Answer:
8 to 10 times
Explanation:
For dry road
u= 15 mph ( 1 mph = 0.44 m/s)
u= 6.7 m/s
Let take coefficient of friction( μ) of dry road is 0.7
So the de acceleration a = μ g
a= 0.7 x 10 m/s ² ( g=10 m/s ²)
a= 7 m/s ²
We know that
v= u - a t
Final speed ,v=0
0 = 6.7 - 7 x t
t= 0.95 s
For snow road
μ = 0.4
de acceleration a = μ g
a = 0.4 x 10 = 4 m/s ²
u= 30 mph= 13.41 m/s
v= u - a t
Final speed ,v=0
0 = 30 - 4 x t'
t'=7.5 s
t'=7.8 t
We can say that it will take 8 to 10 times more time as compare to dry road for stopping the vehicle.
8 to 10 times
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Answer:
(b) False
Explanation:
The specific internal energy of the system does not depend on the path of the process, it is a state function means its depend on only on the initial and the final position it does not depend on the path which it follow in the process.Internal energy is associated with the random motion of the molecules.
So it is false statement as internal energy is not a path function
Answer:
M2 = 0.06404
P2 = 2.273
T2 = 5806.45°R
Explanation:
Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.
Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,
To1 = (1.008)*(1000) = 1008 ºR
R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)
F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga
For the air q = cp(To2– To1)
(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2
Table A.3 of steam table gives P/P* = 2.273,
T/T* = 0.2066,
To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =
F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit
