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2 years ago

A ball with a volume of 3 cm3 and a density of 9.0 g/cm3 increases its velocity from 2 m/s to 6 m/s over a 12

Physics

1 answer:

V125BC [204]2 years ago

4 0

Answer:

M = ρ V = 9 gm/cm^ 3 * cm^3 = 27 gm

a = (V2 - V1) / t = (6 - 2) m/s / 12 s = 1/3 m/s^2 the acceleration

F = M a = 27 gm * 1/3 m/s^2 = 9 dynes net force applied

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3) $27000 m/s^2$

4) 36.3 m/s

Explanation:

The angular acceleration of an object in rotation is the rate of change of angular velocity.

It can be calculated using the following suvat equation for angular motion:

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$\theta=90^{\circ} = \frac{\pi}{2}rad$ is the angular displacement

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$\omega_i = 0$ is the initial angular velocity

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For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

$\omega_f = \omega_i + \alpha t$

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$\omega_i$ is the initial angular velocity

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t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

Therefore, the final angular speed is:

$\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s$

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$a_t = \alpha r$

where

$a_t$ is the tangential acceleration

$\alpha$ is the angular acceleration

r is the distance of the point from the centre of rotation

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$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

$a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2$

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$a=27900 m/s^2$ (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

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$v=0+(27900)(0.0013)=36.3 m/s$

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