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Rainbow [258]
3 years ago
6

If a material were being designed to demonstrate no thermal expansions, how would the energy plot look like?

Physics
1 answer:
bixtya [17]3 years ago
5 0

The characteristics of thermal expansion allow finding that the response for a material without thermal expansion is

  • The length variation is zero
  • In the graph the line is horizontal so there is no change in length with temperature

Thermal expansion is the macroscopic sum of the changes in the length of the bonds when the energy (temperature) changes, it can be written

              ΔL = α L₀ ΔT

Where ΔL is the change in length, α the coefficient of linear expansion, L₀ the initial length and ΔT the change in body temperature

In this case, a material is being designed that the thermal expansion is very small, for this the material must be made up of several compounds where some of them present a contraction with temperature, some examples: water at low temperature, liquefied gases , ceramic tile, quartz, etc.

The thermal expansion measurement processes control the body temperature and measure the change in length, in this case the change in length must be zero, in the attachment we can see a graph of a composite material with these characteristics, an example of this type of material is Invar an alloy of nickel and iron α = 3.7 10⁻⁶ ºC⁻¹

In conclusion, using the characteristics of thermal expansion we can find that the response of material without thermal expansion is

  • The length variation is zero
  • In the graph the line is horizontal so there is no change in length with temperature

Learn more here:  brainly.com/question/18717902

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The rotational inertia of a thin rod about one end is 1/3 ML2. What is the rotational inertia of the same rod about a point loca
zlopas [31]

Answer:

The value is  I = 0.0932 ML ^2  

Explanation:

From the question we are told that

  The rotational inertia about one end is I_R =  \frac{1}{3} ML^2

   The location of the axis of rotation considered is d =  0.4 L

Generally the mass of the portion of the rod from the axis of rotation considered to the end of the rod is  0.4 M

Generally the length of the rod from the its beginning to the axis of rotation consider is

      k = 1 - 0.4 L = 0.6L

Generally the mass of the portion  of the rod from the its beginning to the axis of rotation consider is

    m  =  1- 0.4 M = 0.6 M

Generally the rotational inertia about the axis of rotation consider for the first portion of the rod is

     I_{R1} =  \frac{1}{3} (0.6 M )(0.6L)^2

    I_{R1} =  \frac{1}{3} (0.6 M )L^2 0.6^2

Generally the rotational inertia about the axis of rotation consider for the second  portion of the rod is

     I_{R2} =  \frac{1}{3} (0.6 M )(0.6L)^2

=> I_{R2} =  \frac{1}{3} (0.4 M )(0.4L)^2

=>  I_{R2} =  \frac{1}{3} (0.4 M )L^2 0.4^2

Generally by the principle of superposition that rotational inertia of the rod at the considered axis of rotation is

  I =   \frac{1}{3} (0.6 M )L^2 0.6^2 +   \frac{1}{3} (0.4 M )L^2 0.4^2

=>   I =  \frac{1}{3} ML ^2  [0.6 * (0.6)^2 + 0.4 * (0.4)^2 ]

=>   I = 0.0932 ML ^2  

8 0
3 years ago
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dimaraw [331]

Answer:

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Explanation:

I just went ahead and gave you the five signs of chemical change hoped it helped

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Answer:

c. because A will land first becuase its heavier

and D.

Explanation:

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3 years ago
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Answer:

First answer.

Explanation:

There may be a 5N force, but if the frictional force also equals 5N, than they cancel eachother out, resulting in the brick still staying still, as it is resting on a (perfectly) level surface, but any amount of force would make the brick move.

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