The coefficient of static friction is 0.222
Explanation:
In order for the car to remain in circular motion, the frictional force must be able to provide the necessary centripetal force. Therefore, the car will start skidding when the two forces are equal:
where the term on the left is the frictional force, while the term on the right is the centripetal force, and where
is the coefficient of static friction
m is the mass of the car
g is the acceleration of gravity
v is the speed of the car
r is the radius of the track
In this problem, we have:
r = 564 m
v = 35 m/s
And re-arranging the equation for , we can find the coefficient of static friction:
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Originally there must been
1,4775E6 + 2.25E4 = 147.75E4 + 2.25E4 = 150E4 present at start
% = 2.25 / 150 = 1.5 % of 235 U left
Kinetic Energy = (1/2) (mass) (speed)
First runner: KE = (1/2) (45kg) (49 m/s) = 1,102.5 Joules
Second runner: KE = (1/2) (93kg) (9 m/s) = 418.5 Joules
The <em>first runner </em><em>has 163</em>% more kinetic energy than the second runner has.
Answer:
The Celcius and kelvin scale are related unit for unit. One degree unit on the Celcius scale is equivalent to one degree unit on the kelvin scale. The only difference between these two scales is the zero point.
Expansion work against constant external pressure: w=-pex Δ Δ V 3. The attempt at a solution . I tried following that. Because Vf>>Vi, and Vf=nRT/pex, then w=-pex x nRT/pex=-nRT (im assuming n is number of moles of CO2?). 1 mole of CaCO3 makes 1 mole of CO2, so plugging in numbers, I get 8.9kJ, although I dont use the 1 atm pressure at all
