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3 years ago

7

a 2-kg object is dropped from a height of 1000 m. What is the force of air resistance on the object when it reaches terminal vel

ocity

1 answer:

strojnjashka [21]3 years ago

5 0

It stops accelerating when the air resistance is equal to its weight.
That's (m•g)

= (2 kg) • (9.8 m/s^2)

= 19.6 newtons

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4 years ago

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Answer:

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Explanation:

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$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

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