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ki77a [65]
3 years ago
12

An astronaut orbits planet Y in her spaceship. To remain in orbit at 410km above the planet's center, she maintains a speed of 6

8 m/s. What is the mass of planet Y? Note: the constant of universal gravity (G) equals 6.674 × 10-11 N ⋅ m2/kg2.
Physics
1 answer:
choli [55]3 years ago
3 0
Formula for orbital speed, v =  √(GM/R)

Where G is the universal gravitational constant, M = Central Mass,

R = Distance between centers of Mass.

Given.  v = 68 m/s, M = ? , R = 410 km = 410000 m., G = 6.674 * 10⁻¹¹ Nm²/kg²

68 = √(GM/R)

68 = √(6.674 * 10⁻¹¹ * M/410000)

68² =  (6.674 * 10⁻¹¹ * M)/410000

(68²  * 410000) / 6.674 * 10⁻¹¹    = M

2.84 × 10¹⁹    = M

Mass of Planet Y =  2.84 × 10¹⁹ kg
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melamori03 [73]
To answer this question, we will refer to the attached diagram which represents the complete rock cycle.

Rock 1:
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Rock 2:
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Rock 3:
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3 0
4 years ago
Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope
Kitty [74]

a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J   the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

       

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

  = -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

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b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

                     = 26,754 (0.816)

                     = 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c.  What is the work done by the gravitational force on the sled?

By using  the formula:

Gravity work = mgdsinθ

                    = 97.5 kg * 9.8 m/s² * 28 m * sin 60

                    = 23,170 J

23,170 J   the work, in joules done by the gravitational force on the sled .

       

D. What is the total work done?

By adding all the values

work done =  -1337.3 + 21,835 + 23,170

                 = 43,670 J

The net work done on the sled, in joules is 43,670 J.

Learn more about friction work here:

brainly.com/question/14619763

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