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Yuri [45]
3 years ago
8

Two identical satellites orbit the earth in stable orbits. One satellite orbits with a speed v at a distance r from the center o

f the earth. The second satellite travels at a speed that is less than v. At what distance from the center of the earth does the second satellite orbit? a) At a distance that is less than r b)At a distance equal to r. c)At a distance greater than r
Physics
1 answer:
jolli1 [7]3 years ago
7 0

Answer:

c)At a distance greater than r

Explanation:

For a satellite in orbit around the Earth, the gravitational force provides the centripetal force that keeps the satellite in motion:

G\frac{Mm}{r^2}=m\frac{v^2}{r}

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance between the satellite and the Earth's centre

v is the speed of the satellite

Re-arranging the equation, we write

r = \frac{GM}{v^2}

so we see from the equation that when the speed is higher, the distance from the Earth's centre is smaller, and when the speed is lower, the distance from the Earth's centre is larger.

Here, the second satellite orbit the Earth at a speed less than v: this means that its orbit will have a larger radius than the first satellite, so the correct answer is

c)At a distance greater than r

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I think the correct answer would be that the temperature of the clouds that made the very first stars where thought to be higher since the clouds are made up of hydrogen and helium. Hope this answers the question. Have a nice day.
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A. 2kg<br><br> B. 6kg<br><br> C. 8kg<br><br> D. 14kg
Sati [7]

Answer:

A

Explanation:

We know the force is (16N-12), which is 4N, and we know the acceleration is 2 m/s^2. Meaning, we can solve the formula m = F / a (mass equals force divided by acceleration), and we get 2kg.

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2 years ago
A 5.00 gram bullet moving at 600 m/s penetrates a tree trunk to a depth of 4.00 cm.
Crazy boy [7]
A) Work energy relation;
 Work =ΔKE ; work done = Force × distance, while, Kinetic energy = 1/2 MV²
 F.s = 1/2mv²
F× 4×10^-2 = 1/2 × 5 ×10^-3 × (600)²
 F = 900/0.04
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Therefore, force is 22500 N

b) From newton's second law of motion;
 F = Ma
Thus; a = F/m
             = 22500/(5×10^-3)
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6 0
3 years ago
2. a) A disc rotates about its axis at speed 25 revolutions per minute and takes 15 s to stop. Calculate the
ASHA 777 [7]

The statement shows a case of rotational motion, in which the disc <em>decelerates</em> at <em>constant</em> rate.

i) The angular acceleration of the disc (\alpha), in revolutions per square second, is found by the following kinematic formula:

\alpha = \frac{\omega_{f}-\omega_{o}}{t} (1)

Where:

  • \omega_{o} - Initial angular speed, in revolutions per second.
  • \omega_{f} - Final angular speed, in revolutions per second.
  • t - Time, in seconds.

If we know that \omega_{o} = \frac{5}{12}\,\frac{rev}{s}, \omega_{f} = 0\,\frac{rev}{s} y t = 15\,s, then the angular acceleration of the disc is:

\alpha = \frac{0\,\frac{rev}{s}-\frac{5}{12}\,\frac{rev}{s}}{15\,s}

\alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}

The angular acceleration of the disc is \frac{1}{36} radians per square second.

ii) The number of rotations that the disk makes before it stops (\Delta \theta), in revolutions, is determined by the following formula:

\Delta \theta  = \frac{\omega_{f}^{2}-\omega_{o}^{2}}{2\cdot \alpha} (2)

If we know that \omega_{o} = \frac{5}{12}\,\frac{rev}{s}, \omega_{f} = 0\,\frac{rev}{s} y \alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}, then the number of rotations done by the disc is:

\Delta \theta = 3.125\,rev

The disc makes 3.125 revolutions before it stops.

We kindly invite to check this question on rotational motion: brainly.com/question/23933120

6 0
3 years ago
What is the relationship between the angle of an incline and the acceleration of an object moving down the incline? How would yo
iren2701 [21]

Answer:

See Explanation

Explanation:

The relationship between angle of an incline and the acceleration of an object moving down the incline.

As the angle of an incline increases, so does the acceleration of the body moving down the incline increases, resolving the force acting on an inclined object

Parallel force = mgsin, perpendicular = mgcosΘ

With th weigh component 'mg' of the parallel force accounting for the acceleration of the body down the incline.

mgsinΘ = ma

Fnet = ma

B.) From Fnet = ma

Fnet = ma

a = Fnet / m

Where Fnet = Net force = mgsinΘ, a = acceleration

5 0
3 years ago
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