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kvasek [131]
2 years ago
6

Shanika is an engineer at an amusement park who is experimenting with changes to the setup for a magnetic roller coaster ride. I

n one ride, there are two identical roller coaster cars (orange and green) that start on opposite sides of a large magnet located at the center of a station. Shanika wants to get the largest increase in potential energy she can by moving one car one space to the left or the right.
Shanika can move the orange car to point A or point B, or she can move the green car to point C or point D. Which movement should she make? Why will that movement result in the largest increase in potential energy? Describe the magnetic force that will act on the roller coaster car she moves.

Physics
1 answer:
Whitepunk [10]2 years ago
8 0

Answer:

I believe it might be point A since the question ask what will result in the ln a largest increase in potential energy

Explanation:

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Find the wavelength of the third line in the lyman series, and identify the type of em radiation.
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The wavelength of the third line in the Lyman series, and identify the type of EM radiation

In this series, the spectral lines are obtained when an electron makes a transition from any high energy level (n=2,3,4,5... ). The wavelength of light emitted in this series lies in the ultraviolet region of the electromagnetic spectrum.

1 / lambda = R(h)* ( \frac{1}{(n1)^{2} } -   \frac{1}{(n2)^{2} })

                 = 109678 ( \frac{1}{1^{2} } -  \frac{1}{3^{2} } )

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1 year ago
A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?
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To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

I = MR^2

Here,

M = Mass

R = Radius of the hoop

The precession frequency is given as

\Omega = \frac{Mgd}{I\omega}

Here,

M = Mass

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Replacing the value for moment of inertia

\Omega= \frac{MgR}{MR^2 \omega}

\Omega = \frac{g}{R\omega}

The value for our angular velocity is not in SI, then

\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})

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Therefore the precession period is 5.4s

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3 years ago
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