Liquid Hydrogen is the fuel used by rockets.
Explanation:
- Liquid hydrogen which can be chemically denoted as "
" is often considered as the significant fuels for rocket.
- However rocket in its lower stages uses fuels such as Kerosene and oxygen where as in the higher stages such as second and third stages it uses liquid hydrogen.
- Liquid hydrogen is known to easily cool the nozzle and then also other parts of the rocket before mixing with the oxidizer such as the oxygen.
- Thus liquid hydrogen helps in preventing nozzle erosion and also reduces combustion chamber.
- Liquid hydrogen one the other hand is very expensive as 384,071 gallons of it will cost approximately $376,389.58
.
Thus liquid hydrogen is effectively used as a fuel for rocket.
A broken yellow line on the pavement tells that the adjacent lane is traveling in the opposite direction and passing is permitted.
A broken white line on the pavement show that the adjacent lane is traveling in the same direction and passing is permitted.
<h3>What does pavement markings show?</h3>
Pavement markings are known to be tools that are used to pass infor or messages to roadway users.
Note that they tell the part of the road that one need to use, give information about conditions ahead, and others
Note that A broken yellow line on the pavement tells that the adjacent lane is traveling in the opposite direction and passing is permitted.
Learn more about pavement markings from
brainly.com/question/10179521
#SPJ1
Answer
For isotropic material plastic yielding depends upon magnitude of the principle stress not on the direction.
Tresca and Von Mises yield criteria are the yield model which is widely used.
The Tresca yield criterion stated that yielding will occur in a material only when the greatest maximum shear stress reaches a critical value.
max{|σ₁ - σ₂|,|σ₂ - σ₃|,|σ₃ - σ₁|} = σ_f
under plane stress condition
|σ₁ - σ₂| = σ_f
The Von mises yielding criteria stated that the yielding will occur when elastic energy of distortion reaches critical value.
σ₁² - σ₁ σ₂ + σ₂² = σ²_f
Answer:
0.5 kW
Explanation:
The given parameters are;
Volume of tank = 1 m³
Pressure of air entering tank = 1 bar
Temperature of air = 27°C = 300.15 K
Temperature after heating = 477 °C = 750.15 K
V₂ = 1 m³
P₁V₁/T₁ = P₂V₂/T₂
P₁ = P₂
V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³
For ideal gas, 
= 5/2×R = 5/2*0.287 = 0.7175 kJ
PV = NKT
N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)
N = 9.66×10²⁴
Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles
The average mass of one mole of air = 28.8 g
Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg
∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ
The power input required = The rate of heat transfer = 149.211/(60*5)
The power input required = 0.49737 kW ≈ 0.5 kW.
Answer:
The three types of relearn procedures are auto relearn, stationary and OBD.
Explanation:
In TPMS system, after the direct service like adjustment of air pressure, tire rotation or replacement of sensors etc, is performed then maximum vehicle often needs TPMS system relearn that needs to be performed.
For performing these relearn procedure, there are mainly three types:
- auto relearn
- stationary relearn
- OBD
After applying the relearn process, the TPMS system will again be in proper function.
