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3 years ago

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A gas undergoes two processes. In the first, the volume remains constant at 0.200 m3m3 and the pressure increases from 2.50×105

PaPa to 5.50×105 PaPa . The second process is a compression to a volume of 0.110 m3m3 at a constant pressure of 5.50×105 PaPa.

1 answer:

SVETLANKA909090 [29]3 years ago

5 0

Answer:

$W = -4.95 \times 10^4\ J$

Explanation:

given,

In first case Volume remains constant.

Work done in the first case is zero.

In Second case Volume change

V₁ = 0.2 m³

V₂ = 0.11 m³

Pressure, P = 5.5 x 10⁵ Pa

Work done = Pressure x change in volume

W = P ΔV

$W = P(V_2-V_1)$

$W = 5.5\times10^5\times (0.11 - 0.2)$

$W = -4.95 \times 10^4\ J$

Hence, Work done when volume changes is equal to $W = -4.95 \times 10^4\ J$

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what happens when light travels from one medium to another with a different index of refraction at a 0 degree angle of incidence

tankabanditka [31]

Answer: light will travel in a straight line undeviated. Thus, there will be no refraction.

Explanation:

If the light hits the interface at angle equal to zero, then light movement is perpendicular to the plane of the second medium. And it will pass through undeviated. Therefore, the angle of refraction will also equal to zero. That is, there is no refraction.

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4 years ago

To measure moderately low pressures, oil with a density of 9.0 x 102 kg/m3 is used in place of mercury in a barometer. A change

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Δ P = 13.24 Pa

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Given that

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Δ P = ρ₁ g Δh

Now by putting the values

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Therefor the change in the pressure will be 13.24 Pa.

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4 years ago

A guitar string is 90 cm long and has a mass of 3.5g . The distance from the bridge to the support post is L=62cm, and the strin

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Answer:

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Explanation:

From the question we are told that

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The mass of the string is $m_s = 3.5 \ g =0.0035 \ kg$

The distance from the bridge to the support post $L = 62 \ c m = 0.62 \ m$

The tension is $T = 540 \ N$

Generally the frequency is mathematically represented as

$v = \frac{n}{2 * L } [\sqrt{ \frac{T}{\mu} } ]$

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i.e n = 1 is for fundamental frequency

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Also $\mu$ is the linear density of the string which is mathematically represented as

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=> $\mu = \frac{0.0035 }{ 0.9 }$

=> $\mu = 0.003889 \ kg/m$

So for n = 1

$v_1 = \frac{1}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ]$

$v_1 = 301 \ Hz$

So for n = 2

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$v_2 = 601 \ Hz$

So for n = 3

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Hence ,The size of the disc will Increase

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brainly.com/question/15314981

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