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djyliett [7]
3 years ago
11

In the 1920s what did Edmund hubble notice about the galaxies

Physics
1 answer:
zlopas [31]3 years ago
5 0
Hubble noticed that the galaxies were moving away from us, which meant the universe was expanding.

This is why constellations change over time. In some years, the Big Dipper won't actually look like a dipper anymore.
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In the development of atomic models, it was realized that the atom is mostly empty. Consider a model for the hydrogen atom where
Rama09 [41]

Answer:

Check the explanation

Explanation:

a) in solving the first question, we will be choosing a spherical ball of radius 1 m, Therefore the width of the object is the diameter = 2m.

b)Given that and according to the question, the radius of the electron's orbit = 1x105 x radius of the object = 1x105 x1 = 1x105 m

8 0
3 years ago
The starships of the Solar Federation are marked with the symbol of the Federation, a circle, whereas starships of the Denebian
Llana [10]

Complete question

The complete question is shown on the first uploaded image  

Answer:

The velocity is  v = c* \sqrt{1 -  \frac{1}{n^2} }

Explanation:

From the question we are told that

           a = nb

The length of the minor axis  of  the symbol of the Federation, a circle, seen by the observer at velocity v must be equal to the minor axis(b) of the  Empire's symbol, (an ellipse)

Now this length seen by the observer can be mathematically represented as

        h = t \sqrt{1 - \frac{v^2}{c^2} }

Here t  is the actual length of the major axis of of the  Empire's symbol, (an ellipse)

So t = a = nb

and  b is the length of the minor axis of the symbol of the Federation, (a circle) when seen by an observer at velocity v which from the question must be the length of the minor axis of the of the  Empire's symbol, (an ellipse)

 i.e    h = b

So

    b  =  nb  [\sqrt{1 - \frac{v^2}{c^2} } ]  

     [\frac{1}{n} ]^2 =  1 -  \frac{v^2}{c^2}

      v^2 =c^2 [1- \frac{1}{n^2} ]

       v^2 =c^2 [\frac{n^2 -1}{n^2} ]

        v = c* \sqrt{1 -  \frac{1}{n^2} }

     

     

6 0
3 years ago
The single-pole, single-throw switch is normally wired ? between the source and the load to turn devices on and off.
zzz [600]

In series.

Single-pole and single-throw switch:

A switch with only one input and one output is referred to as a Single Pole Single Throw (SPST) switch. This indicates that it has a single output terminal and a single input terminal.

A single pole, one throw switch functions as an on/off switch in circuits. The circuit is turned on when the switch is closed. The circuit is shut off when the switch is open.

Thus, SPST switches are relatively basic in design.

Circuit for a single-pole, single-throw (SPST) switch

Types:

According to the application, it can be divided into three categories, including:

  • Simple SPST
  • (ON)-OFF, Push-to-close, SPST Momentary
  • ON-(OFF), Push-to-Open, SPST Momentary
  • Inches Switch SPST

Learn more about terminal here:

brainly.com/question/14236970

#SPJ4

7 0
2 years ago
Three +3.0-μC point charges are at the three corners of a square of side 0.50 m. The last corner is occupied by a −3.0-μC charge
kramer

Answer:

E = 440816.32 N/C

Explanation:

Given data:

Three point charge of charge equal to +3.0 micro coulomb

fourth point charge = - 3.0 micro coulomb

side of square = 0.50 m

K =1/4 \pi \epsilon_0 = 8.99 \times 10^9 N.m^2/c^2

Due to having equal charge on center of square, 2 charge produce equal electric field at center and other two also produce electric field at center of same value

So we have

E_1 + E_3 = 0

E =E_2 + E_4

E = 2 E_2

[E_2 =\frac{2\times k \times q}{r^2}

[r= \frac{(0.5^2 + 0.5^2)^2}{2} = 0.35 m]

plugging all value

E = 2 E_2

E = 2 E_2 =\frac{2\times k \times q}{r^2}

E = \frac{2 \times 8.99 \times 10^93\times 10^{-6}}{0.35^2}

E = 440816.32 N/C

3 0
3 years ago
Read 2 more answers
The force of repulsion between two like-charged particles will increase if
svlad2 [7]
If they're brought closer together
5 0
3 years ago
Read 2 more answers
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