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Hitman42 [59]
3 years ago
6

What is the centripetal force that holds planets in orbit?

Physics
1 answer:
Fudgin [204]3 years ago
8 0
<h2>Answer: Gravity force</h2>

If we approximate the orbit of the planets around the Sun to circular orbits with a uniform circular motion, where the velocity \vec{V} is a vector, whose direction is perpendicular to the radius r of the trajectory; the acceleration \vec{a} is directed towards the center of the circumference (that's why it's called centripetal acceleration).  

Now, according to Newton's 2nd law, the force \vec{F} is directly proportional and in the same direction as the acceleration:  

\vec{F}=m.\vec{a}  

Therefore the net force resulting from the movement of a planet orbiting the Sun points towards the center of the circle, this is called Centripetal Force which is a central force that in this case is equal to the gravity force.

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A CAR ACCElerates FROM REST To
Ilia_Sergeevich [38]

Answer:

<u>666.6 kW</u>

Explanation:

<u>Power Formula</u>

  • Power = Force × Velocity

<u>Calculating Force</u>

  • F = ma
  • F = m(v - u / t) [From the equation v = u + at]
  • F = 2,000 (50/7.5)
  • F = 2,000 (20/3)
  • F = 40,000/3
  • F = 13333.3 N

<u>Solving for Power</u>

  • P = 13333.3 × 50
  • P = 666666.6 W
  • P = <u>666.6 kW</u>
5 0
2 years ago
A baseball of mass m = 0.31 kg is spun vertically on a massless string of length L = 0.51m. The string can only support a tensio
natulia [17]

Given data:

* The mass of the baseball is 0.31 kg.

* The length of the string is 0.51 m.

* The maximum tension in the string is 7.5 N.

Solution:

The centripetal force acting on the ball at the top of the loop is,

\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}

For the maximum velocity of the ball at the top of the vertical circular motion,

v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}

where g is the acceleration due to gravity,

Substituting the known values,

\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}

Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.

8 0
1 year ago
If a particle's position is given by x=4-12t+3t^2, where t is in seconds and x is in meters, what is its velocity at t=1 second?
andreyandreev [35.5K]

Answer:

v = -6m/s

Explanation:

x=4-12t+3t^2

\frac{dx}{dt}=-12+6t

For t = 1:

\frac{dx}{dt}=-6

3 0
3 years ago
Will give brainliest, Pleaseee help!!!
puteri [66]

Answer:

below

Explanation:

1.1115 im not sure tho

5 0
2 years ago
When a person falls from certain height on cemented floor, he receives more injuries why​
monitta

Explanation:

When a man falls on a hard cemented floor his momentum reduced to zero in a very short time and hurt the man. Whereas when a man falls on a heap of sand. As sand can compress, it takes longer time for the man to hit the ground (or hard surface)

8 0
2 years ago
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