Answer:1.6 rad/s
Explanation:
Given
moment of Inertia of disk 
radius of disc 
Force 
Torque 


using








Since we're dealing with radial acceleration around a circle, I used the radial acceleration equation a=v²/r. At the top of the hill, the force upward exerted by the hill is less than the weight of the sled. if v is large enough the term (g-v²/r) will become 0 and the sled will fly off the ground as it reaches the peak. Let me know if I can clarify any of my work.
Stephen`s Law:
P = (Sigma) · A · e · T^4
P in = P out
e = 1 for blacktop;
1150 W = (Sigma) · T^4
(Sigma) = 5.669 · 10 ^(-8) W/m²K^4
T^4 = 1150 : ( 5.669 · 10^(-8) )
T^4 = 202.875 · 10^8
![T = \sqrt[4]{202.857 * 10 ^{8} }](https://tex.z-dn.net/?f=T%20%3D%20%20%5Csqrt%5B4%5D%7B202.857%20%2A%2010%20%5E%7B8%7D%20%7D%20)
T = 3.774 · 10² =
377.4 KAnswer: Equilibrium temperature is 377.4 K.
A parallel circuit - there are multiple branches at which current is able to to be pushed around the circuit.
Yhuihoifjhh <span>F = Gm1m2 / r^2
if the masses are doubled then the force is increased by a factor of 4
if the distance is doubled the force is decreased by a factor of 1/ 2^2
the net result is no change in force</span>