g = 9.8m/s²
= μ . m . g
= 0.9 . 80 . 9.8
= 705.6 N
Substituing into the work formula, we have:
W = . d
W = 705.6 . 5
W = 3528 J
The work done to move the crate across a distance of 5m is 3528 J.
Answer:
Explanation:
After time "t" the angular position of A is given as
now we know that B start motion after time t1
so its angular position is also same as that of position of A after same time "t"
so we have
now since both positions are same
Using equation you can solve
Clever problem.
We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks. So if Fork-A is 256 Hz and the beat is 6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz. But which one is it ?
Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz. That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.
If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.
The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz. While it was loaded with wax, it was 261 Hz.