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EleoNora [17]
3 years ago
8

A physics student is conducting an optical experiment using an Optical Bench Kit available in many physics classrooms. A lit can

dle is positioned 20.cm from a concave mirror that produces an image 12cm from the mirror. Calculate the image position if the candle is repositioned 30.cm from the mirror. A. 5.Ocm B. 7.5cm C. 10.cm OD. 11cm E. 15cm
Physics
1 answer:
kobusy [5.1K]3 years ago
7 0

Answer:

C) 10 cm

Explanation:

We shall first calculate the focal length of the mirror.

object distance u = 20 cm ( negative )

Image distance v = 12 cm ( negative )

using mirror formula

1/ v + 1 /u = 1/ f

-1/12  - 1/ 20 =1/ f

f =- 7.5 cm

In the second case

u = - 30 cm

f = - 7.5 cm

i / v + 1 / u = 1 / f

1 / v - 1/30 = - 1/ 7.5

v = -10 cm

Answer is C that is 10 cm

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Answer:

The correct answer is c

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In the Copernicus system, it poses the Sun as the center of the solar system.

Copernicu's system was accepted for giving a simpler and more complete explanation of the problem

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Which voice can produce a pitch that has a speed of 343 m/s and a wavelength of 0.68 m?
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3 years ago
A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider
shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

d=v_x t

where here we have

d = 3.0 m is the horizontal distance covered

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t = 1.3 s is the duration of the fall

Solving for vx,

v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

y(t) = h + v_y t - \frac{1}{2}gt^2

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s

So now we can find the magnitude of the initial velocity:

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4 years ago
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grin007 [14]
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4 years ago
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The electric field strength is 5.50×10^4 N/C inside a parallel-plate capacitor with a 2.50 mm spacing. A proton is released from
valentina_108 [34]

Answer:

v=1.6\times10^{5}m/s

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a=\frac{F}{m}=\frac{qE}{m}

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In our case the proton is released from rest, so v_0=0m/s and we get v=\sqrt{2ad}

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6 0
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