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snow_lady [41]
3 years ago
6

Your answer is incorrect. an empty swimming pool needs to be filled to the top. the pool is shaped like a cylinder with a diamet

er of 8 m and a depth of 1.6 m . suppose water is pumped into the pool at a rate of 11 m3 per hour. how many hours will it take to fill the empty pool? use the value 3.14 for π , and round your answer to the nearest hour. do not round any intermediate computations.
Physics
1 answer:
evablogger [386]3 years ago
4 0

How can my answer be incorrect if I haven't even figured it out yet ? ?

The volume of a cylinder is (pi) (radius²) (height, length, or depth).

The pool's diameter is 8m, so its radius is 4m .

The volume of the pool is (pi) (4m)² (1.6 m) = 25.6 pi m³ .

Filling it at the rate of 11 m³/hr, it will take

(25.6 pi / 11) = 7.308... hours to fill it to the brim.

Rounded to the nearest hour, that's 7 hours.

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     The  radius is r =  3.17 \  mm  =  3.17 *10^{-3} \ m

      The current density is  J =  c\cdot r^2  =  9.00*10^{6}  \ A/m^4 \cdot r^2

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Generally current density is mathematically represented as

          J  =  \frac{I}{A }

Where A is the cross-sectional area represented as

         A  =  \pi r^2

=>      J  =  \frac{I}{\pi r^2  }

=>    I  =  J  *  (\pi r^2 )

Now the change in current per unit length is mathematically evaluated as

        dI  =  2 J  *  \pi r  dr

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

         I  = 2\pi  \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr

         I  = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr

        I  = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ]  | \left    0.001585} \atop 0}} \right.

        I  = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]

substituting values

        I  = 2 *  3.142  *  9.00 *10^6 *   [ \frac{0.001585^4}{4} ]

        I  = 8.9 *10^{-5} \  A

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