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Sergio039 [100]
3 years ago
5

A 500 kg satellite is placed in a circular orbit 6394 km above the surface of the earth. At this elevation, the acceleration of

gravity is 4.09 m/s2. Knowing that its orbital speed is 20,000 km/h, determine the kinetic energy of the satellite. (Round the final answer to two decimal places.)
Physics
1 answer:
Digiron [165]3 years ago
6 0

Answer:

KE  = 7.7160 GJ

Explanation:

given data

mass = 500 kg

radius = 6394 km

acceleration of gravity = 4.09 m/s²

orbital speed = 20,000 km/h  = 20000 × \frac{1000}{3600}  = 5555.56 m/s

solution

we Kinetic energy is express as

KE = 0.5 × m × v²   .................1

here m is mass and v is velocity

put here value and we get

KE = 0.5 × 500 × 5555.56²

KE = 7716061728.400

KE  = 7.7160 GJ

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Answer:

The height of the image of the candle is 20 cm.

Explanation:

Given that,

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Object distance from the candle, u = -6 cm

Focal length of converging lens, f = 15 cm

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Solution,

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\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

v is image distance

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-6)}\\\\v=-10\ cm

If h' is the height of the image. Magnification is given by :

m=\dfrac{h'}{h}=\dfrac{v}{u}

h'=\dfrac{vh}{u}\\\\h'=\dfrac{-10\times 12}{-6}\\\\h'=20\ cm

So, the height of the image of the candle is 20 cm.

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Explanation:

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Explanation:

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