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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.
Answer:
The correct option is;
A. be in compliance with school attendance requirements
Explanation:
The requirements to acquire a special restricted driver's license for driver's license for under under 18 that have had a beginner's for up to 180 days include a minimum of 40 hours driving practice 10 of which should be in the dark. The application for the conditional drivers license is to be signed by the parent or guardian and the application is to be accompanied with proof of acceptable school attendance
At 17, after holding the special restricted drivers licence for a year without issues you can obtain the full drivers license.
Answer:
- the capacity of the pump reduces by 35%.
- the head gets reduced by 57%.
the power consumption by the pump is reduced by 72%
Explanation:
the pump capacity is related to the speed as speed is reduces by 35%
so new speed is (100 - 35) = 65% of orginal speed
speed Q ∝ N ⇒ Q1/Q2 = N1/N2
Q2 = (N2/N1)Q1
Q2 = (65/100)Q1
which means that the capacity of the pump is also reduces by 35%.
the head in a pump is related by
H ∝ N² ⇒ H1/H2 = N1²/N2²
H2 = (N2N1)²H1
H2 = (65/100)²H1 = 0.4225H1
so the head gets reduced by 1 - 0.4225 = 0.5775 which is 57%.
Now The power requirement of a pump is related as
P ∝ N³ ⇒ P1/P2 = N1³/N2³
P2 = (N2/N1)³P1
H2 = (65/100)²P1 = 0.274P1
So the reduction in power is 1 - 0.274 = 0.725 which is 72%
Therefore for a reduction of 35% of speed there is a reduction of 72% of the power consumption by the pump.
Answer:
d. is applied after the ceiling joists are
installed.