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grandymaker [24]
3 years ago
9

) A circular swimming pool has a diameter of 16 m. The circular side of the pool is 4 m high, and the depth of the water is 3.5

m. (The acceleration due to gravity is 9.8 m/s2 and the density of water is 1000 kg/m3.) How much work (in Joules) is required to: (a) pump all of the water over the side
Physics
1 answer:
Ostrovityanka [42]3 years ago
3 0

Complete Question

A circular swimming pool has a diameter of 16 m. The circular side of the pool is 4 m high, and the depth of the water is 3.5 m. (The acceleration due to gravity is 9.8 m/s2 and the density of water is 1000 kg/m3.) How much work (in Joules) is required to:

(a) pump all of the water over the side

(b) pump all of the water out of an outlet 3 m over the side?

Answer:

a

The work required is W_d = 14.25*10^6J

b

The work is  W_{d3} = 3.6 *10^7 J

Explanation:

From the question we are told that

      The diameter is d = 16m

       The height of the circular side is H =4m

      The depth of water is  D = 3.5m

       The acceleration due to gravity is g =9.81m/s^2

        The density of water is  \rho_w = 1000kg/m^3

The radius of the swimming pool is

             r = \frac{16}{2} = 8m

The work is mathematically given as

             W_d = Force * distance

Now force is mathematically given as

            F  = density * area * height \ of  \ pool

               = \rho * (\pi r^2 ) dx

Now the work done to pump all of the water over the side

is

          W_d = \int\limits^D_0 {\rho (\pi r^2 )(H-x)} \, dx

               =\int\limits^D_0 {1000*9.81 *(\pi * 8^2) (4- x)} \, dx

               = 64000 *9.8 \pi \int\limits^{3.5} _0 {(4-x)} \, dx

              = 64000 * 9.8 \pi [4x - \frac{x^2}{2} ]\left {3.5} \atop {0} \right.

              = 64000 *9.8 \pi [4(3.5) - \frac{3.5^2}{2} ]

              =14.25*10^6J

Now for 3 meters over the side the new height would be

             H_{new} = H +3

                       = 4+3

                      =7m

Now applying the formula for work that is  above

           W_{d3 }= \int\limits^D_0 {\rho g (\pi r^2)(H-x)} \, dx

                 = \int\limits^{3.5}_0 {1000 *9.8 * (\pi * 8^2)(7 - x)} \, dx

                 = 64000 * 9.8 \pi [7x - \frac{x^2}{2} ]\left {3.5} \atop {0} \right.

                = 64000 *9.8  \pi[7(3.5) - \frac{(3.5)^2}{2} ]

                =3.6*10^7J

 

                     

             

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