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
Mid-ocean ridges happen along divergent plate boundaries, where new ocean floor is created as the Earth’s tectonic plates spread apart. As the plates separate, molten rock rises to the seafloor, producing large volcanic eruptions of basalt.
it can be said that the speed of the east wind is
v=0.3608m/s
From the question we are told
A small boat sailed <u>straight </u>north out of a harbor in <em>strong </em>east wind (blowing from west to east).
After sailing for 120 minutes, it ended up hitting a buoy 60^\circ60 ∘ to the north-east of the harbor. If the straight-line distance between the buoy and the harbor is 3 km,
- what is the speed of the east wind?.
<h3> the speed of the east wind</h3>
Generally the equation for the distance is mathematically given as
BA=3000sin60
BA=2598.07m
Therefore
the speed of the east wind
v=0.3608
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Answer:
140265.8 C = 1.403 × 10⁵ C
Explanation:
The battery's electric potential energy is used to account for the kinetic and potential work done in moving the car up this hill.
Potential work required to move the 757 kg car up a vertical height of 195 m = mgh
P.E = 757 × 9.8 × 195 = 1446627 J
Kinetic work done = (1/2)(m)(v²)
K.E = (1/2)(757)(25²) = 236562.5 J
Total work done in moving the car up that height = 1446627 + 236562.5 = 1683189.5 J
And this would be equal to the potential of the battery.
For the battery, potential difference = (electric potential energy)/(charges moved)
ΔV = ΔU/q
q = ΔU/ΔV
ΔU = 1683189.5 J
ΔV = 12.0 V
q = 1683189.5/12 = 140265.8 C
