Answer:
See explanation
Explanation:
We have a mass 
revolving around an axis with an angular speed 
, the distance from the axis is 
. We are given:
![\omega = 10 [rad/s]\\r=0.5 [m]\\m=13[Kg]](https://tex.z-dn.net/?f=%5Comega%20%3D%2010%20%5Brad%2Fs%5D%5C%5Cr%3D0.5%20%5Bm%5D%5C%5Cm%3D13%5BKg%5D)
and also the formula which states that the kinetic rotational energy of a body is:
.
Now we use the kinetic energy formula
where 
is the tangential velocity of the particle. Tangential velocity is related to angular velocity by:
After replacing in the previous equation we get:
now we have the following:
therefore:
then the moment of inertia will be:
![I = 13*(0.5)^2=3.25 [Kg*m^2]](https://tex.z-dn.net/?f=I%20%3D%2013%2A%280.5%29%5E2%3D3.25%20%5BKg%2Am%5E2%5D)
Following your push the ball rolls down the lane at 4.2m/s. What is the net force on the ball as it rolls down the lane at the constant speed?
by angular momentum conservation we will have
angular momentum of child + angular momentum of merry go round = 0
angular momentum of child = mvR
m = mass of child
R = radius of child
v = speed = 2 m/s
now let's say moment of inertia of merry go round is I
so we will have
so merry go round will turn in opposite direction with above speed
Ethylene glycol is termed as the primary ingredients in antifreeze.
The ethylene glycol molecular formula is C₂H₆O₂.
Molar mass of C₂H₆O₂ is = (2×12) +(6×1) + (216) = 62g/mol
Now that antifreeze by mass is 50%, then there is 1kg of ethylene glycol which is present in 1kg of water.
ΔTf = Kf×m
ΔTf = depression in the freezing point.
= freezing point of water freezing point of the solution
= O°c - Tf
= -Tf
Kf = depression in freezing constant of water = 1.86°C/m
M is the molarity of the solution.
=(mass/molar mass) mass of solvent in kg
=1000g/62 (g/mol) /1kg
=16.13m
If we plug the value we get
-Tf = 1.86 × 16.13 = 30
Tf = -30°c
