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oksano4ka [1.4K]
3 years ago
10

Which of the following takes place in the combustion chamber of a gas turbine engine?

Physics
2 answers:
11Alexandr11 [23.1K]3 years ago
6 0
The correct option is D.
In the combustion chamber of a gas turbine engine, the fuel oil is mixed with a proper portion of a compressed air, this is then ignited. The major function of the combustion chamber is to add energy to the system to power the turbine. 
Yuliya22 [10]3 years ago
3 0

My test came up that D. was incorrect.

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The ability of solid rock to flow is called _____.
zysi [14]
The ability of solid rock to flow is called Plasticity.

Plasticity is the deformation of a solid mateial which resulted in non reversible changes of shape in response to applied force.

Another example of plasticity is when you bend metal through a forcer force to create an art or kitchen set
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3 years ago
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Batman (95kg) is standing on top of a 50m high building looking out over the city of Gotham. Given that he uses the potential en
Oksanka [162]

Answer:

47 kJoules (kJ)

Explanation:

Potential enegy on Earth is given by the relationship:

P.E. = mgh, where m is mass, g is the acceleration due to Earth's gravity, and h is height. Since we are given metric values, we will look for an answer that is consistent with Joules, the metric measure of energy. 1 Joule is defined as 1 kg*m^2/s^2, so we wnat units of kg, m, and sec.

We are given:

m = 95kg

h = 50 meters

Earth's gravity, g is 9.8 m/s^2

Enter the data:

P.E. = mgh

P.E. = (95kg)(9.8m/s^2)(50m)

P.E. = 46550 kg*m^2/s^2 or 46550 Joules(J)

Since we only have 2 sig figs, and since 1kJ =- 1000J

We can state the potential energy is 47kJ.

Spiderman has 47kJ of potential energy for the start of any dive back to Earth. [He needed that same amount of energy to reach that height, but we don't know from where it came. A jump, helicopter, beamed up by Scotty, or tossed up by Doctor Octopus.]

3 0
1 year ago
Which landform represents the boundary between the land and an ocean or a lake?
Likurg_2 [28]
Coastline or seashore
3 0
3 years ago
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Electrons are added into the outermost what in groups 1 & 2
aivan3 [116]

S orbital.

Group 1 elements have a general configuration ns^{1}, where n represents the highest occupied Principal Energy Level. For example, Lithium has the valence configuration 2s^{1} whereas Cesium has 6s^{1}. Both of them belong to Group 1 of Periodic Table.

Group 2 elements have a general configuration of ns^{2}. For example, Magnesium has 3s^{2} as its outer shell configuration while Strontium has the same as 5s^{2}.

We see that in both the cases, the outermost S orbital is being filled.

3 0
3 years ago
In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu
Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

Explanation:

The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

Distance traveled = 5.31 \times 10^{-13} m

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

  (K.E + P.E)_{initial} = (K.E + P.E)_{final}

 (\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)

where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

     U = (\frac{1}{2}m_{p}v^{2}_{p})

Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

            = (\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})

            = 5.218 \times 10^{-15} J

Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

4 0
3 years ago
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