When a car rolls down a ramp, which force supports the car's weight?
1 answer:
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Answer:
a) 0 J
b) W = nRTln(Vf/Vi)
c) ΔQ = nRTln(Vf/Vi)
d) ΔQ = W
Explanation:
a) To find the change in the internal energy you use the 1st law of thermodynamics:
Q: heat transfer
W: work done by the gas
The gas is compressed isothermally, then, there is no change in the internal energy and you have
ΔU = 0 J
b) The work is done by the gas, not over the gas.
The work is given by the following formula:
n: moles
R: ideal gas constant
T: constant temperature
Vf: final volume
Vi: initial volume
Vf < Vi, then W < 0 and the work is done on the gas
c) The gas has been compressed. Thus, its temperature increases and heat has been transferred to the gas.
The amount of heat is equal to the work done W
d)
Answer:
the ball will fly in AX direction, making angle of 8.84° from the motion of the car
Explanation:
Given the data in the question and as illustrated in the diagram below;
Now, Lets assume line AB represent the movement of the car,
AC is the movement of the ball been thrown back and forth in the back seat
Ax is the motion of the ball it flies off the window
so from the diagram, We can see triangle ABC
where AB is 45 mph and AC = 7 mph
and angle ∠CAB = 90°
using SOH CAH TOA
TOA; tanθ = Opposite / Adjacent
tanθ = Opposite / Adjacent
tan( ∠ ABC ) = AC / AB
we substitute
tan( ∠ ABC ) = 7 / 45
tan( ∠ ABC ) = 0.15555
( ∠ ABC ) = tan⁻¹ 0.15555
( ∠ ABC ) = 8.84°
Therefor, angle ( ∠ ABC ) is 8.84°
Meaning angle ( ∠ XAA' ) is also 8.84°
Therefore, the ball will fly in AX direction, making angle of 8.84° from the motion of the car
