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BlackZzzverrR [31]
2 years ago
6

A convex mirror has a focal length of -13.0 cm. When you place a 6.00 cm tall pencil 60.0 cm in front of that mirror, what is th

e location of the pencil's image?
Physics
1 answer:
wlad13 [49]2 years ago
6 0

The location of the pencil's image is =16.6cm

<h3>Calculation of image location</h3>

The focal length of the convex mirror (f)= -13.0 cm

The object distance (u) = 60cm

The image distance (v) = xcm

Using the formula,1/f= 1/v + 1/u

Make 1/v the subject of formula,

1/v = 1/f -1/u

1/v = 1/13 - 1/60

1/v = 60-13/780

1/v = 47/780

V = 780/47

V = 16.6 cm

Therefore, he location of the pencil's image is = 16.6cm

Learn more about mirror here:

brainly.com/question/13164847

#SPJ1

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First, let’s correct the question. Acceleration is the rate of change in velocity. Its unit therefore is ft/sec/sec. If S is the distance traveled for a given duration, S = Vot + (1/2)at^2 where Vo is the initial velocity, a is the acceleration and t is the time. For Vo = 0, a = 6m/sec/sec and t = 3 sec. The distance traveled is S = 0 + (1/2) x 6 x 3^2 = 27 meters

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1) The law of motion of the projectile is
h(t) = 2+22.5 t-4.9 t^2
To find the velocity, we should compute the derivative of h(t):
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So now we can calculate the speed at t=2 s and t=4 s:
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The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
v(t)=0
So we have
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3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
2+22.5t -4.9 t^2 = 0
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
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And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36  m/s
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