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Basile [38]
3 years ago
7

An electric motor uses 670 kJ of electrical energy to generate 595 kJ of mechanical kinetic energy. What is the efficiency of th

e motor?
Full example, please...
Thank you!
Physics
1 answer:
stellarik [79]3 years ago
4 0
  • Used energy=595KJ
  • Total energy=670KJ

Efficiency:-

\\ \tt\longmapsto \dfrac{595}{\cancel{670}}\times \cancel{100}

\\ \tt\longmapsto \dfrac{595}{67}\times 10

\\ \tt\longmapsto 8.88(10)

\\ \tt\longmapsto 88.8\%

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The closeness of particles of gas and their low speeds allow intermolecular forces to become important at certain pressures and
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This is a limitation of kinetic-molecular energy. Right?
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A 3.50 cm tall object is held 24.8 cm from a lens of focal length 16.0 cm. What is the image height?
Mandarinka [93]

Answer:

<h2>6.36 cm</h2>

Explanation:

Using the formula to first get the image distance

1/f = 1/u+1/v

f = focal length of the lens

u = object distance

v = image distance

Given f = 16.0 cm, u = 24.8 cm

1/v = 1/16 - 1/24.8

1/v = 0.0625-0.04032

1/v = 0.02218

v = 1/0.02218

v = 45.09 cm

To get the image height, we will us the magnification formula.

Mag = v/u = Hi/H

Hi = image height = ?

H = object height = 3.50 cm

45.09/24.8 = Hi/3.50

Hi = (45.09*3.50)/24.8

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6 0
3 years ago
why does diving 30m below sea level affect our bodies more than being in a building 30m above sea level
Genrish500 [490]
Imagine you are in a swimming pool 30m deep. Assuming you know that water is denser than air, you would know that the 30m of water above you will carry more weight, and press down on your body. Say you were in a swimming pool 60m deep, you would be sandwiched between 30m of water pressing down on you, and the upthrust created by the 30m of water below you.

In a building 30m up, the pressure will be regulated, as you are in a building. The floor will be strong enough to support the weight of the body, and the body will not recoil into itself.
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Answer:

The centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².

                     

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The centripetal acceleration is given by:

a_{c} = \frac{v^{2}}{r}

Where:

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r: is the distance = 6.00 m

Hence, the centripetal acceleration is:

a_{c} = \frac{v^{2}}{r} = \frac{(9.50 m/s)^{2}}{6.00 m} = 15.04 m/s^{2}

Therefore, the centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².

I hope it helps you!

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Ke= 1/2 x m x v^2
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