A 0.2-kg stone is attached to a string and swung in a circle of radius 0.6 m on a horizontal and frictionless surface. If the st
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Answer:
The value is the temperature of the air inside the tire 340.54 K
% of the original mass of air in the tire should be released 99.706 %
Explanation:
Initial gauge pressure = 2.7 atm
Absolute pressure at inlet = 2.7 + 1 = 3.7 atm
Absolute pressure at outlet = 3.2 + 1 = 4.2 atm
Temperature at inlet = 300 K
(a) Volume of the system is constant so pressure is directly proportional to the temperature.
340.54 K
This is the value is the temperature of the air inside the tire
(b). Since volume of the tyre is constant & pressure reaches the original value.
From ideal gas equation P V = m R T
Since P , V & R is constant. So
m T = constant
value of the original mass of air in the tire should be released is
⇒ -0.99706
% of the original mass of air in the tire should be released 99.706 %.
Consult the attached free body diagram.
By Newton's second law, the net force on the crate acting parallel to the surface is
∑ F[para] = (370 N) cos(-20°) - f = 0
(this is the x-component of the resultant force)
where
• (370 N) cos(-20°) = magnitude of the horizontal component of the pushing force
• f = magnitude of kinetic friction
The crate is moving at a constant speed and thus not accelerating, so the crate is in equilibrium.
Solve for f :
f = (370 N) cos(-20°) ≈ 347.686 N
The net force acting perpendicular to the surface is
∑ F[perp] = n - 1480 N - (370 N) sin(-20°) = 0
(this is the y-component of the resultant force)
where
• n = magnitude of normal force
• 1480 N = weight of the crate
• (370 N) sin(-20°) = magnitude of the vertical component of push
The crate doesn't move up or down, so it's also in equilibrium in this direction.
Solve for n :
n = 1480 N + (370 N) sin(-20°) ≈ 1606.55 N ≈ 1610 N
Then the coefficient of kinetic friction is µ such that
f = µn ⇒ µ = f/n ≈ 0.216
