Question

The sun rotates once in 25.05 days. That is 601.2 hours. so the sun's rotational speed is 0.0017 rotations per hour. what is the tangential velocity of a sunspot if the spot is 6.955 x 10^5 km from thew center of the sun?
in the part: 10^5 it means 10 to the power of 5, its an exponent, but i dint know how to put it in

Answer 1

Answer:

$v=2019.09\ m.s^{-1}$

Explanation:

Given:

• time taken by the sun to complete one revolution, $t=601.2\ hr$

• radial distance of the sunspot, $r=6.955\times 10^8\ m.s^{-1}$

Therefore, angular speed of rotation of sun:

$\omega=\frac{2\pi}{601.2\times 3600} \ rad.s^{-1}$

Now the tangential velocity of the sunspot can be given by:

$v=r.\omega$

$v=6.955\times 10^8\times \frac{2\pi}{601.2\times 3600}$

$v=2019.09\ m.s^{-1}$