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andreev551 [17]
3 years ago
12

a 0.04kg ball tied to a string moves in a circle that has a radius of 0.70 m. If the ball is accelerating 43.2m/s, what is the t

angential velocity of the ball?​
Physics
1 answer:
Luba_88 [7]3 years ago
4 0

Answer:

Tangential Velocity = 30.24 m/s

Explanation:

Given that,

Mass of ball, m = 0.04 Kg

Length of the string, r = 0.70 m

Acceleration of the ball, a = 43.2 m/s²

The tangential velocity of ball, V = ?

The centripetal force is given by the relation

        Fc = mV²/r  newton

where,   m - mass of body

              V - tangential velocity of body

               r - radius of the trajectory

Force applied on the ball to rotate on a circular path

                F = m x a newton

The applied force is equal to centripetal force.

So, equalizing the force equations

                     m x a = m V²/r

Therefore

                     V² = a x r

                      V = \sqrt{a X r}

Substituting the values

                       V = \sqrt{43.2 X 0.70}

                        V = 30.24 m/s

So, the tangential velocity of the ball is 30.24 m/s

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Answer:

0.1 s

Explanation:

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So, substituting the values of the variables into the equation, we have

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a = 9.5 m/s² - 4.41 m/s²

a = 5.09 m/s²

Since acceleration, a = (v - u)/t where u = initial velocity of log = 0 m/s (since it was a rest before being pulled out of the ditch), v = final velocity of log = 0.5 m/s and t = time taken for the log to reach a speed of 0.5 m/s.

So, making t subject of the formula, we have

t = (v - u)/a

substituting the values of the variables into the equation, we have

t = (v - u)/a

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Answer:

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Kinetic Energy K.E

Solution

From the moment of inertia

I=(1/2)MR^{2}\\ as \\W=mg\\So\\I=(1/2)(W/g)R^{2}\\I=(1/2)(953/9.8)(1.68)^{2}\\I=137.232kg.m^{2}

The angular acceleration is given as

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