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nataly862011 [7]
3 years ago
8

How do inertia and centripetal force combine to keep an object moving in circular motion?

Physics
2 answers:
katen-ka-za [31]3 years ago
6 0
As the centripetal force<span> acts upon an </span>object moving <span>in a </span>circle<span> at constant speed, the </span>force<span> always acts inward as the velocity of the </span>object<span> is directed tangent to the </span>circle. ... In fact, whenever the unbalanced centripetal force<span> acts perpendicular to the direction of </span>motion<span>, the speed of the </span>object will<span> remain constant.</span>
mash [69]3 years ago
5 0

Answer:

As we know that inertia it the property of mass which resist any change in the state of motion of the object. Which means object has tendency to move in its initial state of motion and when we try to change that state of motion then object will resist to change that.

While moving in circular path the object will have tendency to change it's direction of motion at each instant. So this change in the direction of motion at every instant is due to centripetal acceleration of the object.

Now when object is revolving in circular path then the inertia of the object will help to move the object in tangential direction at every moment of the motion while the centripetal acceleration of the object will have tendency to change the direction at each moment of the motion.

So here we will say that inertia and centripetal force combine to keep an object moving in circular motion

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kirza4 [7]

Answer:

The value is U  = 0.06 \  J

Explanation:

From the question we are told that

The value of charge on each three point charge is

q_1 = q_2 = q_3 =q=  1.05 \mu C  =  1.05 *10^{-6} \  C

The length of the sides of the equilateral triangle is r  =  0.500 \

Generally the total potential energy is mathematically represented as

U  = k *  [ \frac{q_1 *  q_2}{r}  +  \frac{q_2 *  q_3}{r}   + \frac{q_3 *  q_1}{r} ]

=> U  = k * 3 * \frac{q^2}{r}

Here k is coulomb constant with value k = 9*10^{9}\  kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.

=>    U  = 9*10^9 * 3 * \frac{(1.05 *10^{-6})^2}{0.5 }

U  = 0.06 \  J

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A small rocket with mass 20.0 kg is moving in free fall toward the earth. Air resistance can be neglected. When the rocket is 80
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Answer:

= 308.5 N

Explanation:

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a = \frac{v_{y}^{2} - v_{0}^{2}  }{2y}

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= 20(5.625 m/s^{2}+ 9.8m/s^{2})

= 308.5 N

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3 years ago
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