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Korvikt [17]
2 years ago
7

An electric motor pumps 200 liter of water to a reservoir of height 6m in 2 s. Take g = 10 m/s?. Calculate the power developed b

y the motor. Take 1 liter of water = 1kg of water.​
Physics
1 answer:
lorasvet [3.4K]2 years ago
7 0

The power developed by the electric motor is 6000 Watts

<h3>Definition of power </h3>

Power is defined as the rate at which work is done. Mathematically, it is expressed as

Power = Work / time

But

Work = mass (m) × acceleration due to gravity (g) × height (h)

Work = mgh

Therefore,

Power = mgh / t

With the above formula, we can obtain the power developed by the motor.

<h3>How to determine the power </h3>
  • Mass (m) = 200 Kg
  • Height (h) = 6 m
  • Acceleration due to gravity (g) = 10 m/s²
  • Time (t) = 2 s
  • Power =?

Power = mgh / t

Power = (200 × 10 × 6) / 2

Power = 12000 / 2

Power = 6000 Watts

Learn more about power:

brainly.com/question/5684937

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Which vector has an x-component with a length of 3?<br> А. С.<br> B. d<br> C. a<br> D. b
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Convert <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%280.779mg%29%28min%29%7D%7BL%7D" id="TexFormula1" title="\frac{(0.779mg)(mi
Orlov [11]

The number converted is 0.0467 \frac{(kg)(s)}{m^3}

Explanation:

In order to convert from the original units to the final units, we have to keep in mind the following conversion factors:

1 kg = 1000 g = 10^6 mg

1 min = 60 s

1 m^3 = 1000 L

The original unit that we have is

\frac{mg\cdot min}{L}

Therefore, it can be rewritten as:

=\frac{mg \frac{1}{10^6 mg/kg}\cdot min\cdot  60 s/min}{L\frac{1}{1000L/m^3}}=0.06 \frac{(kg)(s)}{m^3}

Therefore, since the initial number was 0.779, the final value is

0.779\cdot 0.06 \frac{(kg)(s)}{m^3}=0.0467 \frac{(kg)(s)}{m^3}

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