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Neko [114]
3 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]3 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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Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

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\Delta d=0.8 cm=0.008 m

\Delta V=100 V

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

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Capacitance of capacitor after moving plates

C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

Potential difference between plates after moving

V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}

\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100

\frac{Q\Delta d}{\epsilon_0 S}=100

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