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stepladder [879]
3 years ago
10

Explain the changes in molecular motion of particles as they change from solid to liquid to gas. A) As the phase changes occur,

the freedom of motion of the particles increases. B) As the phase changes occur, the freedom of motion of the particles decreases. C) As the phase changes occur, the freedom of motion of the particles compresses. D) As the phase changes occur, the freedom of motion of the particles is constant.
Physics
1 answer:
Masja [62]3 years ago
4 0

Answer:

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

Explanation:

As the phase changes occur from solid to gas, the freedom of motion of the particles increases

At the solid state (let's say ice), the molecules of the solid are tightly packed and they are unable to move freely. When heat is applied to the state of the solid, ice it melts and the molecules of the solid gains kinetic energy and therefore tends to shift away from their mean fixed position due to heat. This kinetic energy possesses by this molecules allow them to move freely (energy due to motion). When further heat is applied to the liquid (above the boiling point), the molecules of the water vapourizes(turns to gas) and this molecules gain more kinetic energy more than they do in liquid state and were able to move freely more than in liquid. This shows that the freedom of the particles (molecules) increase with change in state from solid to gas.

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Zanzabum

Answer:

T_{1} = 14.88 N

Explanation:

Let's begin by listing out the given variables:

M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,

g = 9.8 m/s²

At equilibrium, the sum of all external torque acting on an object equals zero

τ(net) = 0

Taking moment about T_{1} we have:

(M + m) g * 0.5L - T_{2}(L - d) = 0

⇒ T_{2} = [(M + m) g * 0.5L] ÷ (L - d)

T_{2} = [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)

T_{2}= 59.535 ÷ 2.4

T_{2} = 24.80625 N ≈ 24.81 N

Weight of bar(W) = M * g = 2.7 * 9.8 = 26.46 N

Weight of monkey(w) = m * g = 1.35 * 9.8 = 13.23 N

Using sum of equilibrium in the vertical direction, we have:

T_{1} + T_{2} = W + w   ------- Eqn 1

Substituting T2, W & w into the Eqn 1

T_{1} + 24.81 = 26.46 + 13.23

T_{1} = <u>14.88</u> N

8 0
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A Pitot-static probe is used to measure the speed of an aircraft flying at 3000 m. If the differential pressure reading is 3300
adell [148]

Answer:

   v = 306.76 Km/h

Explanation:

given,

height of the aircraft = 3000 m

differential pressure reading = 3300 N/m²

density of air = 0.909 Kg/m³

speed of aircraft = ?

Assuming the air flowing above air craft is in-compressible, irrotational and steady so, we can use Bernoulli's equation to solve the problem.

using Bernoulli's equation

          \dfrac{v^2}{2} = \dfrac{\Delta P}{\rho}

where ρ is the density of the air at 3000 m

          v= \sqrt{\dfrac{2 \times \Delta P}{\rho}}

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          v = \sqrt{7260.726}

                 v = 85.21 m/s

          v= 85.21 \times \dfrac{3600}{1000}

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4 0
3 years ago
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Nikitich [7]

Answer:

I think the answer is ST. Option B

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Lady_Fox [76]

Answer:

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Explanation:

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Then, the kinetic energy is maximum when the velocity is the highest, and this is at the point where it is thrown, after this point, the velocity will be diminished and at the highest point will be equal to zero.

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steposvetlana [31]

Answer:

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