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Novay_Z [31]
3 years ago
8

An object is placed 45 cm in front of a crystal ball. The crystal ball has a radius of 14 cm, and is made from a glass with an i

ndex of refraction of n = 1.5.
Find the position of the final image once the rays have gone through the two faces, as measured from the second face. Is a virtual object formed anywhere in the analysis? Yes or no
Physics
1 answer:
abruzzese [7]3 years ago
6 0

Answer:

Explanation:

 Using the formula for refraction at curved surface

For refraction at first surface

μ₂ /v -μ₁ /u = ( μ₂ - μ₁ )/R

1.5/v -1/-45 =.5/14

v = 111 cm

For refraction at opposite surface ,

u ( object distance ) = 111 - 28 =83 cm

It will act as virtual object for surface 2 so it will be positive in sign

again using the above equation for 2 nd surface we have

1/v - 1.5/83.17 = -.5-=/-14 = 1/28

v = 18.62 cm from second face.

yes, virtual image will be formed.

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A body is dropped from the roof of a 20 m high building by how much:
USPshnik [31]

Answer:

t = 2.01 s

Vf = 19.7 m/s

Explanation:

It's know through the International System that the earth's gravity is 9.8 m/s², then we have;

Data:

  • Height (h) = 20 m
  • Gravity (g) = 9.8 m/s²
  • Time (t) = ?
  • Final Velocity (Vf) = ?

==================================================================

Time

Use formula:

  • \boxed{t=\sqrt{\frac{2*h}{g}}}

Replace:

  • \boxed{t=\sqrt{\frac{2*20m}{9.8\frac{m}{s^{2}}}}}

Everything inside the root is solved first. So, we solve the multiplication of the numerator:

  • \boxed{t=\sqrt{\frac{40m}{9.8\frac{m}{s^{2}}}}}

It divides:

  • \boxed{t=\sqrt{4.08s}}

The square root is performed:

  • \boxed{t=2.01s}

==================================================================

Final Velocity

use formula:

  • Vf = g * t

Replace:

  • Vf = 9.8 m/s² * 2.01 s

Multiply:

  • Vf = 19.7 m/s

==================================================================

How long does it take to reach the ground?

Takes time to reach the ground in <u>2.01 seconds.</u>

How fast does it hit the ground?

Hits the ground with a speed of <u>19.7 meters per seconds.</u>

7 0
3 years ago
1. A cyclist accelerates from 0 m/s to 9 m/s in 3 seconds. What is his<br>acceleration?​
Lapatulllka [165]

The cyclist accelerates from 0 m/s to 9 m/s in 3 seconds with an acceleration of 3 m/s².

Answer:

Explanation:

Acceleration exerted by an object is the measure of change in speed or velocity of that object with respect to time. So the initial and final velocities play a major role in determining the acceleration of the cyclist. As here the initial velocity of the cyclist is the speed at rest and that is given as 0 m/s. Then after 3 seconds, the velocity of the cyclist changes to 9 m/s.

Then acceleration = change in velocity/Time.

Acceleration = \frac{Change in velocity}{Time taken}

Acceleration = (9-0)/3=9/3=3 m/s².

So the cyclist accelerates from 0 m/s to 9 m/s in 3 seconds with an acceleration of 3 m/s².

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Which of the following objects is accelerating?
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Answer:

A

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