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bulgar [2K]
2 years ago
7

Relate the changes in molecular motion of particles as they change from solid to liquid to gas.

Physics
1 answer:
exis [7]2 years ago
8 0

Answer:

As the phase changes occur, the freedom of motion of the particles increases

Explanation:

In thermodynamics the state of matter: solid, liquid and gaseous.

We can characterize them, the solid state maintains a certain shape, which is why particles are tightly bound and can hardly move from their equilibrium positions.

The liquid state does not hold its shape, so the particles have more energy and can move a lot from their equilibrium position.

The gaseous state does not maintain the shape or the volume, so the interaction between the particles is very small, so they can move almost freely.

When examining the answer, the correct one is: As the phase changes occur, the freedom of motion of the particles increases.

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Which has more heat, a home oven at 500 degrees Celsius, or a one ton commercial oven at 500 degrees celsius?
alisha [4.7K]
A commercial oven because it is more durable plus it’s more professional so they use more heat then we do at home
8 0
2 years ago
Can someone please help me?? i’ll give brainilist
Ksivusya [100]
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7 0
2 years ago
A total charge of 4.70 is distributed on two metal spheres. When the spheres are 10 cm apart, they each feel a repulsive force o
Anna35 [415]

Answer:0.114 C

Explanation:

Given

Total 4.7 C is distributed in two spheres

Let q_1 and q_2 be the charges such that

q_1+q_2=4.7

and Force between charge particles is given by

F=\frac{kq_1q_2}{r^2}

4.7\times 10^11=\frac{9\times 10^9\times q_1\cdot q_2}{0.1^2}

q_1\cdot q_2=0.522

put the value of q_1

q_2\left ( 4.7-q_2\right )=0.522

q_2^2-4.7q_2+0.522=0

q_2=\frac{4.7\pm \sqrt{4.7^2-4\times 1\times 0.522}}{2}

q_2=0.114 C

thus q_1=4.586 C

3 0
3 years ago
A 23.5 g piece of aluminum metal is initially at 100.0°C. It is dropped into a coffee cup-calorimeter containing 130.0 g of wate
vivado [14]

Answer: The molar heat capacity of aluminum is 25.3J/mol^0C

Explanation:

heat_{absorbed}=heat_{released}

As we know that,  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]         .................(1)

where,

q = heat absorbed or released

m_1 = mass of water = 130.0 g

m_2 = mass of aluminiunm = 23.5 g

T_{final} = final temperature = 26.0^oC=(273+26)K=299K

T_1 = temperature of water = 23^oC=(273+23)K=296K

T_2 = temperature of aluminium = 100^oC=273+100=373K

c_1 = specific heat of water= 4.184J/g^0C

c_2 = specific heat of aluminium= ?

Now put all the given values in equation (1), we get

130.0\times 4.184\times (299-296)=-[23.5\times c_2\times (299-373)]

c_2=0.938J/g^0C

Molar mass of Aluminium = 27 g/mol

Thus molar heat capacity =0.938J/g^0C\times 27g/mol=25.3J/mol^0C

5 0
3 years ago
A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s
aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is U_{s} = 0

And, initial potential gravitational potential energy of the rod is U_{g} = \frac{mgL}{2}.

It is given that,

       mass of the bar = 0.795 kg

            g = 9.8 m/s^{2}

           L = length of the rod = 0.2 m

Initial total energy T = \frac{mgL}{2}

Now, when the rod is in horizontal position then final total energy will be as follows.

            T = \frac{1}{2}kx^{2} + I \omega^{2}

where,    I = moment of inertia of the rod about the end = \frac{mL^{2}}{3}

Also,    \omega = \frac{\nu}{L}

where,    \nu = speed of the tip of the rod

              x = spring extension

The initial unstrained length is x_{o} = 0.1 m

Therefore, final length will be calculated as follows.

              x' = \sqrt{(0.2)^{2} + (0.1)^{2}} m

Then,  x = x' - x_{o}

          x = \sqrt{(0.2)^{2} + (0.1)^{2}} m - 0.1 m

             = 0.1236 m

       k = 25 N/m

So, according to the law of conservation of energy

       \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}

      \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

Putting the given values into the above formula as follows.

   \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

  \frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}

          v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0
3 years ago
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