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schepotkina [342]
3 years ago
7

What keeps the particles of a liquid from spreading out to fill an entire container in the same way that gas particles do?

Physics
2 answers:
Sonbull [250]3 years ago
7 0

b. the forces of attraction among them limit their motion.


frutty [35]3 years ago
6 0

Answer:

b. the forces of attraction among them limit their motion.

Explanation:

In solids, the particles are tightly packed. The solid has fixed shape and volume.There is strong inter-particle force of attraction which prevents the motion  of particles in solids. Thus, the particles in solids can vibrate about their fixed position.  Liquids have fixed volume but not shape. Gases don't  have fixed shape and volume. The particles of liquid do not spread out like gases. This is because of the inter particle attraction which prevents them from moving far away. In gases, the inter-particle distance is large and thus, particles have weak force of attraction.

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Which equation represents mass-energy equivalence? E = m2c E = mc2 E = (mc)2 E = mc
motikmotik

Einstein's energy mass equivalence relation say that if the whole given mass is converted to energy then it would be

E = mc^2

where

m = mass in kg

c = speed of light in m/s

this is the origination of quantum physics and by this formula we can relate the dual nature of light and particle

So correct relation above will be

E = mc^2

4 0
3 years ago
Read 2 more answers
The distance between two charges a and b is r, and the force between them is F. What is the force between them if the distancbet
Nadusha1986 [10]
See coulomb's law. Force is inversely proportional to the distance squared. So if you multiply r by 2, the force is multiplied by (½)² = ¼.

a. F/4
7 0
3 years ago
A gas in a piston–cylinder assembly undergoes a compression process for which the relation between pressure and volume is given
hichkok12 [17]

Answer:

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Explanation:

3 0
3 years ago
A ball is thrown upwards at an unknown speed. in a time of
alexandr402 [8]

Answer:

a)  Initial speed of the ball = 14.45 m/s

b) At height 6 m speed of ball = 9.55 m/s

c) Maximum height reached = 10.65 m

Explanation:

a)  We have equation of motion s=ut+\frac{1}{2} at^2, where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8m/s^2

 Substituting

    6=u*0.5-\frac{1}{2} *9.8*0.5^2\\ \\ u=14.45m/s

 Initial speed of the ball = 14.45 m/s

 b) We have equation of motion v^2=u^2+2as, where v is the final velocity

   s = 6 m, u = 14.45 m/s, a = -9.8m/s^2

    Substituting

        v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s

  So at height 6 m speed of ball = 9.55 m/s

c) We have equation of motion v^2=u^2+2as, where v is the final velocity

   u = 14.45 m/s, v =0 , a = -9.8m/s^2

   Substituting

     0^2=14.45^2-2*9.8*s\\ \\ s=10.65 m

  Maximum height reached = 10.65 m

8 0
3 years ago
We start with 5.00 moles of an ideal monatomic gas with an initial temperature of 128 ∘C. The gas expands and, in the process, a
o-na [289]

Answer:

The final temperature of the gas is <em>114.53°C</em>.

Explanation:

Firstly, we calculate the change in internal energy, ΔU from the first law of thermodynamics:

ΔU=Q - W

ΔU = 1180 J - 2020 J = -840 J

Secondly, from the ideal gas law, we calculate the final temperature of the gas, using the change in internal energy:

ΔU=\frac{3}{2} nRΔT

ΔU=\frac{3}{2} nR(T_{2} -T_{1} )

Then we make the final temperature, T₂, subject of the formula:

T_{2} =\frac{2ΔU}{3nR} +T_{1}

T_{2} =\frac{2(-840J)}{(3)(5)(8.314J/mol.K)} +128 deg.C

T_{2} =114.53 deg.C

Therefore the final temperature of the gas, T₂, is 114.53°C.

7 0
3 years ago
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