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Oksanka [162]
3 years ago
12

A box of weight 280 N is being pulled to the right by a pulling force vec F of magnitude 50 N. The box is moving at a constant s

peed of 0.84 m/s . To make an accurate free-body diagram of the box, the letter f in the diagram above should be replaced with the label
Physics
1 answer:
boyakko [2]3 years ago
5 0

Answer:

what diagram. I can't see any

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Softa [21]

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0.010 m

Explanation:

So the equation for a pendulum period is: y=2\pi\sqrt{\frac{L}{g}} where L is the length of the pendulum. In this case I'll use the approximation of pi as 3.14, and g=9.8 m\s. So given that it oscillates once every 1.99 seconds. you have the equation:

1.99 s = 2(3.14)\sqrt{\frac{L}{9.8 m\backslash s^2}}\\

Evaluate the multiplication in front

1.99 s = 6.28\sqrt{\frac{L}{9.8m\backslash s^2}

Divide both sides by 6.28

0.317 s= \sqrt{\frac{L}{9.8 m\backslash s^2}}

Square both sides

0.100 s^2= \frac{L}{9.8 m\backslash s^2}

Multiply both sides by m/s^2  (the s^2 will cancel out)

0.984 m = L

Now now let's find the length when it's two seconds

2.00 s = 6.28\sqrt{\frac{L}{9.8m\backslash s^2}}

Divide both sides by 6.28

0.318 s = \sqrt{\frac{L}{9.8 m\backslash s^2}

Square both sides

0.101 s^2 = \frac{L}{9.8 m\backslash s^2}

Multiply both sides by 9.8 m/s^2 (s^2 will cancel out)

0.994 m = L

So to find the difference you simply subtract

0.984 - 0.994 = 0.010 m

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2 years ago
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