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2 years ago

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A roller coaster car drops a maximum vertical distance of 35.4 m. Determine the maximum speed of the car at the bottom of that d

rop. Ignore work done by friction.

1 answer:

marissa [1.9K]2 years ago

8 0

Answer:

The maximum speed of the car at the bottom of that drop is 26.34 m/s.

Explanation:

Given that,

The maximum vertical distance covered by the roller coaster, h = 35.4 m

We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :

$mgh=\dfrac{1}{2}mv^2$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 35.4}$

v = 26.34 m/s

So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.

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Let us consider two bodies having masses m and m' respectively.

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Let the orbital radius of planet A is $r_{1}$ = r and mass of planet is $m_{1}$ .

Let the mass of central star is m .

Hence the gravitational force for planet A is $f_{1} =G \frac{m_{1}*m }{r^{2} }$

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$=\frac{3}{4}$ [ ans]

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2 years ago

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Answer:

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Explanation:

Convert the individual temperatures to Kelvin (Surface temperature and internal temperature) before calculating the temperature difference of the body,

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Converting the surface and the internal temperature to temperature in Kelvin

Surface Temperature of the body (K) = (X + 273) K

Internal Temperature of the body (K) = (X + 7) + 273 = (X + 280) K.

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Also for Fahrenheit, Convert the individual temperatures (Surface temperature and internal temperature) to Fahrenheit before calculating the temperature difference of the body.

We use , F = 1.8C + 32

Where C = temperature in Celsius.

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Let The Surface temperature Be = X °C

And the internal Temperature of the body will be = (X + 7) °C

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2 years ago

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2 years ago

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