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Rina8888 [55]
3 years ago
13

A city planner needs to make a model of the city. In real life, the tallest tree in the city is 40 feet tall. The shortest tree

in the city is 4 feet tall. If the tallest tree in the model is 10 feet tall, how tall is the shortest tree in the model?
Physics
1 answer:
Verdich [7]3 years ago
6 0
To solve this you must set up what is called a proportion.  A proportion is a way of comparing two comparing values where one of the four values is missing.  In your problem the missing value is the height of the smallest tree in the model.

To set up a proportion, you need all of your values.  The easiest way to do this is to list them:

Highest tree in real life:  40ft
Highest tree in model:  10ft
Smallest tree in real life:  4ft
Smallest tree in model:  x

So know you can set your proportion like this:

40/4 = 4/x

(When setting up a proportion, you always want to have the values belong to each other.  For example don't put the height of the small tree in the model underneath the value of the highest tree in real life.)

So know to find what the x values equals, we need to cross multiply.  And then all that's left after that is to solve for x.

40 times x = 4 times 4

40x = 16

x = 2.5

The smallest tree in the model should equal 2.5 feet.

Hope this helps! :)



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Answer:

The velocity of the particle = -1.92 m/s

The speed of the particle = 5.72 m/s

Explanation:

Given equation of motion;

f(t) = 18 \ + \ \frac{48}{t} \ + \ 1

Velocity is defined as the change in displacement with time.

V = \frac{df(t)}{dt} = -\frac{48}{t^2} \\\\at \ t = 5 \ s\\\\V = -\frac{48}{5^2} = \frac{-48}{25} = - 1.92 \ m/s

The distance traveled by the particle in 5 s:

s = f(5) = 18 + \frac{48}{5} + 1\\\\s= 28.6 \ m

The speed of the particle when t = 5s

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A fair ground ride spins its occupants inside a flying saucer-shaped container. if the horizontal circular path the riders follo
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Below is the solution:

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Answer: The unpolarized light's intensity is reduced by the factor of two when it passes through the polaroid and becomes linearly polarized in the plane of the Polaroid. When the polarized light passes through the polaroid with the plane of polarization at an angle \theta with respect to the polarization plane of the incoming light, the light's intensity is reduced by the factor of \cos^2\theta (this is the Law of Malus).

Explanation: Let us say we have a beam of unpolarized light of intensity I_0 that passes through two parallel Polaroid discs with the angle of \theta between their planes of polarization. We are asked to find \theta such that the intensity of the outgoing beam is I_2. To solve this we follow the steps below:

Step 1. It is known that when the unpolarized light passes through a polaroid its intensity is reduced by the factor of two, meaning that the intensity of the beam passing through the first polaroid is

I_1=\frac{I_0}{2}.

This beam also becomes polarized in the plane of the first polaroid.

Step 2. Now the polarized beam hits the surface of the second polaroid whose polarization plane is at an angle \theta with respect to the plane of the polarization of the beam. After passing through the polaroid, the beam remains polarized but in the plane of the second polaroid and its intensity is reduced, according to the Law of Malus, by the factor of \cos^2\theta. This yields I_2=I_1\cos^2\theta. Substituting from the previous step we get

I_2=\frac{I_0}{2}\cos^2\theta

yielding

\frac{2I_2}{I_0}=\cos^2\theta

and finally,

\theta=\arccos\sqrt{\frac{2I_2}{I_0}}

3 0
3 years ago
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