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IceJOKER [234]
3 years ago
12

The sun appears to move through the background stars. This apparent motion would not exist if:_________.

Physics
1 answer:
otez555 [7]3 years ago
3 0

Answer:

b) the earth did not orbit the sun

Explanation:

The sun appears to move through the background stars due to parallax . Parallax is a phenomenon when the near object appears to move faster than the distant object . If we travel in a train , the near object like electric poles on rail track or trees and fields nearby appear to move faster against the distant background .

Hence when the earth moves around the sun , the sun appears to move against the background the stars which are far away . Had earth remained stationary at a place on its orbit around the sun , the sun would have appeared stationary against the background the   stars .

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What did Newton's book Mathematical Principles of Natural Philosophy argue about the role of mathematics in understanding the na
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Name some of the mediums that sound can travel through
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gas, liquid, and solid

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2 years ago
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Car 1 brakes to a stop on a dry road. Car 2 does the same thing, at the same speed, on a road where it has been raining. Explain
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3 years ago
The electric field between two parallel plates is uniform, with magnitude 628 N/C. A proton is held stationary at the positive p
aliina [53]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Data Given:

Electric Field between two parallel plates = 628 N/C

Separation = 4.22 cm

a) In this part, we are asked to calculate the distance from positive plate at which the electron and proton pass each other.

Solution:

First of all:

Force on proton due to the Electric field between the plates is:

F_{p} = q_{p}E

and, we know that, F = ma

So,

m_{p}a = q_{p}E

a = \frac{q_{p}.E }{m_{p} }      Equation 1

So,

The distance covered by the electron is:

S = ut + 1/2at^{2}

Here, u = 0.

S = 1/2at^{2}

Put equation 1 into the above equation:

S = 1/2 x (\frac{q_{p}.E }{m_{p} }  )t^{2}      Equation 2

So,  

Similarly, the distance covered by electron will be:

(D-S) = 1/2 x (\frac{q_{e}.E }{m_{e} }  )t^{2}    Equation 3

We know that the charge of electron is equal to the charge of proton so,

q_{p} = q_{e} = q

By dividing the equation 2 by equation 3, we get:

\frac{S}{D-S} = \frac{m_{e} }{m_{p} }

Solve the above equation for S,

Sm_{p} = m_{e}D - m_{e}S

So,

S = \frac{m_{e}.D }{(m_{e} + m_{p})  }

Plugging in the values,

As we know the mass of electron is 9.1 x 10^{-31} and the mass of proton is 1.67 x 10^{-27}

S = \frac{9.1 . 10^{-31} . 4.22 }{(9.1 . 10^{-31} + 1.67 . 10^{-27}  }

S = 0.002298 cm (Distance from the positive plate at which the two pass each other)

b) In this part, we to calculate distance for Sodium ion and chloride ion as above.

So,

we already have the equation, we need to put the values in it.

So,

S = \frac{m_{Cl}.D }{(m_{Cl} + m_{Na})  }

As we know the mass of chlorine is 35.5 and of sodium is 23

S = \frac{35.5 . 4.22}{(35.5 + 23)}

S = 2.56 cm

7 0
3 years ago
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