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ycow [4]
3 years ago
5

An ideal gas in a piston is placed in a refrigerator, which removes 413 J of energy from the gas via cooling. The plunger on the

piston is pushed down, which does 70 J of work on the gas.
1)

What is the net change in the internal energy of the gas in the piston?
Physics
1 answer:
maria [59]3 years ago
5 0

Answer:

The net change in the internal energy of the gas in the piston is -343J

Explanation:

Because heat and workdone are the only means of energy transfer between the system and the surrounding, change in internal energy is given by;

∆E = q + w

q = heat transfer

w = workdone

Because heat is lost by the system, the heat transfer is negative

q = -413J

Because work is done on the system, workdone is positive

w = +70J

∆E = -413J + 70J

∆E = -343J

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Due to the wave nature of light, light shined on a single slit will produce a diffraction pattern? Green light (520 nm) is shine
TiliK225 [7]

Answer:

Yes, it will produce a diffraction pattern.

a. 3.9 mm b. 1.95 mm

Explanation:

The light shined from a single slit will produce a diffraction pattern because,  the wavefront act as wavelets which generates its own wave according to Huygens principle. This therefore causes the diffraction pattern.

Given

wavelength of green light, λ = 520 nm = 520 × 10⁻⁹ m = 5.20 × 10⁻⁷ m

width of slit, d = 0.440 mm = 0.44 × 10⁻³ m = 4.4 × 10⁻⁴ m

Distance of slit from central maximum , D = 1.65 m

Distance of first minimum from central maximum, y = ?

a. The relationship between the slit width and wavelength is given by [tex} dsinθ = mλ [/tex]where d = slit width, θ = angular distance from central maximum, λ = wavelength of light and m = ±1, ±2, ±3...

The relationship between y and D is given by tanθ = y/D

Since θ is small, sinθ ≈ θ ≈ tanθ

so, dθ = mλ ⇒ θ = mλ/d = y/D

Therefore, y = mλD/d

Now, for the first minimum above the slit, m = +1 and for the first minimum below the slit, m = -1. So, y₁ =  λD/d and y₋₁ =  -λD/d. So, the width of the central maximum Δy is the difference between the first minima below and above the central maximum. So, Δy = y₁ - y₋₁ = λD/d -(-λD/d) = 2λD/d

Substituting the values from above, Δy= 2 × 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴ =  3900 × 10⁻⁶ m = 3.9 × 10⁻³ m = 3.9 mm

b. The first order fringe is the fringe located between the first minimum and the second minimum. From dsinθ = mλ and tanθ = y/D when θ is small, sinθ ≈ θ ≈ tanθ. So, y = mλD/d. Let m= 1 and m=2 be the first and second minima respectively. So,y₁ =  λD/d and y₂ =  2λD/d. The difference Δy₁ = y₂ - y₁ is the width of the first order fringe. Therefore, Δy₁ = 2λD/d - λD/d= λD/d. Substituting the values from above, we have

λD/d= 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴= 1.95 × 10⁻³ m = 1.95 mm

7 0
4 years ago
Acceleration of a free-falling object in a frictionless environment increases as a function of time.
Digiron [165]

A free-falling object is an object moving under the effect of gravitational forces alone

The correct option to select for the True or False question is False

The reason the above selected option is correct is as follows:

According to Newton's second law of motion, we have;

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Where;

G \cdot \dfrac{M }{r^{2}} = Acceleration \ due \ to \ gravity , \ g \approx 9.81 m/s^2

m  = The mass of the object

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Therefore the acceleration of an object in free fall is the constant acceleration due to gravity, and it therefore, does not change with time

The correct option for the question, acceleration of a free-falling object in a frictionless environment increases as a function of time is <u>False</u>

<u></u>

Learn more about object in free fall here:

brainly.com/question/13712424

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melamori03 [73]

Storing music digitally requires less storage room than analog records or tapes. Digital music is easier to copy and the copies are the same as the original. The quality of the signal does not degrade over long periods of time.

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