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3 years ago

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A space transportation vehicle releases a 470-kg communications satellite while in a circular orbit 350 km above the surface of

the Earth. A rocket engine on the satellite boosts it into an orbit 2350 km above the surface of the Earth. How much energy does the engine have to provide for this boost?

1 answer:

natima [27]3 years ago

8 0

Answer:

E = 3.194 x 10⁹ J = 3.194 GJ

Explanation:

The formula for the absolute potential energy is:

U = - GMm/2r

where,

G = Gravitational Constant = 6.67 x 10⁺¹¹ N m²/kg²

M = mass of Earth = 5.972 x 10²⁴ kg

m = mass of satellite = 470 kg

r = distance between the center of Earth and satellite

Thus, the energy required from engine will be difference between the potential energies.

E = U₂ - U₁

E = - GMm/2r₂ - (- GMm/2r₁)

E = (GMm/2)(1/r₁ - 1/r₂)

where,

r₁ = Radius of Earth + 350 km = 6371 km + 350 km = 6721 km = 6.721 x 10⁶ m

r₂=Radius of Earth + 2350 km=6371 km + 2350 km= 8721 km = 8.721 x 10⁶ m

therefore,

E = [(6.67 x 10⁺¹¹ N m²/kg²)(5.972 x 10²⁴ kg)(470 kg)/2](1/6.721 x 10⁶ m - 1/8.721 x 10⁶ m)

<u>E = 3.194 x 10⁹ J = 3.194 GJ</u>

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Answer:

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Explanation:

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3 years ago

Suppose 310. grams of ethanol (ethyl alcohol) is in an aluminum cup of 90.0 grams. Both of these are at 30.0C. A mass m of ice a

Ira Lisetskai [31]

Answer:

Explanation:

Given

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mass of aluminium cup $m_{al}=90\ gm$

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specific heat of aluminium $c_{al}=0.9\ J/g-K$

specific heat of ice $c_i=2.108\ J/g-K$

specific heat of water $c_w=4.184\ J/g-K$

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suppose m is the mass of ice added

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3 years ago

Work out the current through an electric kettle with a power of 2.2 kW if it uses the 180 V mains supply. Round your answer to 2

weeeeeb [17]

Answer:

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Explanation:

Given,

Power of electric kettle=2.2 kW=2200W (since 1kW=1000W)

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Current = Power ÷ Voltage

=2200W ÷ 180V

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An automobile traveling 92.0 km/h has tires of 64.0 cm diameter. (a) What is the angular speed of the tires about their axles? (

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Answer:

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Explanation:

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Angular speed is given by

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The angular speed of the tires about their axles is 79.861 rad/s.

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-79.861^2}{2\times 2\pi \times 38}\\\Rightarrow \alpha=-13.355\ rad/s^2$

The magnitude of acceleration is 13.355 m/s²

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The distance moved while slowing down is 76.4035 m

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Answer:

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3 years ago