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faust18 [17]
3 years ago
8

When the Earth is closest to the Sun, it is at:

Physics
1 answer:
ahrayia [7]3 years ago
5 0

a). Perihelion . . . the point in Earth's orbit that's closest to the Sun.
                            We pass it every year early in January. 


b). Aphelion . . . the point in Earth's orbit that's farthest from the Sun.
                          We pass it every year early in July. 

c). Proxihelion . . . a made-up, meaningless word

d). Equinox . . . the points on the map of the stars where the Sun
                         appears to be on March 21 and September 21.

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Colored sparks and rusting.
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3 years ago
Which refers to any disturbance that carries energy from one place to another through matter and space?
Black_prince [1.1K]
I think frequency it sounds like the correct answer but I am not completely sure if I am correct
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At its Ames Research Center, NASA uses its large "20-G" centrifuge to test the effects of very large accelerations ("hypergravit
Over [174]

Answer:

v=32.9m/s

Explanation:

The acceleration needed to mantain a circular motion of radius r and speed v is given by the equation a=v^2/r

This is the centripetal acceleration. The person will feel what is called a centrifugal acceleration, of the same value, because he is not in an inertial frame (thus subject to fictitious forces, product of inertia).

We want to know the speed of his head when it is subject to 12.5 times the value of the acceleration of gravity while moving on a 8.84m radius circle, so we must do:

v=\sqrt{ar} = \sqrt{12.5gr}=\sqrt{(12.5)(9.8m/s)(8.84m)}=32.9m/s

7 0
3 years ago
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

\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})

\omega = 104.7rad/s

Replacing our values we have that

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

\Omega = 1.17rad/s

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

Therefore the precession period is 5.4s

7 0
3 years ago
A student finds an unlabeled liquid container in his lab. He notices that the container has two liquids. Since the two liquids h
raketka [301]

Answer:

\rho = 1848.03 kg m^{-3}

Explanation:

given data:

density of water \rho = 1 gm/cm^3 = 1000 kg/m^3

height  of water  = 20 cm  =0.2 m

Pressure  p = 1.01300*10^5 Pa

pressure at bottom

P =  P_{fluid} + P_{h_2 o}

P   = P_{fluid}  + \rho g h

P_{fluid}  = P - \rho g h

                 = 1.01300*10^5 - 1000*0.2*9.8

                 = 99340 Pa

p_{fluid}  = P_{atm} + \rho g h_{fluid}                       h_[fluid} = 0.307m

99340 = 104900 + \rho *9.8*0.307

\rho = 1848.03 kg m^{-3}

5 0
3 years ago
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