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Alinara [238K]
3 years ago
14

A rigid tank whose volume is unknown is divided into two parts by a partition. One side of the tank contains an ideal gas at 927

°C. The other side is evacuated and has a volume twice the size of the part containing the gas. The partition is now removed and the gas expands to fill the entire tank. Heat is now transferred to the gas until the pressure equals the initial pressure. Determine the final temperature of the gas.
Physics
1 answer:
Marianna [84]3 years ago
4 0

Answer:

The final temperature will be 3600 K.

Explanation:

Side 1

T=927°C

T=1200 K

Lets take volume of one side is V then the volume of other side will be 2V. The final volume of system will become 3 V.

The system is closed so the mass of system will remain constant .

We know that for ideal gas  P V = m R T

The final pressure is same as initial pressure so we can say that

\dfrac{T_i}{V_i}=\dfrac{T_f}{V_f}

\dfrac{1200}{V}=\dfrac{T_f}{2V+V}

T_f=3600\ K

So the final temperature will be 3600 K.

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The bob (weight) at the end of a pendulum has a mass of 0.3 kilograms. The bob is pulled to position B and allowed to swing. It
Ivahew [28]

Answer:

0.147 J

Explanation:

The total energy that has been transformed into thermal energy is equal to the loss of gravitational potential energy between the initial situation (bob at h=0.5 m above the ground) and the final situation (bob back but at h=0.45 m above the ground).

Therefore, we have

E_{thermal}=\Delta U=mgh_1 - mgh_2 = mg(h_1 -h_2)

where

m = 0.3 kg is the mass of the bob

g = 9.8 m/s^2

h1 = 0.5 m is the initial height

h2 = 0.45 m is the final height

Substituting, we find the thermal energy

E_{thermal}=(0.3 kg)(9.8 m/s^2)(0.5 m-0.45 m)=0.147 J

Therefore, the energy transformed into thermal energy is 0.147 J.

3 0
3 years ago
The two uniform, slender rods B1and B2, each of mass 2kg, are pinned together at P, and then B1is suspended from a pin at O. (Th
Bezzdna [24]

Answer:

hello the diagram relating to this question is attached below

a) angular accelerations : B1 = 180 rad/sec,  B2 = 1080 rad/sec

b) Force exerted on B2 at P = 39.2 N

Explanation:

Given data:

Co = 150 N-m ,

<u>a) Determine the angular accelerations of B1 and B2 when couple is applied</u>

at point P ; Co = I* ∝B2'

                150  = ( (2*0.5^2) / 3 ) * ∝B2

∴ ∝B2' = 900 rad/sec

hence angular acceleration of B2 = ∝B2' + ∝B1 = 900 + 180 = 1080 rad/sec

at point 0 ; Co = Inet * ∝B1

                  150 = [ (2*0.5^2) / 3  + (2*0.5^2) / 3  + (2*0.5^2) ] * ∝B1

∴ ∝B1 = 180 rad/sec

hence angular acceleration of B1 =  180 rad/sec

<u>b) Determine the force exerted on B2 at P</u>

T2 = mB1g + T1  -------- ( 1 )

where ; T1 = mB2g  ( at point p )

                 = 2 * 9.81 = 19.6 N

back to equation 1

T2 = (2 * 9.8 ) + 19.6 = 39.2 N

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A grating has 470 lines/mm. how many orders of the visible wavelength 538 nm can it produce in addition to the m = 0 order?
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Three complete orders on each side of the m=0 order can be produced in addition to the m=0 order.

The ruling separation is d=1/(470mm-1)

= 2.1 \times 10 {}^{ - 3}mm

Diffraction lines occurs at an angle θ such that dsin=mλ,when λ is the wavelength and m is an integer.

Notice that for a given order,the line associated with a long wavelength is produced at a greater angle than the line associated with shorter wavelength.

we take λ to be the longest wavelength in the visible spectrum (538nm) and find the greatest integer value of m such that θ is less than 90°.

That is,find the greater integer value of m for which mλ<d.

since,d/λ

= 538 \times 10 {}^{ - 9} m/2.1 \times 10 {}^{ - 6}

There are three complete orders on each side of the m=0 order.

The second and third orders overlap.

learn more about diffraction from here: brainly.com/question/28168352

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