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il63 [147K]
3 years ago
12

The slow coning pattern of the spin axis of the Earth's precession will move the end of this axis in a complete circle in a peri

od of_________.
Physics
1 answer:
GREYUIT [131]3 years ago
3 0

Answer:

26000 years

Explanation:

Precession describes the angular motion of the Earth's body. Since the attitude of telescopes relative to the Earth's body can be controlled with high accuracy, and telescopes can measure the direction of incoming light also with high accuracy, the motion of Earth is under permanent high precision monitoring. Thus the basic numerical descriptor of precission, an angular rate of 5029.0966 seconds of arc per Julian century, traditionally denoted p (for precession) is a measured value from observed coordinate changes of thousands of stars over, say, two centuries. The understanding of this value in terms of forces acting on an oblate Earth from the Moon is well understood so that an extrapolation back and forth over a few full cycles contains little uncertainties. Of course, you can find details on the coordinate transformations mentioned above (the direct observational effect of precession) on the net. I was surprised to see that the Wikipedia article on precession covers the astronomical aspect very poorly. You thus better look for other sources.

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Rashid [163]

Answer:

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Explanation:

7 0
2 years ago
Can we see a halo around the half moon?​
Hatshy [7]

Answer:

No we cannot

Explanation:

But what causes a ring to appear around the moon? This phenomenon is called a "moon halo." According to the National Weather Service, this ring of light, which is actually an optical illusion, forms around the moon when moonlight refracts off ice crystals in cirrus clouds, high up in the Earth's atmosphere.

8 0
1 year ago
Read 2 more answers
Two friction disks A and B are brought into contact when the angular velocity of disk A is 240 rpm counterclockwise and disk B i
mel-nik [20]

Answer:

a) αA = 4.35 rad/s²

αB = 1.84 rad/s²

b) t = 3.7 rad/s²

Explanation:

Given:

wA₀ = 240 rpm = 8π rad/s

wA₁ = 8π -αA*t₁

The angle in B is:

\theta _{B} =4\pi =\frac{1}{2} \alpha _{B} t_{1}^{2}  =\frac{1}{2} (\frac{r_{A} }{r_{B} } )^{3} \alpha _{A} t_{1}^{2}=\frac{1}{2} (\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}^{2}

\alpha _{A} =8\pi (\frac{0.2}{0.15} )^{3} =59.57rad

w_{B,1} =\alpha _{B} t_{1}=(\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}=0.422\alpha _{A} t_{1}

The velocity at the contact point is equal to:

v=r_{A} w_{A} =0.15*(8\pi -\alpha _{A} t_{1})=1.2\pi -0.15\alpha _{A} t_{1}

v=r_{B} w_{B} =0.2*(0.422\alpha _{A} t_{1})=0.0844\alpha _{A} t_{1}

Matching both expressions:

1.2\pi -0.15\alpha _{A} t_{1}=0.0844\alpha _{A} t_{1}\\\alpha _{A} t_{1}=16.09rad/s

b) The time during which the disks slip is:

t_{1} =\frac{\alpha _{A} t_{1}^{2}}{\alpha _{A} t_{1}} =\frac{59.574}{16.09} =3.7s

a) The angular acceleration of each disk is

\alpha _{A}=\frac{\alpha _{A} t_{1}}{t_{1} } =\frac{16.09}{3.7} =4.35rad/s^{2} (clockwise)

\alpha _{B}=(\frac{0.15}{0.2} )^{3} *4.35=1.84rad/s^{2} (clockwise)

6 0
2 years ago
Convert 70 mi/h to m/s. 1 mi = 1609 m.<br><br> Answer in units of m/s.<br><br> Plz help me now
melisa1 [442]

Answer:

70mi/h

1mile =1609 meters

1hour=3600 seconds

So,

70×1609/3600

112,630/3600

31.286 m/s

Hope this helps you

6 0
3 years ago
In a certain time period a coil of wire is rotated from one orientation to another with respect to a uniform 0.38-T magnetic fie
Evgesh-ka [11]

Answer:

8.4 V

Explanation:

induced emf, e1 = 5.8 V

Magnetic field, B1 = 0.38 T

magnetic field, B2 = 0.55 T

induced emf, e2 = ?

As we know that the induced emf is directly proportional to the magnetic field strength.

When the other parameters remains constant then

\frac{e_{1}}{e_{2}}=\frac{B_{1}}{B_{2}}

\frac{5.8}{e_{2}}=\frac{0.38}{0.55}

e2 = 8.4 V

Thus, the induced emf is 8.4 V.

4 0
2 years ago
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