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goblinko [34]
3 years ago
15

"an open tank has the shape of a right circular cone. the tank is 6 feet across the top and 5 feet high. how much work is done i

n emptying the tank by pumping the water over the top edge? note: the density of water is 62.4 lbs per cubic foot."
Physics
1 answer:
Nonamiya [84]3 years ago
3 0

Answer:

The amount of workdone by in emptying the tank by pumping the water over the top edge is \mathbf{ 1169.99 \pi \ ft-lbs}

Explanation:

Given that the tank is 6 feet across the top and 5 feet high.

Using the similar triangles.

\dfrac{3}{5}  = \dfrac{r}{y}

5r = 3y

r = \dfrac{3}{5} y

Thus; each disc is a circle with area

A = \pi ( \frac{3}{5} y)^2

The weight of each disc is ;

m = \rho_{water}A

= 62.4 × \pi ( \frac{3}{5} y)^2

= \dfrac{561.6}{25} \pi y^2

The distance pumped is 5-y

Thus; the workdone in pumping the tank by pumping the water over the top edge is :

W = \dfrac{561.6 \pi}{25} \int\limits^5_0 {(5-y)} \, y^2dy

W = \dfrac{561.6 \pi}{25} \int\limits^5_0 {(5y^2-y^3)} dy

W = \dfrac{561.6 \pi}{25}[52.083]

\mathbf{W = 1169.99 \pi \ ft-lbs}

The amount of workdone by in emptying the tank by pumping the water over the top edge is \mathbf{ 1169.99 \pi \ ft-lbs}

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Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

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The value v_2  =  4 \sqrt{10} \  m/s

Explanation:

From the question we are told that

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    The charge on the second sphere is  q_2 =  8 \mu C = 8*10^{-6} \  C

     The  mass of the second charge is m  =  1.50 \  g  =  1.50 *10^{-3} \ kg

      The  distance apart is  d =  0.4 \  m

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Generally the total energy possessed by when q_2 and  q_1 are separated by 0.8 \  m is mathematically represented

     Q =  KE + U

Here KE   is  the kinetic energy which is mathematically represented as

     KE  =  \frac{1 }{2}  m (v_1)^2

substituting value

     KE  =  \frac{1 }{2}  * ( 1.50 *10^{-3}) (20 )^2

     KE  =  0.3 \  J

And  U is  the  potential  energy which is mathematically represented as

        U  =  \frac{k *  q_1 *  q_2  }{d }

substituting values

       U  =  \frac{9*10^9 *  2*10^{-6} * 8*10^{-6}  }{0.8 }

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       U_f  =  \frac{9*10^9 *  2*10^{-6} * 8*10^{-6}  }{0.4 }

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    0.48 =  0.36 +(7.50 *10^{-4} v_2^2)

   v_2  =  4 \sqrt{10} \  m/s

     

   

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