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tatiyna
3 years ago
8

A woman wears bifocal glasses with the lenses 2.0 cm in front of her eyes. The upper half of each lens has power-0.500 diopter a

nd corrects her far vision so that she can focus clearly on distant when looking through that half. The lower half of each lens has power +2.00 diopters and corrects her near vision when she looks through that half What are the far point and near point of her eyes?
Physics
1 answer:
saveliy_v [14]3 years ago
8 0

Answer:

q = -2 m  and  q = -0.5 m

Explanation:

For this exercise we must use the equation of the optical constructor

        1 / f = 1 / p + 1 / q

where f is the focal length, p and q are the distance to the object and the image, respectively

Let's start with the far vision point, in this case the power of the lens is

        P = -0.5D

power is defined as the inverse of the focal length in meter

      f = 1 / D

      f = -1 / 0.5

      f = - 2m

the object for the far vision point is at infinity p = infinity

     1 / f = 1 / p + i / q

      1 / q = 1 / f - 1 / p

      1 / q = -1/2 - 1 / ∞

       q = -2 m

The sign indicates that the image is on the same side as the object

Now let's lock the near view point

D = +2.00 D

f = 1 / D

f = 0.5m

the near mink point is p = 25 cm = 0.25 m

        1 / f = 1 / p + 1 / q

        1 / q = 1 / f - 1 / p

        1 / q = 1 / 0.5 - 1 / 0.25

        1 / q = -2

        q = -0.5 m

the sign indicates that the image is on the same side as the object in front of the lens

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A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet
dimaraw [331]

Answer:

v  = 2.8898 \frac{m}{s}

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

E_{ep} = \frac{1}{2} k (\Delta x)^2

where \Delta x is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

\vec{F} = - k \Delta \vec{x}

as we need 9.12 N to compress 4.60 cm, this means:

k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}

k = 198.26 \ \frac{ N}{m}

So, the elastic energy of the compressed spring is:

E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2

E_{ep} = 0.209759 \ Joules

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

\Delta E_{gp} = m g \Delta h

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m

\Delta E_{gp} = 0.00224 \ Joules

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

E_k = \frac{1}{2} m v^2

so, for the pellet will be

E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2

<h3>All together</h3>

By conservation of energy, we know:

E_{ep} = \Delta E_{gp} + E_k

0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2

So

\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2  = 0.209759 \ Joules - 0.00224 \ Joules

\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2  = 0.207519 \ Joules

v  = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }

v  = 2.8898 \frac{m}{s}

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Answer:

Explanation:

A )

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