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harina [27]
2 years ago
12

Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed of a

satellite in an orbit 900 km above Earth’s surface. (b) Suppose a loose rivet is in an orbit of the same radius that intersects the satellite’s orbit at an angle of 90º relative to Earth. What is the velocity of the rivet relative to the satellite just before striking it?
Physics
1 answer:
Pie2 years ago
6 0

Answer:

Part a)

v = 7407.1 m/s

Part b)

v_{rel} = 1.05 \times 10^4 m/s

Explanation:

Part a)

As we know that orbital velocity at certain height from the surface of Earth is given as

v = \sqrt{\frac{GM}{R+h}}

here we know that

M = 5.98 \times 10^{24} kg

R = 6.37 \times 10^6 m

h = 900 km = 9.0 \times 10^5 m

now we have

v = \sqrt{\frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})}{6.37 \times 10^6 + 9.0 \times 10^5}}

v = 7407.1 m/s

Part b)

When a loose rivet is moving in same orbit but at 90 degree with the previous orbit path then in that case the relative speed of the rivet with respect to the satellite is given as

v_{rel} = \sqrt{2} v

v_{rel} = 1.05 \times 10^4 m/s

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Answer:

1) Hence, the period is 0.33 s.

2) The amplitude is 10 cm.

Explanation:

1) The period is given by:

T = \frac{1}{f}

Where:

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T = \frac{1}{f} = \frac{1}{3 Hz} = 0.33 s

Hence, the period is 0.33 s.

2) The amplitude is the distance between the equilibrium position and the maximum position traveled by the spring. Since the spring is moving up and down over a distance of 20 cm, then the amplitude is:          

A = \frac{20 cm}{2} = 10 cm  

Therefore, the amplitude is 10 cm.          

I hope it helps you!                    

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This DNA strand consists of eight pairs of nitrogenous bases. How many different sequences of eight bases can you make?
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Two racing boats set out from the same dock and speed away at the same constant speed of 104 km/h for half an hour (0.500 h), th
Zinaida [17]

Answer:

The blue boat traveled 6.1 km farther west than the green boat

The green boat traveled 10.7 km farther south than the blue boat

Explanation:

The equation for linear uniform speed movement is

X(t) = X0 + v * t

Since we have two coordinates (X, Y) we use

X(t) = X0 + vx * t

Y(t) = Y0 + vy * t

The dock will be the origin of coordinates so X0 and Y0 will be zero. The X axis will be pointing west and the Y axis south.

The blue boat moves with a direction 24° south of west, so it will have speeds:

vxb = 104 * cos(24) = 95 km/h

vyb = 104 * sin(24) = 42.3 km/h

And the green boat:

vxg = 104 * cos(37.7) = 82.3 km/h

vyg = 104 * sin(37.7) = 63.6 km/h

After half hour the boats will have arrived at positions

Xb = 95 * 0.5 = 47.5 km

Yb = 42.3 * 0.5 = 21.1 km

And

Xg = 82.3 * 0.5 = 41.4 km

Yg = 63.6 * 0.5 = 31.8 km

The difference in positions of the boats

47.5 - 41.4 = 6.1 km

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3 years ago
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