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Tamiku [17]
2 years ago
15

A diver can change his rotational inertia by drawing his arms andlegs close to his body in the tuck position. After he leaves th

ediving board in the straight position (with some unknown angularvelocity), he pulls himself into a ball as closely as possible (thetuck position) and makes 2.00 complete rotations in 1.33 s. If hisrotational inertia decreases by a factor of 3.00 when he goes from the straight to the tuck position, what was his angularvelocity when he left the diving board?
Physics
1 answer:
NeX [460]2 years ago
3 0

Answer:

3.14946 rad/s

Explanation:

I_i = Intial moment of inertia

I_f = Final moment of inertia

\omega_i = Initial angular velocity

\omega_f = Final angular velocity = \dfrac{2}{1.33}\times 2\pi\ rad/s

\dfrac{I_f}{I_i}=\dfrac{1}{3}

In this system the angular momentum is conserved

L_i=L_f\\\Rightarrow I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_i=\dfrac{I_f\omega_f}{I_i}\\\Rightarrow \omega_i=\dfrac{1\times \dfrac{2}{1.33}\times 2\pi}{3}\\\Rightarrow \omega_i=3.14946\ rad/s

The angular velocity when the diver left the board is 3.14946 rad/s

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A. A land speed car can decelerate at 9.8m/s. How long does it take the car to come to a complete stop from a run of 885 km/hr (
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Answer:

A. 25.08 s

B. 3082.53 m

C. 3×10⁵ m/s²

Explanation:

A. Determination of the time.

This can be obtained as illustrated below:

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Time (t) =.?

v = u + at

0 = 245.8 + (–9.8 × t)

0 = 245.8 – 9.8t

Collect like terms

0 – 245.8 = – 9.8t

– 245.8 = – 9.8t

Divide both side by –9.8

t = –245.8 / –9.8

t = 25.08 s

Therefore, it will take 25.08 s for the car to come to a complete stop.

B. Determination of the distance travelled by the car.

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Distance (s) =?

v² = u² + 2as

0² = 245.8² + (2 × –9.8 × s)

0 = 60417.64 – 19.6s

Collect like terms

0 – 60417.64 = – 19.6s

– 60417.64 = – 19.6s

Divide both side by –19.6

s = –60417.64 / –19.6

s = 3082.53 m

Thus, the car travelled a distance of 3082.53 m before stopping completely.

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Initial velocity (u) = 0 m/s

Final velocity (v) = 600 m/s

Distance (s) = 0.6 m

Acceleration (a) =?

v² = u² + 2as

600² = 0² + (2 × a × 0.6)

360000 = 0 + 1.2a

360000 = 1.2a

Divide both side by 1.2

a = 360000 / 1.2

a = 300000 = 3×10⁵ m/s²

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