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3 years ago

An individual is nearsighted; his near point is 18.0 cm and his far point is 40.0 cm.

Physics

1 answer:

swat323 years ago

3 0

Answer

given,

near point = 18 cm

far point = 40 cm

a) The lens should form an upright, virtual image at far point from the distant object.

therefore, f = q = -40 cm = -0.4 m

where f is the focal length.

the required power

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

$P =\dfrac{1}{-0.40}$

P = -2.5 D

b) If the lens is used the Person's near point

The lens should form an upright, virtual image at near point from the distant object should be q = - 18 cm = -0.18 m

$p = \dfrac{qf}{q-f}$

$p = \dfrac{(-0.18)(-0.4)}{-0.18-(-0.4)}$

p = 32.72 cm

The person's near point is 32.72 cm

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A box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surfa

Burka [1]

Answer:

(a) = 0 N

(b) = 2.4 N

Explanation:

given

box of banana weight = 40.0 N

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a). when no horizontal force is applied .

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b) magnitude of friction force

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A sheet of gold leaf has a thickness of 0.125 micrometer. A gold atom has a radius of 174pm. Approximately how many layers of at

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Answer:

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Conversion

1 picometer (pm) is equivalent to $1 × 10^{-12}$ meter

1 micrometer is equivalent to $1 × 10^{-6}$ meter

To find the number of layers, we divide the overal leaf thickness by the thickness of one atom hence dividing tex]0.125 × 10^{-6}[/tex] meter by $174 × 10^{-12}$ meter we get that the number of sheets will be as follows

$\frac {0.125× 10^{-6}}{174\times 10^{-12}}=718.3908045\approx 719$

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Answer:

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