Suppose mercury had 5-to-3 spin-orbit coupling and that its sidereal period remained 88 days, as now. you observe the sun directly overhead in the mercurian sky and then observe again from the same location 88 days later. where will the sun be in the sky? (mercury, like the earth, rotates toward the east.)
a) below the eastern horizon
b) below the western horizon
c) directly overhead again
d) directly on the opposite side of the planet

no

sodium is solid at rt

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