Once you have that value, power is equal to work (which uses energy, measured in Joules) divided by time. As your work here is all due to gravitational potential energy, so P = (mgh)/t. P is your power you just calculated. t is 1.0 seconds, h is 24.0 m, and g is 9.8 m/s^2. Plus all those in and solve for m, the mass of water needed.

you mean E

PE = mgh = 1 x 9/8 x 24 = 235 J

235 / 0.8 = 294 J input.

in one second that is 294 watts, for 1 kg

a) The potential energy for 1.0 kg of water can be calculated using your equation for gravitational potential energy. GPE = mgh. M is mass (1.0 kg), g is 9.8 m/s^2, and h is the height (24.0 m). Plug your variables in and solve. Your answer will be in Joules.

45e6 / 294 = 153000 kg of water.

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Typical Hydroelectric Dam

Once you have that value, power is equal to work (which uses energy, measured in Joules) divided by time. As your work here is all due to gravitational potential energy, so P = (mgh)/t. P is your power you just calculated. t is 1.0 seconds, h is 24.0 m, and g is 9.8 m/s^2. Plus all those in and solve for m, the mass of water needed.

a) U=-mgh=-(1*9.8*24)= 235.2

b) 45.0 MW = 4.5 * 10^7 Watts, a unit of power. If the dam is only 80% efficient, you need to divide that number by 0.8 in order to figure out the total power that needs to be generated in order to create that much usable power.

This post is last updated on hrtanswers.com at Date : 1st of September – 2022