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olganol [36]
3 years ago
15

Is the state of the air in an isolated room completely specified by the temperature and the pressure? Explain

Physics
1 answer:
anyanavicka [17]3 years ago
4 0
I think these two variables are sufficient to completely specify the state.

In an isolated room with air only ,the volume is fixed.Mass ,density and its specific volume can be easily known.

Other thermodynamic properties like entropy, enthalpy etc are also fixed at a given temperature & pressure.
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Four +2 μC charges are placed at the positions (10 cm, 0 cm), (−10 cm, 0 cm), (0 cm, 10 cm), and (0 cm, −10 cm) such that they f
mylen [45]

Answer:

The force will be zero

Explanation:

Due to the symmetric location of the +2μC charges the forces the excert over the +5μC charge will cancel each other resulting in a net force with a magnitude of zero.However in this case it would be an unstable equilibrium, very vulnerable to a kind of bucking. If the central charge is not perfectly centered on the vertical axis the forces will have components in that axis that will add together instead of canceling each other.

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2 years ago
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2 years ago
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asambeis [7]

Answer:

c

Explanation:

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What are Heredity and Punnett Squares?
uysha [10]

Answer: A Punnett square can be used to predict genotype and phenotypes of offspring from genetic crosses. ... In the P generation, one parent has a dominant yellow phenotype and the genotype YY, and the other parent has the recessive green phenotype and the genotype yy.

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8 0
3 years ago
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An airplane traveling at one third the speed of sound (i.e., 114 m/s) emits a sound of frequency 3.72 kHz. At what frequency doe
nlexa [21]

Answer:

f'=5.58kHz

Explanation:

This is an example of the Doppler effect, the formula is:

f'=\frac{(v+v_o)}{(v+v_s)}f

Where f is the actual frequency, f' is the observed frequency, v is the velocity of the sound waves, v_o the velocity of the observer (which is negative if the observer is moving away from the source)  and v_s the velocity of the source  (which is negative if is moving towards the observer). For this problem:

f=3.72kHz\\v=342m/s\\v_o=0m/s\\v_s=-114m/s

f'=\frac{(342+0)}{(342-114)}3.72\times10^3\\f'=\frac{342}{228}3.72\times10^3\\f'=(1.5)3.72\times10^3\\f'=5580Hz=5.58kHz

5 0
2 years ago
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