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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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2 years ago
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A cruise ship is moving at constant speed through the water. The vacationers on the ship are eager to arrive at their next desti
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Answer: 1. higher than it was before they started running

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3 years ago
A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When 3300 kcal o
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Answer

given,

heat added to the gas,Q = 3300 kcal

initial volume, V₁ = 13.7 m³

final volume, V₂ = 19.7 m³

atmospheric pressure, P = 1.013 x 10⁵ Pa

a) Work done by the gas

    W = P Δ V

    W = 1.013 x 10⁵ x (19.7 - 13.7)

    W = 6.029 x 10⁵ J

b) internal energy of the gas = ?

  now,

 change in internal energy

  Δ U = Q - W

    Q = 3300 x 10³ cal

    Q = 3300 x 10³ x 4.186 J

    Q = 1.38 x 10⁷ J

now,

  Δ U = 1.38 x 10⁷  - 6.029 x 10⁵

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6 0
3 years ago
George is applying a downward force of 50N to and object that has a mass of 50kg. What is the normal force (FN) of the object wh
attashe74 [19]

The normal force acting on the object is 500 N in the upward direction

<u>Explanation:</u>

As George is applying a downward force, the normal force will be in the upward direction. The normal force will be exerted due to the acceleration due to gravity exerted on the object.

So, as per Newton's second law, the normal force acting on the object can be measured by the product of mass of the object and the acceleration due to gravity acting on the object.

But as the acceleration due to gravity is a downward acting acceleration and the normal force is a upward acting force, so the acceleration will be having a negative sign in the formula.

Normal\ force = Mass \times Acceleration\ due\ to\ gravity

Here, acceleration due to gravity g = -10 m/s² and mass is given as 50 kg, then

Normal force = 50 × (-10) = -500 N

So, the normal force acting on the object is 500 N in the upward direction.

3 0
3 years ago
A tensile test specimen has a gage length = 50 mm and its cross-sectional area = 100 mm2. The specimen yields at 48,000 N, and t
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Answer:

a) yield strength

   \sigma_y = \dfrac{F_y}{A} = =\dfrac{48000}{100} = 480 MPa

b) modulus of elasticity

strain calculation

\varepsilon_0=\dfrac{L-L_0}{L_0}=\dfrac{50.23-50}{50} = 0.0046

strain for offset yield point

\varepsilon_{new} = \varepsilon_0 -0.002

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now, modulus of elasticity

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d) percentage elongation

\% Elongation = \dfrac{L-L_0}{L_0}\times 100 = \dfrac{67.3-50}{50}\times 100 = 34.6\%

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\% Area\ reduction = \dfrac{A-A_f}{A}\times 100=\dfrac{100-53}{100}= 47 \%                            

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