Four point masses, each of mass 1.9 $kg$ are placed at the corners of a square of side 2.2 $m$. find the moment of inertia of this system about an axis that is perpendicular to the plane of the square and passes through one of the masses.

I = 36.78 kg m^{2}

Explanation:\

Given data:

side of square = 2.2 m

mass =1.9 kg

The moment of inertia i is the total sum of the moments of inertia of the 4 point masses  and it is given as

       =

I = 36.78 kg m^{2}

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer

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