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2 years ago

What have we learned from the work of harlow shapley and others about the location of the sun in the milky way galaxy?.

Physics

1 answer:

marin [14]2 years ago

3 0

We learned that We are in the disk of the Galaxy, about 5/8 of the way from the center.

<h3></h3><h3>What is the work of Harlow Shapley?</h3>

Shapley, who was based in Boulder, Colorado, utilised Cepheid variable stars to determine the Milky Way Galaxy's size and the Sun's location within it through the use of parallax. He put out his "liquid water belt" theory, now referred to as the idea of a livable zone, in 1953.

Milky way Galaxy-

There are many stars, grains of dust, and gas in the Milky Way. It is known as a spiral galaxy because, from the top or bottom, it would appear to be whirling like a pinwheel. About 25,000 light-years from the galaxy's nucleus, the Sun is situated on one of the spiral arms.

Approximately 5/8 of the way from the galaxy's nucleus, we are in the disc. William Herschel believed that the Sun and Earth were about in the middle of the vast cluster of stars known as the Milky Way.

to learn more about Harlow Shapley's original estimate go to - brainly.com/question/28145909

#SPJ4

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Why mass does not affect period of the pendulum?

Strike441 [17]

Because gravity accelerates all objects the same regardless of their mass

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3 years ago

Why do you think we cannot use a common fuel like diesel or gasoline for space travel ?

Ede4ka [16]

Answer:

Explanation:

it take oxygen in the atmosphere to burn it... in space there isn't any air :0

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3 years ago

During an observation, it was noticed that light diffracts as it passes through small slits in a barrier. What does this evidenc

Lynna [10]

Answer:

It reveals that light is a wave

Explanation:

Diffraction is the property of a wave in which there is a bending of the wave about the corners of an obstacle or aperture into the geometrical shadow of the obstacle or aperture.

This simply implies that a wave bends or spreads out when it passes through openings. Since the light diffracts through small slits and diffraction has been shown to occur in water waves and sound waves, this property of diffraction can only be characteristic of a wave and thus, this evidence reveals that light is a wave.

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3 years ago

A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate

Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is $\bf{2.81 \times 10^{4}~n~C^{-1}}$ .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

$C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

$C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF$

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

$&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm$

Part(c):

If we apply Gauss' law of electrostatics, then

$&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}$

3 0

3 years ago

Lauren throws her first pitch of the season for her school's softball team. If the ball travels 80 feet and takes 1.58 seconds t

kow [346]

Answer:

50.63 ft/s (2dp)

Explanation:

Speed = Distance/Time

80/1.58 = 50.63291139

= 50.63 ft/s (2dp)

6 0

3 years ago