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dybincka [34]
3 years ago
10

Ppman2000 Aye Get On Treasure Quest and Join Masons Kingdom :D

Engineering
1 answer:
kondaur [170]3 years ago
5 0

Answer:

Explanation:

who's ppman2000

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What fuel do rockets use
natta225 [31]

Liquid Hydrogen is the fuel used by rockets.

Explanation:

  • Liquid hydrogen which can be chemically denoted as "LH_{2}" 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.

8 0
3 years ago
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1: asha started abusness with 30.000
svetoff [14.1K]

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Explanation:adrive with visual acutity of 20/30 can just decipher asing adistance 20ft from asing determine the maximum destance from the sing which drivers with the flowing visual acuities will able to see the same sing 20/15 20/50

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3 years ago
Market research is a good place to start the design process and usually involves asking questions about consumers.
ElenaW [278]
what is your question?
4 0
3 years ago
Marks
Lina20 [59]

Answer:

i think C

Explanation:

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3 0
2 years ago
Four kilograms of carbon monoxide (CO) is contained in a rigid tank with a volume of 1 m3. The tank is fitted with a paddle whee
Juli2301 [7.4K]

Answer:

a) 1 m^3/Kg  

b) 504 kJ

c) 514 kJ

Explanation:

<u>Given  </u>

-The mass of C_o2 = 1 kg  

-The volume of the tank V_tank = 1 m^3  

-The added energy E = 14 W  

-The time of adding energy t = 10 s  

-The increase in specific internal energy Δu = +10 kJ/kg  

-The change in kinetic energy ΔKE = 0 and The change in potential energy  

ΔPE =0  

<u>Required  </u>

(a)Specific volume at the final state v_2

(b)The energy transferred by the work W in kJ.  

(c)The energy transferred by the heat transfer W in kJ and the direction of  

the heat transfer.  

Assumption  

-Quasi-equilibrium process.  

<u>Solution</u>  

(a) The volume and the mass doesn't change then, the specific volume is constant.

 v= V_tank/m ---> 1/1= 1 m^3/Kg  

(b) The added work is defined by.  

W =E * t --->  14 x 10 x 3600 x 10^-3 = 504 kJ  

(c) From the first law of thermodynamics.  

Q - W = m * Δu

Q = (m * Δu) + W--> (1 x 10) + 504 = 514 kJ

The heat have (+) sign the n it is added to the system.

7 0
3 years ago
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