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Romashka [77]
4 years ago
13

an object 8.25 cm from a lens creates a virtual image of magnification 2.40 what is the focal length of the lens (mind your minu

s sign) (unit=cm)
Physics
1 answer:
murzikaleks [220]4 years ago
5 0

Answer:

Focal length of the lens is 14.14 cm

Explanation:

As we know that the image formed due to lens is virtual image

so here we have

M = 2.40

now we have

\frac{d_i}{d_o} = 2.40

now we have

distance of object is 8.25 cm

so we have

d_i = 8.25(2.40) = 19.8 cm

now by lens formula

\frac{1}{d_i} - \frac{1}{d_o} = \frac{1}{f}

\frac{1}{-19.8} - \frac{1}{-8.25} = \frac{1}{f}

f = 14.14 cm

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Answer:

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Explanation:

Given that

Speed of the shoulder, v(h) = 75 cm/s = 0.75 m/s

Distance moved during the hook, d(h) = 5 cm = 0.05 m

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Average speed of the fist during the hook, v(f) = ? cm/s = m/s

This can be solved by a very simple relation.

d(f) / d(h) = v(f) / v(h)

v(f) = [d(f) * v(h)] / d(h)

v(f) = (1 * 0.75) / 0.05

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v(f) = 15 m/s

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