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user100 [1]
3 years ago
11

Please help!! giving a lot of points

Physics
1 answer:
Readme [11.4K]3 years ago
7 0

Question 1.

  • mass = 4500 kg
  • potential energy (p.e) = 67500 J

now, we know :

=》

p.e =  mgh

=》

67500 = 4500 \times 10 \times h

=》

67500 = 45000 \times h

=》

h =  \dfrac{67500}{45000}

=》

h = 1.5 \: m

note : if we take acceleration due to gravity as 9.8, then height = 1.53 m

Question 2.

  • mass = 4500 kg
  • kinetic energy = 63000 j

we know,

=》

k.e =  \dfrac{1}{2} mv {}^{2}

=》

63000 =  \dfrac{1}{2}  \times 4500 \times  {v}^{2}

=》

{v}^{2}  =  \dfrac{63000 \times 2}{4500}

=》

{v}^{2}  = 28

=》

v =  \sqrt{28}

=》

v = 2 \sqrt{7} \:  \:  ms {}^{ - 1}

or

=》

5.29 \:  \: ms {}^{ - 1}

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The water required to pump all the water to a platform 2 feet above the top of the pool is  is 6061310.32 foot-pound.

Explanation:

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So, the total distance to lift slice is (12-h) instead of of 10-h

We need to calculate the water required to pump all the water to a platform 2 feet above the top of the pool

Using formula of work done

W=\int_{0}^{h}{W}

Put the value into the formula

W=\int_{0}^{10}{27562.5\pi\\times(12-h)}dh

W=27562.5\pi(12h-\dfrac{h^2}{2})_{0}^{10}

W=27562.5\pi(12\times10-\dfrac{100}{2}-0)

W=1929375\pi

W=6061310.32\ foot- pound

Hence, The water required to pump all the water to a platform 2 feet above the top of the pool is  is 6061310.32 foot-pound.

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