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pickupchik [31]
2 years ago
13

A pump is used to lift 100 KG of water from a wel 60 m deep,in 20 S If force of gravity on 1 KG is 10 N,find

Physics
1 answer:
Harlamova29_29 [7]2 years ago
6 0

Explanation:

Given,

  • m = 100 kg
  • g = 10 N/kg¹
  • h = 60 m
  • t = 20 s

To Find:

a) Work done by the pump

b) Potential energy stored in the water

c)Power spent by the pump

d)Power rating of the pump.

Solution:

  • a) Work done by the pump

We know that,

\rm \: Work  \: done = Force * Distance  \: moved

  • f = 100 kg * 10N/kg
  • d = 60 m

\rm \: Work\; Done =(100 \: kg \times  \cfrac{10N}{kg} ) \times 60 \: m

\rm \: Work\; Done =1000 \times 60 \: joule

\boxed{\rm \: Work\; Done =60000 \: joule}

[The unit'll be joule since N×M = J]

  • b) Potential energy stored in the water

\rm \: P.E = m \cdot g \cdot  h

  • m = 100 kg
  • g = 10N/kg
  • h = 60

\rm \: P.E =100 \:kg \:  \times  \cfrac{10 \: N}{kg}  \times 60

\boxed{\rm \: P.E =60000  \: joule}

  • same condition here as well, N×M = J
  • c) Power of the Pump

\rm \: P = W/T

  • where P = Power; W = Work done & T = Time taken
  • As we got the value of work done on question (a),& ATQ time taken is 20 S.

\rm \: P =  \cfrac{60000 \: joule}{20 \: seconds}  =\boxed{\rm { 3000 \: Watts }}\: or \: \boxed{\rm 3 \: kW}

  • d) Power rating of the pump = 3 kW

Assumption: The pump is 100% efficient & works well.

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Answer:

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8 0
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v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-95^2}{2\times 0.055}\\\Rightarrow a=-82045.45\ km/h^2

Converting to m/s²

a=82045.45=\frac{82045.45\times 1000}{3600\times 3600}=-6.33\ m/s^2

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Dividing both the accelerations, we get

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Explanation:

Given that,

Height = 10 ft = 3.048 m

Diameter = 10 ft = 3.048 m

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Using formula of power density

P=\dfrac{E}{V}

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E = energy

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Put the value into the formula

P=\dfrac{835\times10^{6}}{22.23\times10^{3}}

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