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Goryan [66]
3 years ago
7

If a cell phone is dropped from a very tall building, how far has the phone fallen after 2.7 seconds, neglecting air resistance?

Physics
1 answer:
Zielflug [23.3K]3 years ago
6 0
The free fall of the phone is an uniformly accelerated motion toward the ground, with constant acceleration equal to
g=9.81 m/s^2

So, assuming the downward direction as positive direction of the motion, since the phone starts from rest the distance covered by the phone after a time t is given by
y(t) =  \frac{1}{2}gt^2
And if we substitute t=2.7 s, we find the distance covered:
y(t)=  \frac{1}{2}(9.81 m/s^2)(2.7 s)^2=35.8 m
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Answer:

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Explanation:

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we apply the equation

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the length of the bar is L = 6m

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let's calculate

let's use the maximum tension that resists the cable T = 900 N

             x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)

             x = (4676 - 5880) / 6860

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Therefore the block can be up to 0.176m to keep the system in balance.

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