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EastWind [94]
3 years ago
15

. The international space station has a mass of 4.2 x 105 kg and it orbits the Earth at an average altitude of 400 km above the

Earth's surface. The radius of the Earth is 6400 km, and the mass of the Earth is 6.0 x 1024 kg. Assume the only force acting on the space station is the Universal Gravitational Force, Fo, and assume that the orbit is perfectly circular. How long would it take for the space station to make one complete orbit around the Earth given these assumptions? A. 93 minutes B. 3.5 x 10 minutes C. 85 minutes D. 21 minutes E. 1.3 minutes
Physics
1 answer:
Tresset [83]3 years ago
5 0

Answer:

A. 93 min

Explanation:

M = mass of earth = 6.0 x 10²⁴ kg

R = Radius of earth = 6400 km = 6.4 x 10⁶ m

h = Altitude above earth = 400 km = 0.4 x 10⁶ m

Radius of orbit of the space station around earth is given as

r = R + h

r = 6400 + 400

r = 6800 km

r = 6.8 \times 10^{6} m

T = Time period of orbit of space station

Here we can use Kepler's third law which is given as

T^{2} = \frac{4\pi ^{2} r^{3}}{GM}

Inserting the above values

T^{2} = \frac{4(3.14)^{2} (6.8\times10^{6})^{3}}{(6.67\times10^{-11})(6.0\times10^{24})}

T^{2} = 3.1\times 10^{7}

T = 5566.53 sec

T = 5566.53 \left ( \frac{1min}{60sec} \right )= 93 min

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A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 152 om and makes an angl
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Answer:

D₂= 167,21 cm : Magnitude  of the second displacement

β= 21.8° , countercockwise from the positive x-axis: Direction of the second displacement

Explanation:

We find the x-y components for the given vectors:

i:  unit vector in x direction

j:unit vector in y direction

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D₂: Displacement Vector 2

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D_{2} =\sqrt{(D_{x})^{2} +(D_{y} )^{2}  }

D_{2} =\sqrt{(155.22)^{2} +(62.18 )^{2}  }

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Direction of the second displacement

\beta = tan^{-1} \frac{D_{y}}{D_{x} }

\beta = tan^{-1} \frac{62.18}{155.22 }

β= 21.8°

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β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement

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