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Zepler [3.9K]
3 years ago
6

What happens to the particles of the liquid inside a thermometer when the thermometer is heated?

Physics
2 answers:
Ahat [919]3 years ago
6 0

Speed up and spread apart


aev [14]3 years ago
5 0
The particles of the liquid inside a thermometer speed up and spread apart when the thermometer is heated. In short, the particles expand from one another when they're heated, and become condensed and compact when chilled.

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a 2.0-mole sample of an ideal gas is gently heated at constant temperature 330 k. it expands from initial volume 19 l to final v
shutvik [7]
Isothermal Work =  PVln(v₂/v₁)

PV = nRT =  2 mole * 8.314 J/ (k.mol) * 330 k = 5487.24 J

Isothermal Work =  PVln(v₂/v₁)            v₂ = ? v₁ = 19L, 

1.7 kJ = (5487.24)In(v₂/19)

1700 = (5487.24)In(v₂/19)

In(v₂/19) = (1700/5487.24) = 0.3098

In(v₂/19) = 0.3098

(v₂/19) = e^{0.3098}


v₂  =  19* e^{0.3098}

v₂ = 25.8999

v₂ ≈ 26 L        Option b.
6 0
3 years ago
A light wave passes through an aperture (that is, a narrow slit). When it does so, the degree to which the wave spreads out will
crimeas [40]

Explanation:

Single slit diffraction

Diffraction is the phenomenon of spreading out of waves as they pass through an aperture or around objects. Diffraction occurs when the size of the aperture or obstacle is of the same order of magnitude as the wavelength of the incident wave. For very small aperture sizes, the vast majority of the wave is blocked. in case of  large apertures the wave passes by or through the obstacle without any significant diffraction.

7 0
3 years ago
In a compound, the atom does or does not take on a new set of properties
disa [49]
It does take on new set of proerties
7 0
2 years ago
A round or canon uses _______ imitation.
bazaltina [42]
The answer is C. strict
5 0
3 years ago
In a physics lab, Asha is given a 10.7 kg uniform rectangular plate with edge lengths 67.3 cm by 53.5 cm . Her lab instructor re
Leya [2.2K]

Answer:

I=2.6363\ kg.m^2

Explanation:

Given:

dimension of uniform plate, (0.673\times 0.535)\ m^2

mass of plate, m=10.7\ kg

Now we find the moment of inertia about the center of mass of the rectangular plate is given as:

I_{cm}=\frac{1}{12} \times m(L^2+B^2)

where:

L= length of the plate

B= breadth of the plate

I_{cm}=\frac{1}{12} \times 10.7\times(0.673^2+0.535^2)

I_{cm}=0.6591\ kg.m^2

We know that the center of mass of the rectangular plane is at its geometric center which is parallel to the desired axis XX' .

Now we find the distance between the center of mass and the corner:

s=\frac{\sqrt{ (0.673^2+0.535^2)}}{2}

s=0.4299\ m

Now using parallel axis theorem:

I=I_{cm}+m.s^2

I=0.6591+10.7\times 0.4299^2

I=2.6363\ kg.m^2

6 0
3 years ago
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