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3 years ago

9

What is the restoring force of a spring with a spring constant of 4a and a stretched displacement of 3b?

2 answers:

Darina [25.2K]3 years ago

3 0

Here is the correct answer of the given question above.
The restoring force of a spring with a spring constant of 4a and a stretched displacement of 3b would be <span>-12a/b.
It is solved using the formula F= -kx. Where x is the displacement and k is the spring constant. F = -(4a)(3b) and you get -12a/b. Hope this answer helps. </span>

Tatiana [17]3 years ago

3 0

F=-kx

F= - (4a) (3b)

F = -12ab

-kx

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I have a combination of myopia and presbyopia—overall, the power of my visual system is too large, but I also have a very limite

e-lub [12.9K]

Answer:

The range of powers is $- 5 \ D \le P \le - 2.667\ D$

Explanation:

From the question we are told that

The far point of the left eye is $n_f = 20 cm$

The near point of the left eye is $n = 15cm$

The near point with the glasses on is $n_g =25 \ cm$

From these parameter we can see that with the glass on that for near point the

Object distance would be $u = -25 \ cm$

Image distance would be $v = -15 \ cm$

To obtain the focal length we would apply the lens formula which is mathematically represented as

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substituting values

$\frac{1}{f} = \frac{1}{-15} - \frac{1}{-25}$

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converting to meters

$f = - \frac{75}{2} * \frac{1}{100}$

$f = - \frac{75}{200} \ m$

Generally the power of the lens is mathematically represented as

$P = \frac{1}{f}$

Substituting values

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$P = - 2.667 \ D$

From these parameter we can see that with the glass on that for far point the

Object distance would be $u_f = - \infty \ cm$

Image distance would be $v_f = -20 \ cm$

To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

$\frac{1}{f_f} = \frac{1}{v_f} - \frac{1}{u_f}$

substituting values

$\frac{1}{f} = \frac{1}{-20} - \frac{1}{- \infty}$

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converting to meters

$f_f = - \frac{20}{1} * \frac{1}{100}$

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Substituting values

$P = - \frac{100}{20} m$

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This implies that the range of powers of the lens in his glass is

$- 5 \ D \le P \le - 2.667\ D$

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