teresageorge1000
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Consider a compact car that is being driven
at 98 km/h.
The acceleration of gravity is 9.8 m/s2.
From what height would the car have to be
dropped to have the same kinetic energy?
Answer in units of m.

Respuesta :

Answer:

Height h = 37.8 m

Explanation:

Given :

Velocity of car (v) = 98 km / h

Acceleration of gravity = 9.8 m/s²

Computation:

Acceleration of gravity = 9.8 m/s²

Acceleration of gravity = (98)(1,000 m / 3,600 s)

Acceleration of gravity = 27.22 m/s

By using law of conservation of energy ;

(1/2)mv² = mgh

h = v² / 2g

h = 27.22² / 2(9.8)

Height h = 37.8 m

Lanuel

The height from which the car would have to be  dropped to have the same kinetic energy is 37.81 meters.

Given the following data:

  • Velocity = 98 km/h.
  • Acceleration of gravity = 9.8 [tex]m/s^2[/tex]

To find the height from which the car would have to be  dropped to have the same kinetic energy:

First of all, we would convert the value of velocity in km/h to m/s.

Conversion:

[tex]98 = \frac{98}{3600} \times 1000 = 27.22 \;m/s[/tex]

Applying the law of conservation of energy:

[tex]P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\gh = \frac{1}{2}v^2\\\\h= \frac{v^2}{2g}[/tex]

Substituting the given parameters into the formula, we have;

[tex]h = \frac{27.22^2}{2\times 9.8} \\\\h=\frac{741.05}{19.6}[/tex]

Height, h = 37.81 meters

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