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2 years ago

Explain why the temperature is not changing at X

Physics

1 answer:

Scorpion4ik [409]2 years ago

4 0

Answer:

The temperature is constant at 'X'. No increase or decrease. Would be same answer if there were a 'X' at temperature 'D'.

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A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oC. The air undergoes a process to a state where th

vlada-n [284]

Answer:

Explanation:

The process is isothermic, as P V = constant .

work done = 2.303 n RT log P₁ / P₂

= 2.303 x 5 / 29 x 8.3 x 303 log 2 / 1 kJ

= 300.5k J

This energy in work done by the gas will come fro heat supplied as internal energy is constant due to constant temperature.

heat supplied = 300.5k J

specific volume is volume per unit mass

v / m

pv = n RT

pv = m / M RT

v / m = RT / p M

specific volume = RT / p M

option B is correct.

5 0

3 years ago

Review. (c) Assume the dipole is a compass needle-a light bar magnet-with a magnetic moment of magnitude μ . It has moment of in

hammer [34]

The frequency of oscillation is $\frac{1}{2\pi } \sqrt{\frac{\mu B}{I} }$ .

<h3>What is a magnetic moment?</h3>

The magnetic moment is the magnetism of a magnet or other item that creates a magnetic field, as well as its orientation and strength. Electromagnets, permanent magnets, elementary particles like electrons, different compounds, and a variety of celestial objects are examples of things that have magnetic moments (such as many planets, some moons, stars, etc). The phrase "magnetic moment" typically refers to a system's magnetic dipole moment, which can be represented by an analogous magnetic dipole, which has a magnetic north and south pole that are barely separated from one another. For sufficiently small magnets or at sufficiently wide distances, the magnetic dipole component is adequate. For extended objects, additional terms may be required in addition to the dipole moment, such as the magnetic quadrupole moment.

$T = \frac{ Id^{2}\theta }{dt^{2} }$

$-\mu B\theta=\frac{ Id^{2}\theta }{dt^{2} }$

$\frac{d^{2}\theta }{dt^{2} } = -(\frac{\mu BI}{\theta} )$

By Comparing the above equation with the SHM equation

$\frac{d^2 \theta} {dt^{2} } = -\omega^{2} \theta$

$\omega^{2} =\frac{ \mu B}{I}$

Frequency = $\frac{\mu}{2\pi }$

= $\frac{\sqrt{\frac{\mu B}{I} } }{2\pi}$

= $\frac{1}{2\pi } \sqrt{\frac{\mu B}{I} }$

To learn more about a magnetic moment, visit:

brainly.com/question/17000031

#SPJ4

8 0

1 year ago

A neon lamp produces what kind of visible spectrum?

Shalnov [3]

Answer:

Discrete

Explanation:

There can be many kind of visible spectrum. Some of them are as follows :Infrared, Continuous , Blackbody and Discrete.

A neon lamp produces discrete kind of visible spectrum. In this type of spectrum, the spectrum consists of different colors. It means that the neon lamp is made up of different colors or individual frequencies.

Hence, a neon lamp produces discrete visible spectrum.

5 0

3 years ago

Help plz 100 points and guaranteed Brainliest

nadezda [96]

Answer:

Guy above is correct on everything!

Explanation:

I know this because I researched and found the answers, but the guy above has all the right answers, so yes, he is correct! Give him a "Thanks"!

4 0

4 years ago

Read 2 more answers