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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:

1.   3 m/s^{2}

2.   1.5 m/s^{2}

3.   3 seconds

4.   0 m/s^{2}

5.   2.2 seconds

Explanation:

(1)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=0 since it’s at rest, v=30m/s and t=10 seconds

a = \frac {30-0}{10}=3 m/s^{2}

(2)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=10m/s, v=22m/s and t=8 seconds

a = \frac {22-10}{8}=1.5 m/s^{2}

(3)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

t=\frac {v-u}{a}

Substituting u=0m/s since at rest, v=15m/s and a=5 \frac {m}{s^{2}}

= \frac {15-0}{5}=3s

(4)

When initial and final velocity are constant, there’s no acceleration as proven below

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=20 since it’s at rest, v=20m/s and t=10 seconds

a = \frac {20-20}{10}=0 m/s^{2}

(5)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

t=\frac {v-u}{a}

Substituting u=9m/s since at rest, v=0m/s and a=-4.1 \frac {m}{s^{2}}

= \frac {0-9}{-4.1}=2.2s

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3 years ago
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