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blagie [28]
4 years ago
15

Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The

top half of the lens is used to view distant objects and the bottom half of the lens is used to view objects close to the eye. A person can clearly see objects only if they are located between 45 cm and 161 cm away from her eyes. Bifocal lenses are used to correct her vision. What power lens (in diopters) should be used in the top half of the lens to allow her to clearly see distant objects?
Physics
1 answer:
Anarel [89]4 years ago
4 0

Answer:

P= -0.62 D

Explanation:

We know that lens equation

1/f=1/u+1/v

Given that person can see object only when object is located between 45 and 161 cm so we can say that far point of that person is at 161 cm.

So we need to take that objects ray are coming from and infinity and will focus at 161 cm away.

So now by putting the values

1/f=1/∞+1/-161    (negative because image will be virtual ,by sign convention)

So f=-161 cm

We know that inverse of focus(in meter) length is called power of lens.

So power of lens P=1/f

  P=-1/1.61

P= -0.62 D

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No matter what values of m and k you used for the spring, what is the ratio of the period to k m.
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The relationship between the period of an oscillating spring and the attached mass determines the ratio of the period to \sqrt{\dfrac{m}{k} }.

Response:

  • The ratio of the period to  \sqrt{\dfrac{m}{k} } is always approximately<u> 2·π : 1</u>

<u />

<h3>How is the value of the ratio of the period to \sqrt{\dfrac{m}{k} } calculated?</h3>

Given:

The relationship between the period, <em>T</em>, the spring constant <em>k</em>, and the

mass attached to the spring <em>m</em> is presented as follows;

T =  \mathbf{2 \cdot \pi \cdot \sqrt{\dfrac{m}{k} }}

Therefore, the fraction of of the period to \sqrt{\dfrac{m}{k} }, is given as follows;

\mathbf{\dfrac{T}{ \sqrt{\dfrac{m}{k} }}} = 2 \cdot \pi

2·π ≈ 6.23

Therefore;

T :{ \sqrt{\dfrac{m}{k} }} = 2 \cdot \pi : 1

Which gives;

  • The ratio of the period to  \sqrt{\dfrac{m}{k} } is always approximately<u> 2·π : 1</u>

Learn more about the oscillations in spring here:

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7 0
3 years ago
A student takes the elevator up to the fourth floor to see her favorite physics instructor. She stands on the floor of the eleva
never [62]

Answer:

N = 1364 N

Explanation:

given data

accelerate upward = 5.70 m/s²

mass = 88.0 kg

solution

normal force is in upward direction so, weight of the student in downward direction and acceleration is in upward direction so formula is express as

N - mg = ma        ...........................1

N = m × (g+a)

put here value

N = 88.0 × (9.8 + 5.70)

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8 0
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A mxiture of n2 and H2 has mole fraction of 0.4 and 0.6 respectively. Determine the density of the mixture at one bar and 0 c.
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PV = nRT

V = nRT/P = 1×8.314×273/100 = 22.70m^3

Mass of N2 = 0.4×28 = 11.2kg

Mass of H2 = 0.6×2 = 1.2kg

Mass of mixture = 11.2 + 1.2 = 12.4kg

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3 0
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What happens when an electric charge passes through a circuit?
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Explanation:

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