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-Dominant- [34]
3 years ago
13

Hipparchus could have warned us that the dates associated with each of the natal astrology sun signs would eventually be wrong.

Explain why.
Physics
1 answer:
emmasim [6.3K]3 years ago
4 0

Explanation:

Hipparchus had observed and discovered the precession motion of Earth. Due to the precession the axis of the Earth will gradual shift in a cycle of  25772 years. That means the axis will point towards different stars in the celestial sphere and our pole stars will change.

Because of this change in the axis the apparent position of Sun with respect to the background constellation (sun signs) regresses. Because of which the dates associated with the sun signs will be wrong.

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N experiment is performed in deep space with two uniform spheres, one with mass 27.0 and the other with mass 107.0 . They have e
Reptile [31]

Answer:

Explanation:

Apply the law of conservation of energy

KE_i+PE_i=KE_f+PE_f

Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)

from the law of conservation of the linear momentum

m_1v_1=m_2v_2

Therefore,

Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)

=\frac{1}{2} [m_1v_1^2+m_2[\frac{m_1v_1}{m_2} ]^2]\\\\=\frac{1}{2} [m_1v_1^2+\frac{m_1^2v_1^2}{m_2} ]\\\\=\frac{m_1v_1^2}{2} [\frac{m_1+m_2}{m_2} ]

v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]

Substitute the values in the above result

v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]

=[\frac{2(6.67\times 10^-^1^1)(107)^2}{27+107} ][\frac{1}{26} -\frac{1}{41}] \\\\=1.6038\times 10^-^1^0\\\\v_1=\sqrt{1.6038\times 106-^1^0} \\\\=1.2664 \times 10^-^5m/s

B)  the speed of the sphere with mass 107.0 kg is

v_2=\frac{m_1v_1}{m_2}

=[\frac{27}{107} ](1.2664 \times 10^-^5)\\\\=3.195\times 10^-^6m/s

C)  the magnitude of the relative velocity with which one sphere is

v_r=v_1+v_2\\\\=1.2664\times 10^-^5+3.195\times10^-^6\\\\=15.859\times10^-^6m/s

D) the distance of the centre is proportional to the acceleration

\frac{x_1}{x_2} =\frac{a_1}{a_2} \\\\=\frac{m_2}{m_1} \\\\=3.962

Thus,

x_1=3.962x_2

and

x_2=0.252x_1

When the sphere make contact with eachother

Therefore,

x_1+x_2+2r=41\\x_1+0,252x_1+2r=41\\1.252x_1+2r=41\\x_1=32.747-1.597r

And

x_1+x_2+2r=41\\3.962x_2+x_2+2r+41\\4.962x_2+2r=41\\x_2=8.262-0.403r

The point of contact of the sphere is

32.747-1.597r=8.262-0.403r\\\\r=\frac{24.485}{1.194} \\\\=20.506m

3 0
3 years ago
A wave has a wavelength of 20 mm and a frequency of 5 Hz what is the speed?
yKpoI14uk [10]
The answer is 100mm/s. I hope this helps :)

7 0
3 years ago
An alcohol functional group is called what ?
Ksenya-84 [330]
An alcohol functional group is called a hydroxyl group.
6 0
4 years ago
2500000000000 in scientific notation
Alex_Xolod [135]
Your answer is 2.5 x 10^(11).

Please take the time to rate and thank me if you feel I helped you! Thanks, it means a lot to me. :)
4 0
3 years ago
Read 2 more answers
what is the approximate weight of a 20-kg cannonball on the moon if the acceleration due to gravity is 1.6m/s^2
monitta
On Earth, a cannonball with a mass of 20 kg would weigh 196 Newtons.
With the formula F=mg, where F is the weight in Newtons, m is the mass, and g is the acceleration due to gravity on the Earth which is 9.8m/s^2.
F=20kg x 9.8m/s^2= 196 Newtons

BUT on the moon, acceleration due to gravity is 1.6 m/s^2,
so F=mg=20kgx1.6m/s^2= 32 N
5 0
3 years ago
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