1 year ago

Water falls off a cliff of height 20m at a rate of 50kg per second

Physics

1 answer:

docker41 [41]1 year ago

5 0

Answer:

M = 50 kg / sec * 60 sec = 3000 kg water that falls

PE = M g h initial potential energy of water

PE = 3000 kg * 9.8 m/s^2 * 20 m = 588,000 joules of energy lost

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What is the length a rubberband was stretched if it has a spring constant of 5700N/m and is currently holding 8600J OF POTENTIAL

lozanna [386]

Answer:

$\displaystyle \Delta x=1.74\ m$

Explanation:

<u>Elastic Potential Energy </u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

$\displaystyle PE = \frac{1}{2}k(\Delta x)^2$

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

$\displaystyle \Delta x=\swrt{\frac{2PE}{k}}$