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TiliK225 [7]
2 years ago
8

What will happen if there is no oxygen for 2mins???​

Physics
1 answer:
Natasha_Volkova [10]2 years ago
5 0

Without oxygen, there would not any fire and the combustion process in our vehicles would stop. Every mode of transport except electric would fail instantly. Planes flying high in the sky would fall on earth and millions of cars running on petrol and diesel would stop on the roads.

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The apartment’s explosion, reportedly caused by a gas leak, produced a violent release of gas and heat. the heat increased the _
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<h2>Apartment Explosion Reported </h2>

The apartment’s explosion, reportedly caused by a gas leak, produced a violent release of gas and heat. The heat increased the temperature of the air in the room, which means an increase in the air's molecular kinetic energy.

When heat is provided then temperature increases and the molecules of substances move rapidly by increase of kinetic energy (K.E) temperature increases. It is understood that heat increases temperature.

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4 years ago
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Crest : trough :: compression : _____ A. frequency B. amplitude C. rarefaction D. wavelength
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Rarefraction.

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3 0
3 years ago
The question ask:
abruzzese [7]

Answer:

\theta=28.07^{\circ}

Explanation:

Speed of van, v = 28 m/s

Radius of unbanked curve, r = 150 m

Let \theta is the angle with the vertical. In case of banking of road,

T\ cos\theta=mg.............(1)

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T\ sin\theta=\dfrac{mv^2}{r}..........(2)

From equation (1) and (2) :

tan\theta=\dfrac{v^2}{rg}

tan\theta=\dfrac{(28)^2}{150\times 9.8}

\theta=28.07^{\circ}

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4 0
3 years ago
5. A stream of air at 12 bar and 900 K is mixed with another stream of air at 2 bar and 400 K with 2 times the mass flowrate. If
Softa [21]

Answer:

The anserrs to the question are

(a)  The temperature would be 566.67  K and

(b) The pressure of the resulting air stream is 14 bar

Explanation:

Here the two streams of gases meet ad form a single stream

The steady-flow energy equation can be implemented at the mixing point of the to streams as follows

mₐhₐ₁ + mₙ hₙ₁ + Q° + W° = mₐhₐ₂ + mₙhₙ₂

Where the flow is adiabatic, we have

Q = 0 and  W = 0  hence

mₐhₐ₁ + mₙ hₙ₁ = mₐhₐ₂ + mₙhₙ₂ where h = cp×T we have

mₐcpₐ×Tₐ + mₙcpₙ×Tₙ  = mₐcpₐ×T + mₐcpₙ×T

Therefore the final temperature T is given by

T = \frac{m_ac_{pa}T_a + m_nc_{pn}T_n}{m_ac_{pa} + m_nc_{pn}}  for the same kind of gas cpₐ =cpₙ therefore

T = \frac{m_aT_a + m_nT_n}{m_a + m_n} where mₐ and mₙ are mass flow rate

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T = \frac{m_a*900 + 2*m_a*400}{m_a + 2*m_a} = T = \frac{900 + 800}{1 + 2} = 1700/3 = 566.67 K  

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= 12 bar + 2 bar = 14 bar

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