Answer:
Explanation:
Momentum before collision = momentum after collision
If
m₁ = 300 kg
v₁= 10 m/s
m₂ =1000 kg
v₂ = 0 as the car was parked
momemum before collision =
m₁ v₁ + v₂ m₂= 3000kgm/sec
momentum after collision= m₁ v₃+ v₄m₂=
300v₃+15m/sx 1000kg
now from the law of conservation of momentum
momentum before collision= momentum after collision
3000kgm/s= 300v₃ +15000kgm/s
-15000+3000= 300v₃
-12000=300v₃
v₃= -40m/s
the velocity of truck after collision is 40m/s and the negative sign indicates that its a recoil velocity.
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
Answer: According to http://www.scifun.org, "Bernoulli's Principle says that the pressure decreases inside a stream of flowing air. When the balloon begins to move out of this low pressure stream, the higher pressure of the air in the room pushes it back into the moving stream"
Answer:
3.48076 m
Explanation:
= Pressure difference = 35 kPa
= Density of water = 1025 kg/m³
g = Acceleration due to gravity = 9.81 m/s²
h = Depth of water
Pressure difference is given by
The depth at which this pressure difference occurs is 3.48076 m