Explanation:
A worker picks up the bag of gravel. We need to find the speed of the bucket after it has descended 2.30 m from rest. It is case of conservation of energy. So,
h = 2.3 m
So, the speed of the bucket after it has descended 2.30 m from rest is 6.71 m/s.
A) Work energy relation;
Work =ΔKE ; work done = Force × distance, while, Kinetic energy = 1/2 MV²
F.s = 1/2mv²
F× 4×10^-2 = 1/2 × 5 ×10^-3 × (600)²
F = 900/0.04
= 22500 N
Therefore, force is 22500 N
b) From newton's second law of motion;
F = Ma
Thus; a = F/m
= 22500/(5×10^-3)
= 4,500,000 m/s²
But v = u-at
0 = 600- 4500,000 t
t = 1.33 × 10^-4 seconds
I would think that you would multiply then divide
When a crest-trough meet the interference produced will be destructive in nature hence they both will cancel out and the amplitude produced will be equal to zero hence the loudness will reduce to zero.
