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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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3 years ago
A comet fragment of mass 1.96 × 1013 kg is moving at 6.50 × 104 m/s when it crashes into Callisto, a moon of Jupiter. The mass o
Andrej [43]

Answer:

Recoil speed, 1.17\times 10^{-5}\ m/s                          

Explanation:

Given that,

Mass of the comet fragment, m_1=1.96\times 10^{13}\ kg

Speed of the comet fragment, v_1=6.5\times 10^4\ m/s

Mass of Callisto, m_2=1.08\times 10^{23}\ kg

The collision is completely inelastic. Assuming for this calculation that Callisto's initial momentum is zero. So,

m_1v_1=(m_2+m_2)V

V is recoil speed of Callisto immediately after the collision.

V=\dfrac{m_1v_1}{(m_2+m_2)}\\\\V=\dfrac{1.96\times 10^{13}\times 6.5\times 10^4}{(1.96\times 10^{13}+1.08\times 10^{23})}\\\\V=1.17\times 10^{-5}\ m/s

So, the recoil speed of Callisto immediately after the collision is 1.17\times 10^{-5}\ m/s

6 0
3 years ago
The rate at which energy is transferred is called a. joules. b. power. c. work. d. time.
Marrrta [24]
B. Power
I hope this helps you, have a great day!
7 0
3 years ago
Read 2 more answers
A moon orbits a planet every 42 hours with a mean orbital radius of .002819 AU. The mass of the moon is 8.932 x 1022 kg. Using N
Pepsi [2]

Answer:

The mass of the planet  is 1.9407\times10^{27}\ kg

Explanation:

Given that,

Time period = 42 hours = 151200 sec

Orbital radius = 0.002819 AU = 421716397.5 m

Mass of moon m=8.932\times10^{22}\ kg

We need to calculate the mass of the planet

Using Kepler’s third law

T^2\propto a^3

T^2=\dfrac{4\pi^2}{G(M+m)}\times a^3

Where, a = orbital radius

T = time period

G = gravitational constant

M = mass of moon

m = mass of planet

Put the value into the formula

(151200)^2=\dfrac{4\pi^2}{6.673\times10^{-11}(8.932\times10^{22}+m)}\times(421716397.5)^3

(8.932\times10^{22}+m)=\dfrac{4\pi^2}{6.673\times10^{-11}}\times\dfrac{(421716397.5)^3}{(151200)^2}

(8.932\times10^{22}+m)=1.94087\times10^{27}

m=1.94087\times10^{27}-8.932\times10^{22}

m=1.9407\times10^{27}\ kg

Hence, The mass of the planet  is 1.9407\times10^{27}\ kg

8 0
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