A 11,000-kg train car moving due east at 21.0 m/s collides with and couples to a 23,000-kg train car that is initially at rest.
1 answer:
Answer:
v = 6.79 m/s
Explanation:
It is given that,
Mass of a train car, m₁ = 11000 kg
Speed of train car, u₁ = 21 m/s
Mass of other train car, m₂ = 23000 kg
Initially, the other train car is at rest, u₂ = 0
It is a case based on inelastic collision as both car couples each other after the collision. The law of conservation of momentum satisied here. So,
V is the common velocity after the collisions
So, the two car train will move with a common velocity of 6.79 m/s.
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Answer:
A) ≥ 325Kpa
B) ( 265 < Pe < 325 ) Kpa
C) (94 < Pe < 265 )Kpa
D) Pe < 94 Kpa
Explanation:
Given data :
A large Tank : Pressures are at 400kPa and 450 K
Throat area = 4cm^2 , exit area = 5cm^2
<u>a) Determine the range of back pressures that the flow will be entirely subsonic</u>
The range of flow of back pressures that will make the flow entirely subsonic
will be ≥ 325Kpa
attached below is the detailed solution
<u>B) Have a shock wave</u>
The range of back pressures for there to be shock wave inside the nozzle
= ( 265 < Pe < 325 ) Kpa
attached below is a detailed solution
C) Have oblique shocks outside the exit
= (94 < Pe < 265 )Kpa
D) Have supersonic expansion waves outside the exit
= Pe < 94 Kpa
Answer:
11.625
Explanation:
L = length of the ladder = 16 ft
= rate at which top of ladder slides down = - 3 ft/s
= rate at which bottom of ladder slides
y = distance of the top of ladder from the ground
x = distance of bottom of ladder from wall = 4 ft
Using Pythagorean theorem
L² = x² + y²
16² = 4² + y²
y = 15.5 ft
Also using Pythagorean theorem
L² = x² + y²
Taking derivative both side relative to "t"
= 11.625 ft/s
Answer:
Using V= IR
I= 0.2307 Ampere
