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Ostrovityanka [42]
3 years ago
9

A satellite orbiting Earth at an orbital radius r has a velocity v. Which represents the velocity

Physics
1 answer:
babunello [35]3 years ago
8 0

Answer:

v / √5

Explanation:

Centripetal acceleration = acceleration due to gravity

v² / r = GM / r²

v² = GM / r

If the radius is increased to 5r:

u² = GM / (5r)

u² = v² / 5

u = v / √5

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The image shows a model of the gravitational field around Earth. Which best describes the model?
GuDViN [60]

<u>Answer: The model is described by :</u>

The strength of the field increases as you get closer to Earth, as shown by the smaller spaces between the circles.

Explanation :

According to universal gravitational law, the force acting in the universe between two objects is given by :

F=G\dfrac{m_1m_2}{r^2}

It is clear that the gravitational force depends directly on the product of masses and inversely on the square of the distance between them.

The given image shows a model of the gravitational field around Earth. The arrows in the image show the direction of the force. Gravitational field depends on the position of objects.

As we move closer to the earth, the force increases and hence the field increase.

So, the correct option is (D) " The strength of the field increases as you get closer to Earth, as shown by the smaller spaces between the circles ".

3 0
3 years ago
Read 2 more answers
You create a ramp using two text books and a 0.50m board. Using a timer you determine that a cart can roll down the ramp in 0.55
ahrayia [7]

Answer:

The velocity of the cart at the bottom of the ramp is 1.81m/s, and the acceleration would be 3.30m/s^2.

Explanation:

Assuming the initial velocity to be zero, we can obtain the velocity at the bottom of the ramp using the kinematics equations:

v=v_0+at\\\\v^2=v_0^2+2ad

Dividing the second equation by the first one, we obtain:

v=\frac{v_0^2+2ad}{v_0+at}

And, since v_0=0, then:

v=\frac{2ad}{at}\\\\v=\frac{2d}{t}\\\\v=\frac{2(0.50m)}{0.55s}\\\\v=1.81m/s

It means that the velocity at the bottom of the ramp is 1.81m/s.

We could use this data, plus any of the two initial equations, to determine the acceleration:

v=v_0+at\\\\\implies a=\frac{v}{t}\\\\a=\frac{1.81m/s}{0.55s}\\\\a=3.30m/s^2

So the acceleration is 3.30m/s^2.

7 0
3 years ago
The catapults on the uss george h.w. bush can launch aircraft from rest to a speed of 150 mph over a distance of 270 feet. find
vlabodo [156]
Hope this helps, have a great day ahead!

8 0
3 years ago
Professional Application. A 96 kg football player catches a 0.900 kg ball with his feet off the ground with both of them moving
Zarrin [17]

To solve this problem it is necessary to apply the equations related to the conservation of momentum.

This definition can be expressed as

m_1u_1+m_2u_2 = (m_1+m_2)V_f

Where

m_{1,2} = Mass of each object

u_{1,2} = Initial Velocity of each object

V_f= Final velocity

Rearranging the equation to find the final velocity we have,

V_f = \frac{m_1u_1+m_2u_2}{(m_1+m_2)}

Our values are given as

m_1 = 96Kg\\m_2 = 0.9Kg\\u_1 = 6.3m/s\\u_2 = 27.4m/s

Replacing we have,

V_f = \frac{(96)(6.3)+(0.9)(27.4)}{(96+0.9)}

V_f = 6.4959m/s

Therefore the final velocity is 6.5m/s

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3 years ago
Consider an aircraft powered by a turbojet engine that has a pressure ratio of 12. The aircraft is stationary on the ground, hel
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hocico vlgcogfv ljfouclh la hospice

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