if an asteroid is travelling at 5320 mis around the sun at a distance of 4.8 x 10 meters from the sun, what is the period (t) of the asteroid?

answer; ///(74) degrees is your answer..

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s