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MatroZZZ [7]
3 years ago
8

You put a 3 kg block in the box, so the total mass is now 8 kg, and you launch this heavier box with an initial speed of 4 m/s.

How long does it take to stop
Physics
2 answers:
Eduardwww [97]3 years ago
7 0

Answer:

Explanation:

The box stops at zero speed.

Final Velocity = 0 ,Initial speed (s)= -4 m/s

Therefore=  change in velocity =  Vf - Vi.  ( 0 m/s- 4 m/s) =  -4 m/s

Change in velocity  = -0.4 m/s

Gravity  g = 9.8 m/s^2

Mass= 0.8 g

-4 ms divided by 9.8 ms^2 * (0.8) = 0.51 s

It takes 0.51 seconds to stop

____ [38]3 years ago
3 0

Answer:

0.7 secs

Explanation:

In this question, the speed does not change as the mass changes. So we can use

Δ<em>t=Δ∨x/χgμ............................equ 1</em>

To stop, the final speed will be 0

Therefore,

<em>Δvx=vf-vt</em>

Δvx=0-4m/s

=     -4m/s

Now substitute the various values in equ 1

Δ<em>t=Δ∨x/χgμ</em>

Δ<em>t=  -</em>4m/s/(9.8m/s∧2) (0.6)

Δ<em>t=</em>0.7 secs

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Vilka [71]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is  \alpha =2.538 \  rad/s^2

Explanation:

From the question we are told that

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   The radius is  r =  0.69 \  m

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=>   \alpha =2.538 \  rad/s^2

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<h3>Way to do : </h3>

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