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aleksley [76]
2 years ago
13

Find the speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth. (Radiu

s of Earth = 6.37×103 km, mass of Earth = 5.98×1024 kg, G = 6.67×10-11 Nm2/kg2.)
Physics
1 answer:
Gnesinka [82]2 years ago
8 0

The orbiting velocity of the satellite is 4.2km/s.

To find the answer, we need to know about the orbital velocity of a satellite.

<h3>What's the expression of orbital velocity of a satellite?</h3>
  • Mathematically, orbital velocity= √(GM/r)
  • r = radius of the orbital, M = mass of earth
<h3>What's the orbital velocity of a satellite orbiting earth with a radius 3.57 times the earth radius?</h3>
  • M= 5.98×10²⁴ kg, r= 3.57× 6.37×10³ km = 22.7×10⁶m
  • Orbital velocity= √(6.67×10^(-11)×5.98×10²⁴/22.7×10⁶)

=4.2km/s

Thus, we can conclude that the orbiting velocity of the satellite is 4.2km/s.

Learn more about the orbital velocity here:

brainly.com/question/22247460

#SPJ1

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ONE EASY QUESTION
Brrunno [24]

Answer:

its tension force which acts in a string

Explanation:

need a thanks and thats it

8 0
3 years ago
The tires of a car support the weight of a stationary car. If one tire has a slow leak, the air pressure within the tire will __
Kipish [7]

Answer:

The tires of a car support the weight of a stationary car. If one tire has a slow leak, the air pressure within the tire will decrease with time, the surface area between the tire and the road will increase with time, and the net force the tire exerts on the road will be constant with time.

Explanation:

when a wheel has an air leak, it means that the inside of the tire has less air, which means that there will be less air pushing the walls of the tire so that the air pressure decreases.

On the other hand, the tire begins to deform due to lack of air which increases the area of ​​contact with the floor.

As the weight of the car remains constant and the air has a negligible mass the force towards the road will be the same

4 0
4 years ago
A girl delivering newspapers covers her route by traveling 5.00 blocks west, 5.00 blocks north, and then 7.00 blocks east. What
FromTheMoon [43]

Answer:

P = 2i + 5j

Therefore she is 2 blocks east and 5 blocks north.

Resultant P = √(2^2 + 5^2) = √(4+25) = √29 = 5.4 blocks

Angle = taninverse (5/2)

Angle = 68.2°

Explanation:

Given:

Let west be negative and east be positive x axis.

Let north be the positive y axis.

5.00blocks west = -5.00 i

5.00 blocks north = 5.00 j

7.00 blocks east = 7.00i

Addition of the vector form of hee position is;

P = -5i +7i -5j

P = 2i + 5j

Therefore she is 2 blocks east and 5 blocks north.

Resultant P = √(2^2 + 5^2) = √(4+25) = √29 = 5.4 blocks

Angle = taninverse (5/2)

Angle = 68.2°

3 0
3 years ago
A 45.0-kg person steps on a scale in an elevator. The scale reads 460 n. What is the magnitude of the acceleration of the elevat
WINSTONCH [101]

Answer:

The magnitude of the acceleration of the elevator is 0.422 m/s²

Explanation:

Lets explain how to solve the problem

Due to Newton's Law ∑ Forces in direction of motion is equal to mass

multiplied by the acceleration

We have here two forces 460 N in direction of motion and the weight

of the person in opposite direction of motion

The weight of the person is his mass multiplied by the acceleration of

gravity

→ W = mg , where m is the mass and g is the acceleration of gravity

→ m = 45 kg and g = 9.8 m/s²

Substitute these values in the rule above

→ W = 45 × 9.8 = 441 N

The scale reads 460 N

→ F = 460 N , W = 441 N , m = 45 kg

→ F - W = ma

→ 460 - 441 = 45 a

→ 19 = 45 a

Divide both sides by 45

→ a = 0.422 m/s²

<em>The magnitude of the acceleration of the elevator is 0.422 m/s²</em>

4 0
4 years ago
A 200 kg boulder is 1000m above the ground. What is the speed of the object right before it hits the ground?
dem82 [27]

Answer:

140.07m/s

Explanation:

we use the equation for the final speed:

v_{f}^2=v_{0}^2+2gh

where v_{f} is the final speed, v_{0} is the initial speed, g is the gravitational acceleration (g=9.81m/s^2) and h is the height.

In this case we don't have an initial velocity indicated so: v_{0}=0, and we are told that the boulder is at a height: h=1,000m.

We substitute this values into the equation:

v_{f}^2=2gh\\v_{f}^2=2(9.81m/s^2)(1,000m)\\v_{f} ^2=19,620m^2/s^2\\\\v_{f}=\sqrt{19,620m^2/s^2} \\v_{f}=140.07m/s

the speed of the object right before it hits the ground is 140.07m/s

6 0
3 years ago
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