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jeka94
4 years ago
5

A swimming pool has the shape of a box with a base that measures 30 m by 10 m and a uniform depth of 2.5 m. How much work is req

uired to pump the water out of the pool when it is​ full? Use 1000 kg divided by m cubed for the density of water and 9.8 m divided by s squared for the acceleration due to gravity.

Physics
2 answers:
alisha [4.7K]4 years ago
8 0

Explanation:

Below is an attachment containing the solution.

pav-90 [236]4 years ago
4 0

Answer:

18,375 kJ

Explanation:

Parameters given:

Density of water: 1000 kg/m³

Acceleration due to gravity = 9.8 m/s²

Dimensions of pool = 30m × 10m × 2.5m

Depth of pool = 2.5 m

First we need to find the mass of the water in the pool:

Density = mass/volume

Mass = density * volume

Since the pool is full of water, the volume of the water is equal to the volume of the pool:

Volume = 30 * 10 * 2.5 = 750 m³

Mass = 1000 * 750

Mass = 750,000 kg

We can now find the force required to pump the water out of the pool:

F = m * g

Where m = mass of water

g = acceleration due to gravity

F = 750,000 * 9.8 = 7,350,000 N

The work needed to be done is the product of the force required by the distance the water would need to move to be pumped out (depth of pool):

Work = F * d

Work = 7,350,000 * 2.5

Work = 18,375,000 J = 18,375 kJ

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The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde
Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

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                                 S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m

Now

The intensity at O when both speakers are on is given by

I=4I_1 cos^2(\pi \frac{\delta}{\lambda})

Here

  • I is the intensity at O when both speakers are on which is given as 6 W/m^2
  • I1 is the intensity of one speaker on which is 6  W/m^2
  • δ is the Path difference which is given as

                                           \delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m

  • λ is wavelength which is given as

                                             \lambda=\frac{v}{f}

      Here

              v is the speed of sound which is 320 m/s.

              f is the frequency of the sound which is to be calculated.

                                  16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )

where k=0,1,2

for minimum frequency f_1, k=1

                                  {f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz

So the minimum frequency is 702.22 Hz

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