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blondinia [14]
3 years ago
12

A roller-coaster car rolls down a frictionless track, reaching speed v0 at the bottom. If you want the car to go twice as fast a

t the bottom, by what factor must you increase the height of the track
Physics
1 answer:
NeTakaya3 years ago
3 0

Answer:

h should become four times.

Explanation:

The speed at the bottom depends only on the vertical height difference from the top ... if there is no friction ... then the path you take from top to bottom doesn't matter in any way, only the difference in height between the two

mgh at top = 1/2 mv^2 at bottom, so that v=sqrt[2gh]

You see, the final velocity, v, depends on the square root of h, so double v, quadruple h, or four times higher

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A ball is thrown vertically upward from the ground. Its distance in feet from the ground in t seconds is s equals negative 16 t
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t₁ = 3 s

Explanation:

In this exercise, the vertical displacement equation is not given

        y = 240 t + 16 t²

Where y is the displacement, 240 is the initial velocity and 16 is half the value of the acceleration

Let's replace

      864 = 240 t + 16 t²

Let's solve the second degree equation

    16 t² + 240 t - 864 = 0

Let's divide by 16

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The solution of this equation is

     t = [-15 ± √(15 2 - 4 1 (-54)) ] / 2 1

     t = [-15 ±√(225 +216)] / 2

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We have two solutions.

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