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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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The distance between two planets is 1600 km. How much time would the light
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5.33*10^-3 seconds

Explanation:

c = d/t

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t = ?

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6 0
3 years ago
A high-speed flywheel in a motor is spinning at 450 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and
alexira [117]

Answer:

A) \omega_f=17.503\ rad.s^{-1}

B) t=55.6822\ s

C) \theta=1312\ rad

Explanation:

Given:

  • mass of flywheel, m=40\ kg
  • diameter of flywheel, d=0.72\ m
  • rotational speed of flywheel, N_i=450\ rpm \Rightarrow \omega_i=\frac{450\times 2\pi}{60} =15\pi\ rad.s^{-1}
  • duration for which the power is off, t_0=35\ s
  • no. of revolutions made during the power is off, \theta=180\times 2\pi=360\pi\ rad

<u>Using equation of motion:</u>

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2

\alpha=-0.8463\ rad.s^{-2}

Negative sign denotes deceleration.

A)

Now using the equation:

\omega_f=\omega_i+\alpha.t

\omega_f=15\pi-0.8463\times 35

\omega_f=17.503\ rad.s^{-1} is the angular velocity of the flywheel when the power comes back.

B)

Here:

\omega_f=0\ rad.s^{-1}

Now using the equation:

\omega_f=\omega_i+\alpha.t

0=15\pi-0.8463\times t

t=55.6822\ s is the time after which the flywheel stops.

C)

Using the equation of motion:

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2

\theta=1312\ rad revolutions are made before stopping.

3 0
3 years ago
The formation of condensation on a glass of ice water causes the ice to melt faster than it would otherwise. If 8.55 g of conden
SSSSS [86.1K]

Answer:

m = 62.14 g

Explanation:

Energy used to melt the ice is the energy released by the condensation of the water forms on the glass

so here we have

energy for the condensation of water is given as

let mass of water condensed = m

E = m_1 L_f

now the energy of vaporization is given as

E = m_2 L_v

here we know that

L_f = 79.8 kCal/kg

L_v = 580 k Cal/kg

Now we have

8.55 \times 580 = m \times 79.8

m = 62.14 g

3 0
3 years ago
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