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Norma-Jean [14]
3 years ago
11

The ATT Building (IE Batman Building) in Nashville is 614 feet tall. A model of the building is 60 inches tall. What is the rati

o of the height of the model to the height of the actual ATT Building?
SAT
1 answer:
Ludmilka [50]3 years ago
7 0

Answer:

Ratio of height of model to height of the actual Building = 5 : 614

Explanation:

Given:

ATT Building height = 614 feet tall

Model of building = 60 inches tall

Find:

Ratio of height of model to height of the actual Building

Computation:

ATT Building height = 614 feet tall = 614 x 12 = 7,368 inch

Ratio of height of model to height of the actual Building = Model of building / ATT Building height

Ratio of height of model to height of the actual Building = 60 / 7,368

Ratio of height of model to height of the actual Building = 5 : 614

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a helicopter flies parallel to the ground at an altitude of 1/2 kilometer and at a speed of 2 kilometers per minute. if the heli
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The Pythagoras' theorem and the uniform motion allows to find the answer for the rate of change of distance is

               v = 32.3 m / s

The distance is the length of the segment that a white house wing with the helicopter, this can be found using the Pythagoras' theorem

               R = \sqrt{x^2 + y^2}

Where R is the distance, x is the horizontal distance where the helicopter flies and y is the vertical distance that corresponds to the flight height

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Since the helicopter flies at constant speed, we can use the uniform motion relation

              v_h = \frac{x}{t}

              x = v_ht

Where v_h is the speed of the helicopter and t is the time

We substitute  

                r = \sqrt {(V_h \ t)^2 + y^2}

Kinematics defines velocity as the change of position in the unit of time

            v = \frac{dR}{dt}

           v = \frac{1}{2} \  \frac{2 ( v_h t) v_h}{ \sqrt{(v_h t)^2 + y^2} }

           v = \frac{v_h^2 \ t}{ \sqrt{(v_h t)^2 + y^2} }

This expression is the change in distance as a function of time for a given speed. In the exercise, this speed is requested for a time of one minute, for the helicopter speed of

        v = 2km/min (\frac{1000m}{1km} ) ( \frac{1 min}{60s}) = 33.33 m / s

Let's calculate

            v = \frac{33.33^2 \ 60 }{ \sqrt{ )33.3 \ 60)^2 +500^2} }

         

            v = 32.3 m / s

In conclusion using the Pythagoras' theorem and the uniform motion we can find the answer for the rate of change in distance is

               v = 32.3 m / s

Learn more here: brainly.com/question/343682

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