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3 years ago

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A specimen of some metal having a rectangular cross section 10.4 mm × 12.8 mm is pulled in tension with a force of 15900 N, whic

h produces only elastic deformation. Given that the elastic modulus of this metal is 79 GPa, calculate the resulting strain.

1 answer:

marishachu [46]3 years ago

8 0

Answer:

$\sigma=0.00151$

Explanation:

We apply Hooke's Law as follows :

$\sigma= E\varepsilon$

where $\sigma$ is the applied stress

#The applied stress is also equal to :

$\sigma=\frac{F}{A_o}=\frac{F}{lw}$

Where $l\$ and $w$ are the cross-sectional dimensions.

=>We equate the two stress equations:

$E\varepsilon=\frac{F}{lw}\\\\\varepsilon=\frac{F}{Elw}$

#We substitute our values in the equation as:

$\varepsilon=\frac{15900\ N}{(79\times 10^9\ N/m^2)(10.4\ m\times 10^{-3}\times 12.8\ m\times 10^{-3})}}\\\\\\=0.00151$

Hence, the resulting strain is 0.00151

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A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 a

IrinaVladis [17]

Answer:

The original length of the specimen $l_{o} = 104.7 mm$

Explanation:

Original diameter $d_{o}$ = 30 mm

Final diameter $d_{1}$ = 30.04 mm

Change in diameter Δd = 0.04 mm

Final length $l_{1}$ = 105.20 mm

Elastic modulus E = 65.5 G pa = 65.5 × $10^{3}$ M pa

Shear modulus G = 25.4 G pa = 25.4 × $10^{3}$ M pa

We know that the relation between the shear modulus & elastic modulus is given by

$G = \frac{E}{2(1 + \mu)}$

$25.5 = \frac{65.5}{2 (1 + \mu)}$

$\mu = 0.28$

This is the value of possion's ratio.

We know that the possion's ratio is given by

$\mu = \frac{\frac{0.04}{30} }{\frac{change \ in \ length}{l_{o} } }$

${\frac{change \ in \ length}{l_{o} } = \frac{\frac{0.04}{30} }{0.28}$

${\frac{change \ in \ length}{l_{o} } =$ 0.00476

$\frac{l_{1} - l_{o} }{l_{o} } = 0.00476$

$\frac{l_{1} }{l_{o} } = 1.00476$

Final length $l_{o}$ = 105.2 m

Original length

$l_{o} = \frac{105.2}{1.00476}$

$l_{o} = 104.7 mm$

This is the original length of the specimen.

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3 years ago

A 1 m wide continuous footing is designed to support an axial column load of 250 kN per meter of wall length. The footing is pla

creativ13 [48]

Answer:

correct option is (A) 0.5

Explanation:

given data

axial column load = 250 kN per meter

footing placed = 0.5 m

cohesion = 25 kPa

internal friction angle = 5°

solution

we know angle of internal friction is 5° that is near to 0°

so it means the soil is almost cohesive soil.

and for a pure cohesive soil

$N_{\gamma }$ = 0

and we know formula for $N_{\gamma }$ is

$N_{\gamma }$ = (Nq - 1 ) × tan(Ф) ..................1

so here Ф is very less $N_{\gamma }$ should be nearest to zero

and its value can be 0.5

so correct option is (A) 0.5

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3 years ago

Carnot heat engine A operates between 20ºC and 520ºC. Carnot heat engine B operates between 20ºC and 820ºC. Which Carnot heat en

nikklg [1K]

Answer:

engine B is more efficient.

Explanation:

We know that Carnot cycle is an ideal cycle for all working heat engine.In Carnot cycle there are four processes in which two are constant temperature processes and others two are isentropic process.

We also kn ow that the efficiency of Carnot cycle given as follows

$\eta =1-\dfrac{T_1}{T_2}$

Here temperature should be in Kelvin.

For engine A

$\eta =1-\dfrac{T_1}{T_2}$

$\eta =1-\dfrac{273+20}{520+273}$

$\eta =0.63$

For engine B

$\eta =1-\dfrac{T_1}{T_2}$

$\eta =1-\dfrac{273+20}{820+273}$

$\eta =0.73$

So from above we can say that engine B is more efficient.

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4 years ago

List the main activities of exploration??

Trava [24]

Answer: Exploration includes plethora of activities and depend upon the kind of exploration a person is doing. But most include some of the basic activities like research , investigation, planning and execution.

Suppose we want to explore new petroleum sites then we would have to start with studying the geography of that area, then according to our research we will analyse the hot spots or the sector where probability of finding of oil field is highest, post that appropriate man power is skilled professionals, tools and machinery will be brought at the site so that execution can take place.

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3 years ago

.a. What size vessel holds 2 kg water at 80°C such that 70% is vapor? What are the pressure and internal energy? b. A 1.6 m3 ves

vesna_86 [32]

Answer:

Part a: The volume of vessel is 4.7680 $m^3$ and total internal energy is 3680 kJ.

Part b: The quality of the mixture is 90.3% or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

Explanation:

Part a:

As per given data

m=2 kg

T=80 °C =80+273=353 K

Dryness=70% vapour =0.7

<em>From the steam tables at 80 °C</em>

Specific volume of saturated vapours=v_g=3.40527 $m^3/kg$

Specific volume of saturated liquid=v_f=0.00102 $m^3/kg$

Now the relation of total specific volume for a specific dryness value is given as

$v=v_f+x(v_g-v_f)$

Substituting the values give

$v=v_f+x(v_g-v_f)\\v=0.00102+0.7(3.40527-0.00102)\\v_f=2.38399 m^3/kg$

Now the volume of vessel is given as

$v=\frac{V}{m}\\V=v \times m\\V=2.38399 \times 2\\V=4.7680 m^3$

So the volume of vessel is 4.7680 $m^3$ .

Similarly for T=80 and dryness ratio of 0.7 from the table of steam

Pressure=P=47.4 kPa

Specific internal energy is given as u=1840 kJ/kg

So the total internal energy is given as

$u=\frac{U}{m}\\U=u \times m\\U=1840 \times 2\\U=3680 kJ$

The total internal energy is 3680 kJ.

So the volume of vessel is 4.7680 $m^3$ and total internal energy is 3680 kJ.

Part b

Volume of vessel is given as 1.6

mass is given as 2 kg

Pressure is given as 0.2 MPa or 200 kPa

Now the specific volume is given as

$v=\frac{V}{m}\\v=\frac{1.6}{2}\\v=0.8 m^3/kg$

So from steam tables for Pressure=200 kPa and specific volume as 0.8 gives

Temperature=T=120 °C

Quality=x=0.903 ≈ 90.3%

Specific internal energy =u=2330 kJ/kg

The total internal energy is given as

$u=\frac{U}{m}\\U=u \times m\\U=2330 \times 2\\U=4660 kJ$

So the quality of the mixture is 90.3% or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

5 0

3 years ago