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Phantasy [73]
3 years ago
11

How is the pressure of a gas related to its concentration of particles?

Physics
2 answers:
AURORKA [14]3 years ago
6 0

Answer:

Explanation:

Pressure will expand a gas, enlarging its volume and reducing its density and concentration of particles. Pressure will magnify a gas, developing its volume and multiplying its density and concentration of particles.

Increasing the pressure of a gas is exactly the same as increasing its concentration. If you have a given mass of gas, the way you increase its pressure is to squeeze it into a smaller volume. If you have the same mass in a smaller volume, then its concentration is higher.

Nat2105 [25]3 years ago
3 0
Increasing the pressure of gas is like exactly the same as increasing its concentration. If you have a given mass of gas, the way you increase its pressure is to squeeze it into a smaller volume.
Hope this helps!
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There are 5510 lines per centimeter in a grating that is used with light whose wavelegth is 467 nm. A flat observation screen is
Mademuasel [1]

Answer:

1.696 nm

Explanation:

For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,

dsinθ = mλ = (1)λ = λ

dsinθ = λ

sinθ = λ/d.

Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m

From trig ratios 1 + cot²θ = cosec²θ

1 + (1/tan²θ) = 1/(sin²θ)

substituting the values of sinθ and tanθ we have

1 + (D/w)² = (d/λ)²

(D/w)² = (d/λ)² - 1

(w/D)² = 1/[(d/λ)² - 1]

(w/D) = 1/√[(d/λ)² - 1]

w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹  = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.

w is also the distance from the center to the other principal maximum on the other side.

So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm

So, the minimum width of the screen must be 1.696 nm

4 0
3 years ago
A 1100 kgkg safe is 2.4 mm above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compres
Drupady [299]

Answer:

191.36 N/m

Explanation:

From the question,

The Potential Energy of the safe = Energy of the spring when it was compressed.

mgh = 1/2ke²............... Equation 1

Where m = mass of the safe, g = acceleration due to gravity, h = height of the save above the heavy duty spring , k = spring constant, e = compression

Making k the subject of the equation,

k =2mgh/e²................ Equation 2

Given: m = 1100 kg, h = 2.4 mm = 0.0024 m, e = 0.52 m

Constant: g = 9.8 m/s²

Substitute into equation 2

k = 2(1100)(9.8)(0.0024)/0.52²

k = 51.744/0.2704

k = 191.36 N/m

Hence the spring constant of the heavy-duty spring = 191.36 N/m

3 0
3 years ago
What is the frequency of a wave whose wavelength is 0.7m and whose velocity is 120m/s?
ANEK [815]

Answer:

f=171.43Hz

Explanation:

Wave frequency is the number of waves that pass a fixed point in a given amount of time.

The frequency formula is: f=v÷λ, where <em>v</em> is the velocity and <em>λ</em> is the wavelength.

Then replacing with the data of the problem,

f=\frac{120\frac{m}{s} }{0.7m}

f=171.43\frac{1}{s}

f=171.43 Hz (because \frac{1}{s} = Hz, 1 hertz equals 1 wave passing a fixed point in 1 second).

5 0
3 years ago
Question 4 of 10
ch4aika [34]

Habitat fragmentation is a cost of urban development.

Option: A

Explanation:

Though from the view point or perspective of up gradation and development urban development is much needed but in cost of habitat fragmentation which feels very bitter. As habitat fragmentation leads to the loss of habitat, disruption of ecological cycle and environmental equilibrium.

Actually in the name of urban development we the human use our bread giver environment in a wrong way which causes natural disasters in long run. Animals become endangered , vulnerable and extinct with passage of time. Because they forced to enter into human settlements.

4 0
3 years ago
A ball is tossed straight up from the surface of a small, spherical asteroid with no atmosphere. The ball rises to a height equa
alukav5142 [94]

Answer:

d. equal to one-fourth the acceleration at the surface of the asteroid.

Explanation:

The explanation is attached as a picture with this answer

Newton's law of universal gravitation is being used to compare the accelerations at the surface and at the top of the ball's path.

as it can be seen in the explanation that the proportional form of the equation is used because we do not need to necessarily use to final form with "G" for comparison calculations.

As per the given scenario only difference between the two points in the gravitational field is the distance from center of the spherical asteroid, i.e. r.

It is  taken 2r for the top is the path. hence we obtain (1/4)g as our answer.

4 0
3 years ago
Read 2 more answers
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