An object is in circular motion. how will the object behave if the centripetal force is removed?

Physics

2 answers:

Law Incorporation [45]4 years ago

7 0

You have two choices with the same letter. The correct choice is the last one ... Move along a tangent.

Inessa [10]4 years ago

4 0

The correct answer is

B. it would move in a line tangent to the circular path

In fact, the centripetal force is the only force that keeps the object in a circular trajectory, pulling the object towards the centre of the circle. When this force is removed, there are no other forces acting on the object, therefore for the law of inertia, the object will continue to move at constant velocity, therefore in the same direction (tangent to the circular path) at constant speed.

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irina1246 [14]

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3 years ago

True or False: Translations are rigid transformations.

GaryK [48]

Answer:

Explanation:

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3 years ago

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What force is described as the attraction between a sample of matter and all other matter in the universe?

Vinil7 [7]

Answer:

Gravitational Force

Explanation:

Gravitational force also called gravity or gravitation is an attractive force that keeps two objects in space. Gravitational force is an attractive force that tends to pull matters together. Every objects in the universe experience gravitational pull. Planets, stars, galaxies, are held together by gravity. It is a weak force. The weight of an object is the product of gravitational force acting on its mass.

Newton's Law of Universal Gravitation states the force of attraction between two masses m₁ and m₂ in the universe is directly proportional to the product of their masses and inversely proportional to the square of their distance apart.

$F = G\frac{m_{1}m_{2}}{r^2}$

Where;

F is the gravitational force,

G is the gravitational constant = 6.67 × 10¹¹ m³/kg/s,

m1 and m2 are the masses of the objects,

r is the distance between the centers of the masses

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4 years ago

Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial spe

iris [78.8K]

Answer:

Explanation:

1. $V_{x}$ = $V_{0}$ * cos $\alpha$ ⇒ 16*cos32 ≈ 13.6 m/s (13.56)

2. $V_{y}$ = $V_{0}$ * sin $\alpha$ ⇒ 16* sin32 ≈ 9.4 m/s

3. $y_{max}$ = $\frac{v_{0}^2*sin^2\alpha}{2g}$ = $\frac{16^2*sin^232}{2*9.8}$ (the g (gravity) depends on the country but i'll take the average g which is 9.2m/s^2)