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:
Oxygen cycle
Explanation:
The components of the reservoirs of oxygen that are exchange in our environment is the oxygen cycle
It suggests the movement of oxygen between the living and non-living parts.
- The cycle does not account for oxygen that is trapped and cannot be exchanged in nature.
- Oxygen is important component of the atmosphere.
- Gaseous exchange between living organisms and atmosphere involves oxygen to a very large extent.
Explanation:
1.
We use the equation
h = 
, where
h is the height traveled,
g is the acceleration due to gravity and
t is the time taken to reach height h.
We can now calculate t to be
= 0.495 s
Let v be the initial velocity of the player.
The player deaccelarates from v m/s to 0 m/s in 0.495 s at the rate of 9.81 m/s^2.
v = 9.81 m/s^2 x 0.495 s = 4.85 m/s
2.
The player takes 0.3 s to increase his velocity from 0 m/s to 4.85 m/s. So his average accelaration is
4.85 m/s / 0.3 s = 16.2 m/s^2
Explanation:
Q = mc∆T
= (0.34 kg)(94 J/kg-°C)(25°C)
= 799 J
Answer:
Final Velocity = 4.9 m/s
Explanation:
We are given;. Initial velocity; u = 2 m/s
Constant Acceleration; a = 0.1 m/s²
Distance; s = 100 m
To find the final velocity(v), we will use one of Newton's equations of motion;
v² = u² + 2as
Plugging in the relevant values to give;
v² = 2² + 2(0.1 × 100)
v² = 4 + 20
v² = 24
v = √24
v = 4.9 m/s
