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TiliK225 [7]
3 years ago
5

Define extensive and intensive properties of thermodynamic system.

Engineering
2 answers:
Flura [38]3 years ago
7 0

Answer:

An intense property is a physical attribute of a system that is independent of the size of the system or the quantity of material it contains. An extensive property of a system, on the other hand, is dependent on the size of the system or the amount of material in it.

Explanation:

kow [346]3 years ago
3 0
Thermodynamic properties can be divided into two general classes, intensive and extensive properties. An intensive property is independent of the amount of mass. The value of an extensive property varies directly with the mass. Thus, if a quantity of matter in a given state is divided into two equal parts, each part will have the same value of intensive property as the original and half the value of the extensive property. Temperature, pressure, specific volume,and density are examples of intensive properties. Mass and total volume are examples of extensive properties.
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An air mass is a body of air with horizontally uniform temperature, humidity, and pressure.

Explanation:

Because it is

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A kernel-level thread wishes to acquire a mutex lock declared as global in the process. True or False: the function call used be
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What steps would you take to design an improved toothpaste container?
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Answer:

A. Identify the need, recognize limitations of current toothpaste containers, and then brainstorm ideas on how to improve the existing

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Strands of materials A and B are placed under a tensile force of 10 Newtons. Material A deforms more than Material B.
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At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho
victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2}  }{2})^{2}+\tau _{xy}^{2}    }

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2}  }

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}

Where

σp1 = 20 ksi

σp2 = -30 ksi

\sigma _{p1}  +\sigma _{p2}=-10 ksi (equation 1)

\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi equation 2

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σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

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3 years ago
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