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2 years ago

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You ride a roller coaster with a loop-the-loop. Compare the normal force that the seat exerts on you to the force that Earth exe

rts on you when you are passing the bottom of the loop. Express your answer in terms of R (radius of the loop), vb (speed at the bottom of the loop), and constant g. Nbottom/mg

1 answer:

timofeeve [1]2 years ago

4 0

Answer:

$N = mg + \frac{mv^2}{R}$

Explanation:

At the bottom of the loop, the normal force is opposite to my weight.

I am making a circular motion. So,

$F_{net} = \frac{mv^2}{R}$

The relationship between the normal force, my weight, my speed and the radius of the loop is

$N - mg = \frac{mv^2}{R}\\mg = N - \frac{mv^2}{R}\\ N = mg + \frac{mv^2}{R}$

Here, my weight (mg) is constant. But the normal force is inversely proportional to my speed.

If my speed is zero, the normal force would be maximum and equal to my weight. If my speed is to much, then the normal force would be equally high too.

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A (1.25+A) kg bowling ball is hung on a (2.50+B) m long rope. It is then pulled back until the rope makes an angle of (12.0+C)o

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Answer:

F = 0.535 N

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Let's use the concepts of energy, at the highest and lowest point of the trajectory

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2 years ago

The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on th

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Answer:

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

Explanation:

Statement is incomplete. Complete description is presented below:

<em>A freight train has a mass of </em> $1.83\times 10^{7}\,kg$ <em>. The wheels of the locomotive push back on the tracks with a constant net force of </em> $7.50\times 10^{5}\,N$ <em>, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?</em>

If locomotive have a constant net force ( $F$ ), measured in newtons, then acceleration ( $a$ ), measured in meters per square second, must be constant and can be found by the following expression:

$a = \frac{F}{m}$ (1)

Where $m$ is the mass of the freight train, measured in kilograms.

If we know that $F = 7.50\times 10^{5}\,N$ and $m = 1.83\times 10^{7}\,kg$ , then the acceleration experimented by the train is:

$a = \frac{7.50\times 10^{5}\,N}{1.83\times 10^{7}\,kg}$

$a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$

Now, the time taken to accelerate the freight train from rest ( $t$ ), measured in seconds, is determined by the following formula:

$t = \frac{v-v_{o}}{a}$ (2)

Where:

$v$ - Final speed of the train, measured in meters per second.

$v_{o}$ - Initial speed of the train, measured in meters per second.

If we know that $a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$ , $v_{o} = 0\,\frac{m}{s}$ and $v = 22.222\,\frac{m}{s}$ , the time taken by the freight train is:

$t = \frac{22.222\,\frac{m}{s}-0\,\frac{m}{s} }{4.098\times 10^{-2}\,\frac{m}{s^{2}} }$

$t = 542.265\,s$

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

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Answer:

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(A) is correct option.

Explanation:

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