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3 years ago

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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Answer:

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$v=\sqrt{\dfrac{2\times1.6\times10^{-19}\times700}{9.1\times10^{-31}}}$

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(b). We need to calculate the radius of electron

Using formula of centripetal force

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$r=\dfrac{mv}{qB}$

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m = mass of electron

v = speed of electron

r = radius

q = charge of electron

B = magnetic field

Put the value into the formula

$r=\dfrac{9.1\times10^{-31}\times1.56\times10^{7}}{1.6\times10^{-19}\times2.1\times10^{-3}}$

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Answer:

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