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 %.
Answer:
Engine A is able to accelerate a car faster.
Explanation:
(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.
(b) The net work done on the crate while it is on the rough surface is 23.7 J.
(c) The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.
<h3>Magnitude of net force on the crate</h3>
F(net) = F - μFf
F(net) = 280 - 0.351(92 x 9.8)
F(net) = -36.46 N
<h3>Net work done on the crate</h3>
W = F(net) x L
W = -36.46 x 0.65
W = - 23.7 J
<h3>Acceleration of the crate</h3>
a = F(net)/m
a = -36.46/92
a = - 0.396 m/s²
<h3>Speed of the crate</h3>
v² = u² + 2as
v² = 0.845² + 2(-0.396)(0.65)
v² = 0.199
v = √0.199
v = 0.45 m/s
Thus, the magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.
The net work done on the crate while it is on the rough surface is 23.7 J.
The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.
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