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OLga [1]
3 years ago
11

What is the squirrels escape velocity in mph if the squirrel accelerates at a constant 1.5 m/s squared from rest for 2.5s

Physics
1 answer:
Zina [86]3 years ago
7 0

Answer:

3.75 m/s

Explanation:

From the question given above, the following data were obtained:

Acceleration (a) = 1.5 m/s²

Initial velocity (u) = 0 m/s

Time (t) = 2.5 s

Final velocity (v) =?

a = (v – u) / t

1.5 = (v – 0) / 2.5

1.5 = v / 2.5

Cross multiply

v = 1.5 × 2.5

v = 3.75 m/s

Hence, the escape velocity of the squirrel is 3.75 m/s

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\:\:\:\:\:\:\:= \left[\dfrac{3(13.8\:\text{m/s}^2)}{4\pi (6.674×10^{-11}\:\text{Nm}^2\text{/kg}^2)(5500\:\text{kg/m}^3)}\right]

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