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Neko [114]
4 years ago
14

imagine you are working as a rollercoaster designer. you are building a ride the top speed of a 65m/s at the bottom of the first

hill. you estimate that the efficiency of the tracks and car are using roughly 50%. how high must your first hill be to reach that top speed?
Physics
1 answer:
polet [3.4K]4 years ago
5 0

Rollercoasters generate speed by converting gravitational potential energy into kinetic energy by taking the cart to the top of a large hill and letting it go. The conversion of a perfectly efficient system would be like so:

PE = KE

and using the formulas for potential and kinetic energy:

mgh = 1/2mv^2.

However, the efficiency of this system is 50%, meaning that the kinetic energy obtained from this conversion would appear as so:

PE=0.5 KE

mgh=0.5(1/2mv^2)

mgh=1/4mv^2.

The masses cancel out, leaving:

gh=1/4v^2

The goal is to achieve 65 m/s, and with Earth’s innate gravity of 9.806 m/s^2, we have:

gh=1/4v^2

(9.806)h=1/4(65)^2

h=107.71 meters

The height of the first hill must be 107.7 meters to generate a speed of 65 m/s with a conversion efficiency of 50%.

Hope this helps!

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3 years ago
#1 Not sure where to start. This is for AP Physics!
yaroslaw [1]

First,

\rho=\dfrac mV

where \rho is density, m is mass, and V is volume. We can compute the volume of the roll:

2.7\,\dfrac{\mathrm g}{\mathrm{cm}^3}=\dfrac{1275\,\mathrm g}V

\implies V\approx472.22\,\mathrm{cm}^3\approx4.72\,\mathrm m^3

When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness x. Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).

So we have

V=Ax

where A is the given area, so

4.72\,\mathrm m^3=\left(18.5\,\mathrm m^2\right)x

\implies x\approx0.255\,\mathrm m=255\,\mathrm{mm}

If we're taking significant digits into account, the volume we found would have been V=470\,\mathrm m^3, in turn making the thickness x=250\,\mathrm{mm}.

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Below!

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c) Newton

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