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```
What is pumped steadily out of a flooded basement at a speed of 5.0m/s through a uniform hose of radius 1.0 cm. The hose passes out through a window 3.0m above the water line. What is the power of the pump?
What is pumped steadily out of a flooded basement at a speed of 5.0m/sthrough a uniform hose of radius 1.0 cm. The hose passes out through awindow 3.0m above the water line. What is the power of the pump?

```
6 years ago

```							Suppose that a mass delta m of water is pumped in time delta t. The pump
increases the potential energy of the water by Delta mgh, where h is
the vertical distance through which it is lifted, and increases its
kinetic energy by 1 /2 delta mv^2 , where v is its final speed. The
work it does isDelta W = delta m gh +1/2 delta mv^2power P = Delta W / delta tP = delta m gh +1/2 delta mv^2 / delta tP = delta m / deta t (gh+1/2v^2)Now
the rate of mass delta m /delta t = rho *A*v where rho is density of
water and A is the area of the hose. The area of the hose A = pi*r^2 =
3.14*(0.01)^2 = 3.14*10^-4 m^2rho*A*v = 1000*3.14x10^-4x5 = 1.57 kg/secThus P = P = delta m / deta t (gh+1/2v^2) = 1.57*[9.8*3 + 5^2 / 2] P = 66 WThanks & Regards,Nirmal SinghAskiitians faculty
```
6 years ago
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