All categories

3 years ago

How do we know the sun rotates?

2 answers:

6 0

Hi pupil Here's your answer ::::

➡➡➡➡➡➡➡➡➡➡➡➡➡

The sun Doesnt rotates as we know all . But it rotates with the wjole our Milky Way Galaxy. This movement is so slow that we cant even recognise that we are shifted.

This we know because we all had studied about the solar system and space.

By this study we can say that Sun Rotates.

⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅

Hope this helps ......

oee [108]3 years ago

6 0

Answer: One side is blue-shifted and the other side is red-shifted.

Explanation: Correct Apex Answer 12/31/19

You might be interested in

Which of the following is true about scientific models? (2 points) Select one: a. Models are used to simplify the study of thing

Vanyuwa [196]

Question: Which of the following is true about scientific models? Select one: A) Models are used to simplify the study of things. B) Computer models are the most reliable kind of model. C) Models explain past, present and future information. D) A model is accurate if it does not change over time.

Answer: A) Models are used to simplify the study of things.

Hope this helps!.

~~~~~~~~~~~~~~~~~~~~~~

~A.W~ZoomZoom44

5 0

3 years ago

Read 2 more answers

An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea

melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, $v=1.37\times10^7\ \rm m/s$

Explanation:

Given:

Linear charge density of the ring= $0.1\ \rm \mu C/m$
Radius of the ring R=0.2 m
Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

$Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C$

Potential due the ring at a distance x from the centre of the rings is given by

$V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\$

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

$\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V$

Let $\Delta U$ be the change in potential Energy given by

$\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J$

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

$\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s$

So the electron will be moving with $v=1.37\times10^7\ \rm m/s$

5 0

3 years ago

What is the difference between gravitational force and the force of gravity

statuscvo [17]

This might help and it might not:

Gravitation is the acting force between two bodies. On the other hand, gravity is the force occurring between an object and the very big object earth. Every object with some mass exerts the gravitational force on every other object having some mass. This force and its strength depend on the masses of the objects under consideration. Gravity helps to keep the planets to move in their orbit around the sun.

Gravitation is the force of attraction between any two bodies in the universe. In our universe, each object attracts each other with a certain amount of force. The large distance of separation is the main reason for its weak nature.

Gravity is the weakest type of fundamental force in nature. Still, it holds together the entire solar systems and galaxies.

Gravity has the existence with unlimited range.

3 0

2 years ago

An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr

Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m