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Nastasia [14]
3 years ago
9

William Herschel thought that the Sun and Earth were roughly at the center of the great grouping of stars we call the Milky Way.

Today we know this is not the case. What was a key reason that Herschel did not realize our true position in the Milky Way?
Physics
2 answers:
balu736 [363]3 years ago
3 0

Answer:

Explained

Explanation:

William Herschel thought that the Sun and Earth were roughly at the center of the great grouping of stars we call the Milky Way. He could say so because he was only able to see a small part of the galaxy. This was because of the fact that dust extended throughout the disk of the galaxy, making visibility very low.

Serjik [45]3 years ago
3 0

Answer:

William Herschel's Theory (Heliocentrism)

Explanation:

It was in 1783, when William Herschel examined the stars using his handmade telescope to figure out the shape of the galaxy. As a result of this examination, he concluded that the shape of our Milky Way was that of a 'disk' and that the sun was positioned in the center of this disk.

The reason he came to such a conclusion was that to him the Earth appeared to be 'amidst of several other stars' and when he worked out for the magnitudes of these stars the resulted figures were same in every direction. Therefore, based on these findings he concluded that the Earth is positioned somewhat near to the 'center' of the Milky Way.

Later on it was revealed that there were two errors in his findings;

  1. The magnitude is not considered as a reliable means to measure the distance of the stars.
  2. He was unable to keep in consideration the dark areas containing nebulae which blocked the view to the center of the galaxy and he thought of them as the vacant spaces.

<em>Nebulae:  </em>Plural of Nebula (Latin for cloud or fog). A nebula is initially referred to any dispersed astronomical object, including galaxies beyond Milky Way.

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A state in which an atom has more energy than it does at its ground state.

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3 years ago
(b) 360 days into seconds.
Angelina_Jolie [31]

Explanation:

(b) We know that,

1 day = 24 hours

1 hour = 3600 s

So, we found that, 1 day = 86400 s

We need to find the 360 days into seconds. So,

1 day = 86400 s

360 days = 86400×360

360 days = 31104000 seconds

(d) Weight of a body, W = 600 N

Acceleration due to gravity on mars is 3.7 m/s²

Weight, W = mg

m is mass of body

m=\dfrac{W}{g}\\\\m=\dfrac{600}{3.7}\\\\m=162.16\ kg

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8 0
3 years ago
A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 3.05 m/s. If the roof is pitched at
vekshin1

1) Time in the air: 0.78 s

The motion of the ball is a projectile motion, which consists of two independent motions:

- A horizontal motion with constant horizontal velocity

- A vertical motion with constant downward acceleration of

g=-9.8 m/s^2 (acceleration of gravity)

The initial vertical velocity is

u_y = u sin \theta = (3.05)(sin(-40^{\circ}))=-1.96 m/s

where the negative sign means the direction is downward.

The vertical position of the ball is given by

y(t) = h + u_y t + \frac{1}{2}gt^2

where

h = 4.50 m is the initial heigth of the ball when it starts falling down

The ball reaches the ground when y = 0, so we have:

0 = 4.50 -1.96t-4.9t^2

This is a second-order equation; solving for t, we get

t = -1.18 s

t = 0.78 s

We discard the negative solution since it has no physical meaning, so we can say that the ball spent 0.78 s in the air.

2) Horizontal distance: 1.83 m

For this second part of the problem, we just have to consider the horizontal motion of the ball.

As we said previously, the motion of the ball along the horizontal direction is a uniform motion with constant velocity, which is given by

v_x = u cos \theta = (3.05)(cos (-40.0^{\circ}))=2.34 m/s

where u = 3.05 m/s is the initial speed and \theta the angle of projection.

For a uniform motion, we can use the following relationship between distance covered and velocity:

d=v_x t

and substituting t = 0.78 s, we find the total distance travelled along the horizontal direction by the ball before reaching the ground:

d=(2.34)(0.78)=1.83 m

7 0
3 years ago
How many coulombs of charge do 50 * 10^31 electrons possess
Angelina_Jolie [31]
Quantity of Charge , Q = ne
Where n = number of electrons
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             n = 50 * 10^31  electrons

Q =    (50 * 10^31)*( -1.6 * 10 ^-19 ) =  -8 * 10^13 C.

Note that the minus sign indicates that the charge is a negative charge.
7 0
3 years ago
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Tomtit [17]
The answer to this question is A. 



6 0
3 years ago
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