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Doss [256]
3 years ago
5

Explain why the pressure exerted by a gas does not depend on the type of the gas.

Physics
2 answers:
Phantasy [73]3 years ago
5 0

Explanation: the pressure exerted by a gas does not depend on the type of the gas because pressure is a physical force its has not chemical composition.

So, pressure depends upon the numbers of molecules of the gas because gas molecules collide with the wall of the vessel. They exert force on the wall and we know the pressure is force applied per unit area .

According to Gay - Lussac's law,  the temperature of the gas is direct proportional to the pressure  at a constant volume.

charle [14.2K]3 years ago
4 0
The pressure exerted by a gas depends on the number of gas particles present, according to the Kinetic Molecular Theory of gases. Other factors like volume of the particles or the intermolecular forces between molecules are assumed zero. So 50 particles of Helium gas exerts the same pressure of 50 particles of Oxygen gas.
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1. A plane travels a distance of 500
Yanka [14]

Answer:

Option C, 139 m/s

Explanation:

s = d/t

s = 500*10^3/3600

s = 138.8 = 139 m/s

5 0
3 years ago
A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant inten
Lyrx [107]

Answer:

a) the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

b) the angle of twist between the ends of the bar is 16tL² / πGd⁴  

Explanation:

Given the data in the question, as illustrated in the image below;

d is the diameter of the prismatic bar of length AB

t is the intensity of distributed torque

(a) Determine the maximum shear stress tmax in the bar

Maximum Applied torque  T_max = tL

we know that;

shear stress τ = 16T/πd³

where d is the diameter

so

τ_{max} = 16T_{max} /πd³

Therefore, the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

(b) Determine the angle of twist between the ends of the bar.

let theta (\theta) be the angle of twist

polar moment of inertia I_p} = πd⁴/32

now from the second image;

lets length dx which is at distance of "x" from "B"

Torque distance x

T(x) = tx

Elemental angle twist = d\theta = T(x)dx / GI_{p}

so

d\theta = tx.dx / G(πd⁴/32)

d\theta = 32tx.dx / πGd⁴

so total angle of twist \theta will be;

\theta =  \int\limits^L_0  \, d\theta

\theta =  \int\limits^L_0  \, 32tx.dx / πGd⁴

\theta = 32t / πGd⁴  \int\limits^L_0  \, xdx

\theta = 32t / πGd⁴ [ L²/2]

\theta = 16tL² / πGd⁴  

Therefore,  the angle of twist between the ends of the bar is 16tL² / πGd⁴  

7 0
3 years ago
Two transverse waves travel along the same taut string. Wave 1 is described by y1(x, t) = A sin(kx - ωt), while wave 2 is descri
Vadim26 [7]

Answer:

6) Wave 1 travels in the positive x-direction, while wave 2 travels in the negative x-direction.

Explanation:

What matters is the part kx \pm \omega t, the other parts of the equation don't affect time and space variations. We know that when the sign is - the wave propagates to the positive direction while when the sign is + the wave propagates to the negative direction, but <em>here is an explanation</em> of this:

For both cases, + and -, after a certain time \delta t (\delta t >0), the displacement <em>y</em> of the wave will be determined by the kx\pm\omega (t+\delta t) term. For simplicity, if we imagine we are looking at the origin (x=0), this will be simply \pm \omega (t+\delta t).

To know which side, right or left of the origin, would go through the origin after a time \delta t (and thus know the direction of propagation) we have to see how we can achieve that same displacement <em>y</em> not by a time variation but by a space variation \delta x (we would be looking where in space is what we would have in the future in time). The term would be then k(x+\delta x)\pm\omega t, which at the origin is k \delta x \pm \omega t. This would mean that, when the original equation has kx+\omega t, we must have that \delta x>0 for k\delta x+\omega t to be equal to kx+\omega\delta t, and when the original equation has kx-\omega t, we must have that \delta x for k\delta x-\omega t to be equal to kx-\omega \delta t

<em>Note that their values don't matter, although they are a very small variation (we have to be careful since all this is inside a sin function), what matters is if they are positive or negative and as such what is possible or not .</em>

<em />

In conclusion, when kx+\omega t, the part of the wave on the positive side (\delta x>0) is the one that will go through the origin, so the wave is going in the negative direction, and viceversa.

4 0
3 years ago
23) A high school website directs parents to not allow student access to blue laser pointers or allow the students to bring them
vekshin1
Mainly because of the higher energy of blue light than red light.

In fact, light is made of photons, each one carrying an energy equal to
E=hf
where h is the Planck constant while f is the frequency of the light.
The frequency of red light is approximately 450 THz, while the frequency of blue light is about 650 Hz. Higher frequency means higher energy, so blue light is more energetic than red light and therefore it can cause more damages than red light.
6 0
3 years ago
When drawing ray diagrams involving thin lenses, how many rays (at a minimum) are needed show the image distance and magnificati
Rufina [12.5K]

Answer:

GGG he fggfggfufbvg I Rd cbh

6 0
3 years ago
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