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Neporo4naja [7]
3 years ago
7

The level of water in an olympic size swimming pool (50.0 meters long, 25.0 meters wide, and about 2.00 meters deep) needs to be

lowered 6.50 cm. if water is pumped out at a rate of 4.20 liters per second, how long will it take to lower the water level 6.50 cm
Physics
1 answer:
Goshia [24]3 years ago
4 0
<span>We need to start by finding the surface area of the pool.
50 meters multiplied by 25 meters gives us 1250 square meters.
1250 square meters multiplied by .065 (6.5 cm in meters) gives us a volume of 81.25 cubic meters of water that needs to be pumped out of the pool.

There are 1000 liters in a cubic meter so this is 81250 liters. Divide by 4.2 to find the number of seconds required to pump out this much water and we get 19345.2 seconds. This equals approximately 5.37 hours.</span>
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The average power dissipated in a 47 ω resistor is 2.0 w. what is the peak value i 0 of the ac current in the resistor?
pishuonlain [190]
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P=I_{rms}^2 R
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I_{rms} =  \frac{I_0}{\sqrt{2}}
where I_0 is the peak value of the current. Substituting the second formula into the first one, we find
P=( \frac{I_0}{\sqrt{2} } )^2 R =  \frac{1}{2} I_0^2 R
and if we re-arrange this formula and use the data of the problem, we can find the value of the peak current I0:
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3 years ago
Whose contributions to astronomy explained how planets were held in their orbits?
ankoles [38]

Answer:

D newton

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anyanavicka [17]

Answer:

<h2>44 m/s</h2>

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In this problem we are expected to calculate the velocity of Georges movements.

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Potential energy changes along with kinetic energy. At any time, your total energy is equal to your potential energy plus your kinetic energy. As we go up, the kinetic energy converts into potential energy.

Hooke's law is another form of potential energy. Just as the trampoline is about to propel us up, your kinetic energy is 0 but your potential energy is maximized, even though we are at a minimum height. This is because our potential energy is related to the spring constant and Hooke's Law.

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