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NeTakaya
3 years ago
10

12. The small space above the piston in which fuel is burned is called the

Engineering
1 answer:
BaLLatris [955]3 years ago
8 0

Answer:

D. Combustion Chamber

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19. A circuit contains four 100 S2 resistors connected in series. If you test the circuit with a digital VOM,
yanalaym [24]

Answer:

D

Explanation:

in series circuit the resistance is divided Total resistance is equal to the sum of resistances

6 0
3 years ago
50 points
Burka [1]

Answer:

Water vapor

Explanation:

When water is in a vapor it tends to rise to a higher point. Because of this it would be able to reach the top of a building.

4 0
3 years ago
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
Read 2 more answers
A PMMA plate with a 25 mm (width) x 6.5 mm (thickness) cross-section has a contained crack of length 2c = 0.5 mm in the center o
victus00 [196]

Answer:

LAOD = 6669.86 N

Explanation:

Given data:

width= 25 mm = 25\times 10^{-3} m

thickness = 6.5 mm = 6.5\times 10^{-3} m

crack length 2c = 0.5 mm at centre of specimen

\sigma _{applied} =  1000 N/cross sectional area

stress intensity factor  =  k  will be

\sigma_{applied} = \frac{1000}{25\times 10^{-3}\times 6.5\times 10^{-3}}

                   = 6.154\times 10^{6} Pa

we know that

k =\sigma_{applied} (\sqrt{\pi C})

  =6.154\sqrt{\pi (2.5\times 10^{-04})}          [c =0.5/2 = 2.5*10^{-4}]

K = 0.1724 Mpa m^{1/2} for 1000 load

ifK_C = 1.15 Mpa m^{1/2} then load will be

Kc = \sigma _{frac}(\sqrt{\pi C})

1.15 MPa = \sigma _{frac}\times \sqrt{\pi (2.5\times 10^{-04})}

\sigma _{frac} = 41.04 MPa

load = \sigma _{frac}\times Area

load = 41.04 \times 10^6 \times 25\times 10^{-3}\times 6.5\times 10^{-3} N

LAOD = 6669.86 N

3 0
2 years ago
The greater the force applied to an object, the _____ the change in speed or direction of the object.
storchak [24]

Answer:

b

Explanation:

8 0
3 years ago
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