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goblinko [34]
3 years ago
14

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest

2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness
Physics
1 answer:
shutvik [7]3 years ago
5 0

This is note the complete question, the complete question is:

One of the lousy things about getting old (prepare yourself!) is that you can be both near-sighted and farsighted at once. Some original defect in the lens of your eye may cause you to only be able to focus on some objects a limited distance away (near-sighted). At the same time, as you age, the lens of your eye becomes more rigid and less able to change its shape. This will stop you from being able to focus on objects that are too close to your eye (far-sighted). Correcting both of these problems at once can be done by using bi-focals, or by placing two lenses in the same set of frames. An old physicist instructor can only focus on objects that lie at distance between 0.47 meters and 5.4 meters.

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest 2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness?

Answer:  3.04 D

Explanation:

when an object is held 21 cm away from the instructor's eyes, the spectacle lens must produce 0.47m ( the near point) away.

An image of 0.47m from the eye will be ( 47 - 2 )

i.e 45 cm from the spectacle lens since the spectacle lens is 2cm away from the eye.

Also, the image distance will become negative

gap between lense and eye = 2cm

Therefore;

image distance d₁ = - 45cm = - 0.45m

object distance  d₀ = 21 - 2 = 19cm = 0.19m

P = 1/f = 1/ d = 1/d₀ + 1/d₁ = 1/0.19 + (-1/0.45)

P = 1/f =  5.26315789 - 2.22222222

P = 1/f = 3.04093567 ≈ 3.04 D

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3 0
3 years ago
From the window of a house that is placed 15 m
kow [346]

Answer:

a) 52.915 m

b) The vertical velocity is approximately 21.092 m/s

The resultant velocity is approximately 26.5 m/s

Explanation:

a) The height of the window in the house from which the water was thrown = 15 m

The speed of the stream of water thrown = 20 m/s

The angle at which the water was thrown = 37° over the horizontal

The acceleration due to gravity, g = 10 m/s²

a) The distance from the base of the house at which the water will fall is given as follows;

y = y₀ + u·t·sin(θ) + 1/2·g·t²

Where;

y = The vertical height reached    

u = The initial velocity

t = Time of flight

From the point the steam of water is thrown, we get;

y₀ = 15 m

Therefore;

y = 15 + 20 × t × sin(37°) - 1/2 × 10 × t²

y = 15 + 20 × t × sin(37°) - 5 × t²

When y = 0, Ground level, we get

0 = 15 + 20 × t × sin(37°) - 5 × t²

5·t² - 20×sin(37°)×t -15 = 0

∴ t = (20 ×sin(37°) ± √((-20 × ·sin(37°))² - 4 × (5) × (-15)))/(2 × 5)

t ≈ 3.3128302, or t ≈ 0.906

Therefore, the time of flight of the water, t ≈ 3.3128302 seconds

The distance from the base of the house at which the water will fall = The horizontal distance travelled by the water, x

x = u·cos(θ)×t

∴ x = 20 × cos(37°) × 3.3128302 ≈ 52.915

The distance from the base of the house at which the water will fall = x ≈ 52.915 m

b) The velocity at which the water will reach the ground, 'v', is given as follows;

The vertical velocity, v_y = u·sin(θ)·t - g·t

At the ground, t ≈ 3.3128302 seconds

∴ v_y = 20 × sin(37) - 10 × 3.3128302 ≈ -21.092

The vertical velocity at which the water will reach the ground, v_y ≈ 21.092 m/s (downwards)

The resultant velocity, v = √(v_y² + vₓ²)

∴ v = √(21.092² + (0 × cos(37°))²) ≈ 26.5

The resultant velocity at which the water will reach the ground, v ≈ 26.5 m/s.

5 0
2 years ago
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