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Vera_Pavlovna [14]
3 years ago
9

What is the restoring force of a spring with a spring constant of 4a and a stretched displacement of 3b? A. –7 ab B. `-7 a/b ` C

. –12 ab D. `-12 a/b`
Physics
1 answer:
algol133 years ago
3 0

Answer:

C. -12 ab

Explanation:

The restoring force on a spring is given by Hooke's law:

F=-kx

where

k is the spring constant

x is the stretched (or compressed) displacement of the spring

In this problem we have:

k = 4a

x = 3b

Substituting into the equation, we find:

F=-(4a)(3b) = -12 ab

And the negative sign means that the direction of the force (negative) is opposite to the direction of the displacement (positive).

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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}}]  

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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)}

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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.

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