As per the question the initial speed of the car [ u] is 42 m/s.
The car applied its brake and comes to rest after 5.5 second.
The final velocity [v] of the car will be zero.
From the equation of kinematics we know that
[ here a stands for acceleration]



Here a is taken negative as it the car is decelerating uniformly.
We are asked to calculate the stopping distance .
From equation of kinematics we know that
[here S is the distance]
![= 42*5.5 +\frac{1}{2} [-7.64] [5.5]^2 m](https://tex.z-dn.net/?f=%3D%2042%2A5.5%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5B-7.64%5D%20%5B5.5%5D%5E2%20m)
[ans]
Answer:
a) 
b) 
c) Towards the center of the centrifuge
Explanation:
a)
Becuse the centrifuge rotates in circular motion, there's an angular acceleration tha simulates high gravity accelerations

with r the radius and ω the amgular velocity, in or case
so:
and g=9.8
solving for ω:


b) Linear speed (v) and angular speed are related by:


c) The apparent weigth is pointing towards the center of the circle, becuse angular acceleration is pointing in that direction.
Atomic number is equal to the number of protons and electrons
Atomic mass - protons = neutrons
protons + neutrons = atomic mass
I hope this helps
Formula for distance is d=st
so for speed is s=d/t
48 km per hour