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choli [55]
3 years ago
6

10. Mario is making chocolate. In part of the chocolate-making process, he changed the phase from a liquid to a solid. What happ

ened to the molecules of chocolate when it changed phase?
a
Before changing phase, the molecules were moving around each other. After changing phase, they were moving away from each other.

b
Before changing phase, the molecules were moving around each other. After changing phase, they were moving in place.

c
Before the changing phase, the molecules were moving away from each other. After changing the phase, they were moving around each other.

d
Before changing phase, the molecules were moving in place. After changing phase, they were moving around each other.
Physics
1 answer:
Naily [24]3 years ago
6 0

Answer:

B

Explanation:

in a liquid the particles are widespread and move around each other but in a solid they move in place and are tightly packed

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A rope of length L has circular cross-sectional area A and density rho = m/V , where m is the mass of the rope and V = A · L is
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Answer: µ = ρ¹ * A¹

Where x=1 and y=1

Explanation: According to the question, the mass per unit length (µ) is related to the density (ρ) and area A are related by the formulae below

µ = ρ * A

The dimension for each of these quantities is given below

Since µ is mass per unit length, unit is Kg/m and the dimension is ML^-1

ρ is density with unit kg/m³ and the dimension is ML^3

A is area with unit m², thus the dimension is M^2

Note that using dimensional analysis means we will be using the 3 fundamental quantities (mass, length and time) in our analysis.

Their dimensions below

Mass = M

Length = L

Time = T

Since the mass per unit length is related to density and area, we have a mathematical equation to provide a solution as shown below

µ = ρ^x * A^y.

By getting the power of x and y we will be able to get the formula that relates the quantities.

This is done by slotting in the dimensions of the respective quantities.

ML^-1 = (ML^-3)^x * (L²) ^y

By using law of indices on the right hand side of the equation, we have that

ML^-1 = (M^x * L^-3x) * (L^2y)

Also applying law of indices on the right hand side, we have that

ML^-1 = (M^x) * (L^-3x +2y)

The next step is to relate equal variables on both sides

For the M variable

M¹ = M^x which results to

x = 1

For the L variable

L^-1 = L^-3x+2y which results to

-1 = - 3x +2y

But x = 1

We have that

-1 = - 3(1) +2y

-1 = - 3 + 2y

-1 +3= 2y

2 = 2y

y = 1

Thus x=1 and y=1 and the formulae that relates the quantities is

µ = ρ¹ * A¹

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While accelerating at 5.22 m/s/s an object changes its velocity from 6.73 m/s to 29.88 m/s. Over what
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Answer:

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