A 35-ft³ rigid tank has propane at 25 psia, 540 R and is connected by a valve to another tank of 20 ft³ with propane at 40 psia,
700 R. The valve is opened and the two tanks come to a uniform state at 580 R.
What is the final pressure?
What is the final pressure?
1 answer:
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Answer:
1.
2. 163.3 m
Explanation:
Static friction between road and rubber, μs =0.06
The maximum speed of the car, v = 50 km/h
= (50)(1000/3600) m/s
= 13.89 m/s
The acceleration due to gravity,
The frictional force, f = μsN ...... (1)
The component mg cosθ which balance the normal reaction N
The component mg sinθ acts in an opposite direction to the frictional force f.
ΣF = mg sinθ-f = 0 ...... (2)
Substitute the equation (1) in equation (2), we get
ΣF = mgsinθ-μsN = 0
mgsinθ-μsmgcosθ =0
μs = sinθ/cosθ
tanθ = μs
θ = tan-1( μs) = tan-1(0.06) =
(b)The vertical component of the force is
N cosθ = fsinθ+mg
N cosθ = μsNsinθ+mg
N[cosθ- μs sinθ] = mg ...... (3)
The horizontal component of the force along the motion of the car is
Nsinθ+fcosθ = ma (Centripetal acceleration,
Nsinθ+fcosθ =
Nsinθ+μsNcosθ =
N[sinθ+μs cosθ] = ...... (4)
Dividing the equation (4) with equation (3),
[sinθ+μscosθ]/[cosθ- μs sinθ] =
cosθ[sinθ/cosθ+μs]/cosθ[1- μs sinθ/cosθ] =
[tanθ+μs]/[1-μs tanθ] =
From part (1), tanθ = μs
Then the above equation becomes
Therefore, the minimum radius of the curvature of the curve is
=
=
= 163.3 m