You can estimate one more digit past the smallest division on the measuring device. If you look at a 10mL graduated cylinder, for example, the smallest graduation is tenth of a milliliter (0.1mL). That means when you read the volume, you can estimate to the hundredths place (0.01mL).
m = mass of the person = 82 kg
g = acceleration due to gravity acting on the person = 9.8 m/s²
F = normal force by the surface on the person
f = kinetic frictional force acting on the person by the surface
μ = Coefficient of kinetic friction = 0.45
The normal force by the surface in upward direction balances the weight of the person in down direction , hence
F = mg eq-1
kinetic frictional force on the person acting is given as
f = μ F
using eq-1
f = μ mg
inserting the values
f = (0.45) (82) (9.8)
f = 361.6 N
This one is simple :)
use the equation F = ma and then re-arrange for a.
a = F / m
But there is a trick, so be careful. the question gives you wind resistance. Simply subtract the wind resistance from the thrust of the rocket to get the net force upward.

So,
Answer:
Explanation:
Velocity of Top most point of wheel is twice the Velocity of centre of mass of wheel

Thus angular velocity is given by



![K.E.=\frac{1}{2}m_{stone}V^2+2\left [ \frac{1}{2}m_{cyl}\left [ \frac{V}{2}\right ]^2\right ]+2\left [ \frac{1}{2}I\omega ^2\right ]](https://tex.z-dn.net/?f=K.E.%3D%5Cfrac%7B1%7D%7B2%7Dm_%7Bstone%7DV%5E2%2B2%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7Dm_%7Bcyl%7D%5Cleft%20%5B%20%5Cfrac%7BV%7D%7B2%7D%5Cright%20%5D%5E2%5Cright%20%5D%2B2%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7DI%5Comega%20%5E2%5Cright%20%5D)
![K.E.=\frac{V^2}{2}\left [ m_{stone}+\frac{m_{roller}}{2}\right ]+I\omega ^2](https://tex.z-dn.net/?f=K.E.%3D%5Cfrac%7BV%5E2%7D%7B2%7D%5Cleft%20%5B%20m_%7Bstone%7D%2B%5Cfrac%7Bm_%7Broller%7D%7D%7B2%7D%5Cright%20%5D%2BI%5Comega%20%5E2)
![K.E.=\frac{0.319^2}{2}\left [ 672+\frac{82}{2}\right ]+\frac{82\times 0.343^2\times 0.465^2}{2}](https://tex.z-dn.net/?f=K.E.%3D%5Cfrac%7B0.319%5E2%7D%7B2%7D%5Cleft%20%5B%20672%2B%5Cfrac%7B82%7D%7B2%7D%5Cright%20%5D%2B%5Cfrac%7B82%5Ctimes%200.343%5E2%5Ctimes%200.465%5E2%7D%7B2%7D)

Elastic collisions<span> are </span>collisions<span> in which both momentum and kinetic energy are conserved. I think the correct answer from the choices listed above is option A. Two </span><span>glass marbles bounce off each other is an example of an elastic collision. Hope this answers the question.</span>