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polet [3.4K]
3 years ago
13

how do you isolate t (time) in this equation? the question is asking me to derive an equation for time to hit the ground for a d

ropped object in terms of height of the drop, gravity, and initial velocity and that’s what i have so far.

Physics
1 answer:
expeople1 [14]3 years ago
8 0

Fine, so we have the equation

S = So + Vo.t + at²/2

We have the final height = 0 and since its dropped, Vo = 0, our a will be the gravity, negative since its falling

0 = So + 0.t - gt²/2

0 = So - gt²/2

Since So is our h, lets change the letter

0 = h - gt²/2

gt²/2 = h

gt² = 2h

t² = 2h/g

t = \sqrt{\frac{2h}{g}}

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A force of constant magnitude pushes a box up a vertical surface, as shown in the figure.
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The work done on the box by the applied force is zero.

The work done by the force of gravity is 75.95 J

The work done on the box by the normal force is 75.95 J.

<h3>The given parameters:</h3>
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  • Distance moved by the box, d = 2.5 m
  • Coefficient of friction, = 0.35
  • Inclination of the force, θ = 30⁰

<h3>What is work - done?</h3>
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The work done on the box by the applied force is calculated as;

W = Fd cos(\theta)\\\\W = (ma)d \times cos(\theta)

where;

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The acceleration of the box is zero since the box moved at a constant speed.

W = (0) d \times cos(30)\\\\W = 0 \ J

The work done by the force of gravity is calculated as follows;

W = mg \times d\\\\W = 3.1 \times 9.8 \times 2.5 \\\\W = 75.95 \ J

The work done on the box by the normal force is calculated as follows;

W = (F_n) \times d\\\\W = (mg + F sin\theta) \times d\\\\W = (mg + 0) \times d\\\\W = mgd\\\\W = 3.1 \times 9.8 \times 2.5\\\\W = 75.95 \ J

Learn more about work done here: brainly.com/question/8119756

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Answer:

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