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3 years ago

12

Does centripetal force change in uniform circular motion?

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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2 years ago

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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2 years ago

Should i make my pet shark bark like a dog woof woof

mr_godi [17]

Answer: YES !!

Explanation: THAT WOULD BE AMAZING!!!

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3 years ago

How much work is done by a forklift raising a 125 kg object a distance of 10 meters?

lozanna [386]

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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