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Ne4ueva [31]
2 years ago
11

A horizontail rod (oriented in the east -west direction) is moved northward at a constant velocity through a magnetic field that

points straight down. Make a statement concerning the potential induced across the rod.
A) The east end of the rod is at higher potential than the west end.
B) The bottom surface of the rod is at higher potential than the top surface.
C) The top surface of the rod is at higher potential than the bottom surface.
D) The west end of the rod is at higher potential than the east end.
E) The potential is uniform.
Physics
1 answer:
Ede4ka [16]2 years ago
6 0

Answer:

a

Explanation:

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Answer:

B. 7.5 m/s^2

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What is the pressure exerted by a column of fresh water 10m high<br>(p=1000kg/m3 and g=10m/s)​
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  • Height (h) = 10 m
  • Density (ρ) = 1000 Kg/m^3
  • Acceleration due to gravity (g) = 10 m/s^2
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<u>Answer</u><u>:</u>

<u>1000</u><u>0</u><u>0</u><u> </u><u>Pa</u>

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A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?
dolphi86 [110]

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

g= Acceleration due to gravity

d = Distance of center of mass from pivot

I = Moment of inertia

\omega= Angular velocity

Replacing the value for moment of inertia

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

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\omega = 104.7rad/s

Replacing our values we have that

\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}

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The precession frequency is

\Omega = \frac{2\pi rad}{T}

T = \frac{2\pi rad}{\Omega}

T = \frac{2\pi}{1.17}

T = 5.4 s

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