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NeTakaya
3 years ago
8

A basketball player spins the ball with an angular acceleration of 10 rad/s2. What is the ball’s final angular velocity if the b

all starts from rest and the acceleration lasts for 3 seconds
Physics
1 answer:
Lena [83]3 years ago
8 0

Explanation:

It is given that,

The angular acceleration of the basketball, \alpha=10\ rad/s^2

Time taken, t = 3 seconds

We need to find the ball’s final angular velocity if the ball starts from rest. It can be calculated using definition of angular acceleration i.e.

\alpha=\dfrac{\omega_f-\omega_i}{t}

\omega_i=0\ (rest)

\omega_f=\alpha t

\omega_f=10\ rad/s^2\times 3\ s

\omega=30\ rad/s

So, the ball's final angular velocity is 30 rad/s. Hence, this is the required solution.

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A cow wanders 30 m North, turns 22 degrees right of its original path, and wanders another 40 m. Find its total displacement.
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Answer:OB=58.3m

Explanation:

So here cow wanders 30m in north and turns 22 degrees in right side and moves 40m more, as shown in figure given.

now take the starting point as a origin such that cow moves in x-y co-ordinate axis.

As shown in figure length OA is the length when cow moves in north or y direction. Later she takes 22 degrees turn to right and moves 40m more.

So the final displacement is the length of cow from the origin that is length OB.

now co-ordinates of B are [40cos22°,40sin22°+30] i.e [37.084,44.984]

now displacement of cow= length of OB

                                           = \sqrt{[37.084]^{2}+[44.984]^{2}  }

                                           =\sqrt{3398.78}

                                     OB =58.3

                                         

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