Answer:
The work done on the hose by the time the hose reaches its relaxed length is 776.16 Joules
Explanation:
The given spring constant of the of the spring, k = 88.0 N/m
The length by which the hose is stretched, x = 4.20 m
For the hose that obeys Hooke's law, and the principle of conservation of energy, the work done by the force from the hose is equal to the potential energy given to the hose
The elastic potential energy, P.E., of a compressed spring is given as follows;
P.E. = 1/2·k·x²
∴ The potential energy given to hose, P.E. = 1/2 × 88.0 N/m × (4.20 m)²
1/2 × 88.0 N/m × (4.20 m)² = 776.16 J
The work done on the hose = The potential energy given to hose, P.E. = 776.16 J
A. The sound will decrease in volume
The ideal mechanical advantage of the pulley system is 3
Answer:
I don’t understand it
Explanation:
If anyone can the. Let me know
Answer:
Momentum after collision will be 6000 kgm/sec
Explanation:
We have given mass of the whale = 1000
Initial velocity v = 6 m/sec
It collides with other mass of 200 kg which is at stationary
Initial momentum of the whale = 1000×6 = 6000 kgm/sec
We have to find the momentum after collision
From conservation of momentum
Initial momentum = final momentum
So final momentum = 6000 kgm/sec
