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malfutka [58]
2 years ago
10

A boy whirls a ball on a string in a horizontal circle of radius 1 m. How many revolutions per minute must the ball make if its

acceleration towards the center of the circle is to have the same magnitude as the acceleration due to gravity
Physics
1 answer:
spayn [35]2 years ago
7 0

Answer:

Nearest, the revolutions per minute will be 29.

Explanation:

Given that,

Radius of circle = 1 m

Acceleration a =g

We know that,

Angular frequency is defined as,

\omega=2\pi n

Where, n = number of revolutions in one second

We need to calculate the revolutions in one second

Using formula of centripetal acceleration

a=\omega^2r

Put the value of a and ω

g=(2\pi n)^2r

n=\sqrt{\dfrac{g}{r}}\times\dfrac{1}{2\pi}

Put the value into the formula

n=\sqrt{\dfrac{9.8}{1}}\times\dfrac{1}{2\pi}

n=0.49

We need to calculate the revolutions per minute

Using value for the revolutions per minute

n=0.49\times60

n=29.4

Hence, Nearest, the revolutions per minute will be 29.

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Answer:

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  • Let m_{\rm a} and m_{\rm b} denote the mass of the two objects.
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  • Let \vec{v}_{\rm a}(\text{final}) and \vec{v}_{\rm b}(\text{final}) denote the velocity of the two objects right after the interaction.
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The sum of the momentum of the two objects would be:

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