Drag each item to show if it is an element or not an element.
1 answer:
You might be interested in
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:
A) 1040 steel is not a possible candidate for this application
B) 35.94%
Explanation:
Initial length = 100 mm = 0.1 m
Initial diameter ( d ) = 7.5 mm = 0.0075 m
Tensile load ( p ) = 18,000 N
Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m
<u>A) would the 1040 steel be a possible candidate for this application</u>
<em>Yield strength of 1040 steel < stress ( in order to be a possible candidate )</em>
stress = p / A0 = ( 18000 ) / ( ) * 0.0075^2
= 18,000 / (4.418 * 10^-5 ) = 407.424 MPa
Yield strength of 1040 steel = 450 MPa
stress = 407.424 MPa
∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )
Therefore 1040 steel is not a possible candidate for this application
<u>B) Determine How much cold work would be required to reduce the diameter of the steel to 6.0 mm</u>
Area1 = ( ) ( 0.006 )^2 = 2.83 * 10^-5 m^2
therefore % of cold work done = ( A0 - A1 ) / A0 * 100 = 35.94%