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Mariana [72]
3 years ago
12

Which of the following statements are true about the lenses used in eyeglasses to correct nearsightedness and farsightedness? (T

here may be more than one correct choice.)A: They produce a real image.B: They produce a virtual image.C: Both nearsightedness and farsightedness are corrected with a converging lens.D: Both nearsightedness and farsightedness are corrected with a diverging lens
Physics
1 answer:
Slav-nsk [51]3 years ago
4 0

Answer:

A: They produce a real image.

Explanation:

The images formed on the retina of the eye for a normal visibility must always be real.

Only a real image can be physically projected on any physical object whereas the virtual images are visible due to reflections.

  • The nearsightedness is corrected with the help of a concave lens since it is the condition of the eye lens remaining thick and curved to converge the rays entering the eyes after a shorter distance which results in their image formation even before the retinal surface so to initially diverge them a bit so that they converge on the retinal surface and form the image there we use concave lens. Vice-versa of the above justification in  the case of farsightedness.
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(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

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(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s

<h3>Resultant speed of the ball and crosswind</h3>

V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

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6) A sociocultural psychologist might study how people express love in in various cultures across the world.

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