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notsponge [240]
2 years ago
7

Which of the following sets of properties leads to a high degree of thermal shock resistance? (A) High fracture strength High th

ermal conductivity High modulus of elasticity High coefficient of thermal expansion (B) Low fracture strength Low thermal conductivity Low modulus of elasticity Low coefficient of thermal expansion (C) High fracture strength High thermal conductivity Low modulus of elasticity Low coefficient of thermal expansion (D) Low fracture strength Low thermal conductivity High modulus of elasticity High coefficient of thermal expansion
Physics
1 answer:
Rina8888 [55]2 years ago
4 0

Answer:  

The correct Answer is C) <u>High fracture strength,</u> <u>High Thermal Conductivity,</u> <u>Low modulus of elasticity,</u> <u>Low coefficient of thermal</u>                                                                                                                                                                                                                                              

Explanation:

The ability of a solid to withstand sudden changes in temperature either during heating or cooling is known or referred to as Thermal Shock Resistance (TSR).

Thermal shock resistance is one of the most crucial factors of performance in solids for high temperature environments that can cause thermal stresses and risks for thermal shock damage.

Examples are as of such environments are energy conversion systems, electronic devices and cutting tools.

A common way to evaluate TSR is to look for maximum jump in surface temperature which a material can sustain without cracking.  This is known as thermal conductivity.

Failure due to thermal shock can be prevented by;

  • Reducing the thermal gradient seen by the object, by changing its temperature more slowly or increasing the material's thermal conductivity
  • Reducing the material's coefficient of thermal expansion
  • upping its strength
  • Introducing built-in compressive stress, as for example in tempered glass and in some cases tempered plastic
  • reducing its Young's modulus
  • increasing its toughness, by crack tip blunting (i.e., plasticity or phase transformation) or crack deflection

Thermal conductivity is an intensive physical property of a material that relates the heat flow through the material per unit area to temperature gradient across the material. The thermal conductivity of a material is basically a measure of its ability to conduct heat.    

The other factor that contributes to a high degree of thermal shock is:

Fracture Strength: This is the ability of a material containing a crack to resist fracture or resist becoming brittle. For example, glass has a high strength, but the presence of a small fracture reduces the strength. Therefore, glass has low fracture resistance. Fracture toughness is an important consideration in hydraulic fracture design.

Modulus of Elasticity:

An object or substance's resistance to being deformed elastically (i.e., non-permanently) when a stress is applied to it is calculated  or measured by a quantity known as Elastic Modulus (also known as Young modulus of elasticity)

A stiff material has a high Young's modulus and changes its shape only slightly under elastic loads (e.g. steel or diamond). A flexible material has a low Young's modulus and changes its shape considerably (e.g. rubbers).

Coefficient of thermal expansion (CTE) this refers to how the rate of change in the  size of an object with respect of every degree change in temperature assuming that pressure remains the same. An object with low CTE is Fine Ceramics or Advanced Ceramics.

Cheers!

               

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Answer:

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F = Gmm'/r² .............................. Equation 1

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Given: m = 70 kg, m' = 200 kg, r = 0.5 m

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F = (6.67×10⁻¹¹×70×200)/0.5²

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Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time (when they are not slee
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Answer:

I₁ = (7.78 i ^ - 6.71 j ^) 10⁻³ J s ,  I₂ = (-12.5 i ^ -14.6 j ^) 10⁻³ J s ,  I₃ = (19.1i ^ + 18.6 j ^) 10⁻³ J s  and I₄ = (-9.14i ^ + 7.24 j ^) 10⁻³ J s

Explanation:

The impulse is equal to the variation of the moment, to apply this relationship to our case, we will assume that initially the mouse was at rest

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Bold indicates vector quantities, let's calculate the momentum of each mouse in for the x and y axes

We recommend bringing all units to the SI system

Mouse 1.

It has a mass of 22.3 g = 22.3 10⁻³ kg, a final velocity of (v = 0.349 i ^ - 0.301 j ^) m / s with an initial velocity of zero

    Iₓ = m (v_{fx}  - v₀ₓ)

    Iₓ = 22.3 10⁻³ (0.349 -0)

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   I_{y} = m (v_{fy}  -v_{oy} )

   I_{y} = 22.3 10⁻³ (-0.301)

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Mouse 2

Mass 17.9 g = 17.9 10⁻³ kg

Speed ​​(-0.699 i ^ - 0.815 j ^) m / s

    Iₓ = m (v_{fx}  - v₀ₓ)

    Iₓ = 17.9 10⁻³ (-0.699 -0)

    Iₓ = -12.5 10⁻³ J s

    I_{y} = 17.9 10⁻³ (-0.815 - 0)

    I_{y} = -14.6 10⁻³ J s

   I₂ = (-12.5 i ^ -14.6 j ^) 10⁻³ J s

Mouse 3

Mass 19.1 g = 19.1 10⁻³ kg

Speed ​​(0.745i ^ + 0.975 j ^) m / s

    Iₓ = 19.1 10⁻³ (0.745 -0)

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Mouse 4

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Therefore the correct answer is "all of the above".

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I could really use some help on this question guys! Will give brainliest!
aev [14]

Answer:

I think it’s the third one

4 0
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