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3 years ago

36. What is mass? (4.4)

Engineering

1 answer:

KatRina [158]3 years ago

3 0

Answer:

Mass, in physics, quantitative measure of inertia, a fundamental property of all matter.

Explanation:

Mass is the matter that makes up objects

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Define the Artist class in Artist.py with a constructor to initialize an artist's information. The constructor should by default

Sunny_sXe [5.5K]

Answer:

Explanation:

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3 years ago

a) A 10-mm-diameter Brinell hardness indenter produced an indentation 2.50 mm in diameter in a steel alloy when a load of 1000 k

Vinvika [58]

Answer:

a) $HB = 200.484$ , b) $d \approx 1.453\,mm$

Explanation:

The Brinell hardness can be determined by using this expression:

$HB = \frac{2\cdot P}{\pi\cdot D^{2}}\cdot \left(\frac{1}{1-\sqrt{1-\frac{d^{2}}{D^{2}} } } \right)$

Where $D$ and $d$ are the indenter diameter and the indentation diameter, respectively.

a) The Brinell hardness is:

$HB = \frac{2\cdot (1000\,kgf)}{\pi\cdot (10\,mm)^{2}} \cdot \left[\frac{1}{1-\sqrt{1-\frac{(2.50\,mm)^{2}}{(10\,mm)^{2}} } } \right]$

$HB = 200.484$

b) The diameter of the indentation is obtained by clearing the corresponding variable in the Brinell formula:

$d = D\cdot \sqrt{1-(1-\frac{2\cdot P}{\pi\cdot HB \cdot D^{2}} )^{2}}$

$d = (10\,mm)\cdot\sqrt{1-\left[1-\frac{2\cdot (500\,kgf)}{\pi\cdot (300)\cdot (10\,mm)^{2}} \right]^{2}}$

$d \approx 1.453\,mm$

5 0

3 years ago

Read 2 more answers

The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pres

kirza4 [7]

Answer:

ΔP=19.76 KPa

$\Delta m=10^{-3}\ gm$

Explanation:

Given that

$T_1= 25C, P_1= 210 \KPa \gauge$

atmospheric pressure = 100 kPa.

So absolute pressure = Atmospheric pressure + gauge pressure

$P_1=210+100\ KPa$ (absolute)

$P_1=310\ KPa$ (absolute)

Here volume of air is constant .We know that for constant volume pressure

$\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$

here

$T_2= 44C$

$T_1=273+25=298K$

$T_2=273+44=317K$

$\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$

$\dfrac{310}{298}=\dfrac{P_2}{317}$

$P_2=329.76\ KPa$ (absolute)

So rise in pressure

$\Delta P=P_1-P_2$

ΔP=329.76-310 KPa

ΔP=19.76 KPa

$m_1=\dfrac{P_1V}{RT_1}$

$m_1=\dfrac{310\times 0.025}{0.287\times 298}$

$m_1=0.090615\ kg$

$m_2=\dfrac{P_2V}{RT_2}$

$m_2=\dfrac{329.76\times 0.025}{0.287\times 317}$

$m_2=0.090614\ kg$

Δm=0.090615 - 0.090614 kg

$\Delta m=10^{-3}\ gm$

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3 years ago

(a) Please show that the heat capacity at constant volume and constant pressure for air are given by,Cv=macvmwherecvm= (1 + 0.92

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Answer:

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3 years ago

For the storage tank of Prob. 7.104, determine the maximum normal stress and the maximum shearing stress in the cylindrical wall

lara31 [8.8K]

Answer:

The unpressurized cylindrical storage tank shown has a 5-mm wall thickness and is made of steel having a 400-MPa ultimate strength in tension. Determine the maximum height h to which it can be filled with water if a factor of safety of 4.0 is desired.

Explanation:

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2 years ago