True because heat energy moves to the warmer areas.
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:
72.54 degree west of south
Explanation:
flow = 3.9 m/s north
speed = 11 m/s
to find out
point due west from the current position
solution
we know here water is flowing north and ship must go south at an equal rate so that the velocities cancel and the ship just goes west
so it become like triangle with 3.3 point down and the hypotenuse is 11
so by triangle
hypotenuse ×cos(angle) = adjacent side
11 ×cos(angle) = 3.3
cos(angle) = 0.3
angle = 72.54 degree west of south
Answer:
D) 21
Explanation:
When gas absorbs light , electron at lower level jumps to higher level .
and the difference of energy of orbital is equal to energy of radiation absorbed.
Here energy absorbed is equivalent to wavelength of 91.63 nm
In terms of its energy in eV , its energy content is eual to
1243.5 / 91.63 = 13.57 eV. This represents the difference the energy of orbit .
Electron is lying in lowest or first level ie n = 1.
Energy of first level
= - 13.6 / 1² = - 13.6 eV.
Energy of n th level = - 13.6 / n². Let in this level electron has been excited
Difference of energy
= 13.6 - 13.6 / n² = 13.57 ( energy of absorbed radiation)
13.6 / n² = 13.6 - 13.57 = .03
n² = 13.6 / .03 = 453
n = 21 ( approx )
