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AlladinOne [14]
3 years ago
15

an object that is 8 cm tall is located 2.5 m in front of a plane mirror. The image formed by the mirror appears to be. Find di a

nd whether behind or in front of mirror​
Physics
1 answer:
UkoKoshka [18]3 years ago
6 0

The image is at 2.5 m behind the mirror (virtual image)

Explanation:

We can solve the problem by using the mirror equation:

\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}

where

f is the focal length of the mirror

d_o is the distance between the object and the mirror

d_i is the distance between the image and the mirror

For a plane mirror, the focal length is infinite:

f=\infty

And in this problem, the distance of the object from the mirror is

d_o = 2.5 m

Substituting and solving for d_i, we find the location of the image:

0=\frac{1}{d_o}+\frac{1}{d_i}\\d_i = -d_o = -2.5 m

And the negative sign means that the image is located behind the mirror, so it is virtual.

Learn more about reflection and refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

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3. A football is kicked with a speed of 35 m/s at an angle of 40°.
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a) 22.5 m/s

The initial vertical velocity is given by:

u_y = u sin \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

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u_y = (35)(sin 40)=22.5 m/s

b) 26.8 m/s

The initial horizontal velocity is given by:

u_x = u cos \theta

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u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

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c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

v_y = u_y + at

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v_y is the vertical velocity at time t

u_y = 22.5 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, v_y =0; substituting, we find the time t at which this happens:

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d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

s=u_y t + \frac{1}{2}gt^2

where

s is the vertical displacement at time t

u_y = 22.5 m/s

g=-9.8 m/s^2

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

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v_x = 26.8 m/s

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d=v_x t = (26.8)(4.60)=123.3 m

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x = 3 sin .5t / .5

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