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Anna71 [15]
3 years ago
5

An object propelled upwards with an acceleration of 2.0 m / s ^ 2 is launched from rest. After 6 seconds the fuel runs out. Dete

rmine the speed at this time and the maximum height at which it reaches.
Physics
1 answer:
dezoksy [38]3 years ago
7 0

Answer:43.34 m

Explanation:

Given

acceleration(a)=2 m/s^2

Initial Velocity(u)=0 m/s

After 6 s fuel runs out

Velocity after 6 s

v=u+at

v=0+2\times 6=12 m/s

After this object will start moving under gravity

height reached in first 6 s

s=ut+\frac{at^2}{2}

s=0+\frac{2\times 6^2}{2}

s=36 m

After fuel run out distance traveled in upward direction is

v^2-u^2=2as_0

here v=0

u=12 m/s

a=9.8 m/s^2

0-12^2=2(-9.8)(s)

s_0=\frac{144}{2\times 9.8}=7.34 m

s+s_0=36+7.34=43.34 m

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Answer:

<em>A) the moment of inertia of the system decreases and the angular speed increases. </em>

Explanation:

The complete question is

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Answer:

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Part B

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