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2 years ago

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A navy seal of mass 80 kg parachuted into an enemy harbor. At one point while he was falling, the resistive force of air exerted

on him was 520 N. What can you determine about the motion?

1 answer:

Aleonysh [2.5K]2 years ago

5 0

Answer:

<em>The motion of the parachute = 3.3 m/s²</em>

Explanation:

Weight of the parachute - Resistive force of air = ma

W - Fₐ = ma.................... Equation 1

making a the subject of formula in equation 1

a = (W- Fₐ)/m.................. Equation 2

Where W = weight of the parachute, Fₐ = resistive force of air, m = mass of the parachute, a = acceleration of the parachute

<em>Constant: g = 9.8 m/s²</em>

<em>Given: Fₐ = 520 N, m = 80 kg</em>

<em> W = mg = 80 × 9.8 = 784 N, </em>

<em>Substituting these values into equation 2</em>

<em>a = (784-520)/80</em>

<em>a = 264/80</em>

<em>a = 3.3 m/s²</em>

<em>Therefore the motion of the parachute = 3.3 m/s²</em>

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A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice

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$4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}$

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Taking squares out, we get,

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I hope this helps!

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2 years ago