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erastova [34]
3 years ago
12

Does centripetal force change in uniform circular motion?

Physics
1 answer:
HACTEHA [7]3 years ago
7 0
It depends on how fast you are going and in orincipal no
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Assuming you exert a constant force on the wagon, how fast is it moving after 5 seconds?
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I would help but I’m also taking this
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If Manuel exerts a force of 10 N to push a desk to the right at the same time Lynn exerts a force of 15 N to push the desk to th
wolverine [178]

Answer:

The desk will move to the left.

Explanation:

Since Manuel is exerting a force of 10 N to the right, and Lynn is exerting a force of 15 N to the left, it means that Lynn is exerting a force of 5 N more than Manuel, meaning that the desk will move to the left, the direction Lynn is pushing it too.

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<em>Hope </em><em>it </em><em>helps</em><em>!</em>

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Should i make my pet shark bark like a dog woof woof​
mr_godi [17]

Answer: YES !!

Explanation:  THAT WOULD BE AMAZING!!!

8 0
3 years ago
How much work is done by a forklift raising a 125 kg object a distance of 10 meters?
lozanna [386]

W = Fd

W = 1225 N x 10 m = 12250

6 0
3 years ago
A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa
bearhunter [10]

Explanation:

Formula which holds true for a leans with radii R_{1} and R_{2} and index refraction n is given as follows.

          \frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

Since, the lens is immersed in liquid with index of refraction n_{1}. Therefore, focal length obeys the following.  

            \frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]  

             \frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

and,       \frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}

or,          f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}

              f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}

                          = 32.4 cm

Using thin lens equation, we will find the focal length as follows.

             \frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}

Hence, image distance can be calculated as follows.

       \frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}

              s_{i} = \frac{fs_{o}}{s_{o} - f}

             s_{i} = \frac{32.4 \times 100}{100 - 32.4}

                       = 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0
3 years ago
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