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bogdanovich [222]
2 years ago
10

A 100-mm-long strip of metal is stretched in two steps, first to 200 mm and then to 400 mm. Show that the total true strain is t

he sum of the true strains in each step; in other words, the true strains are additive. Show that, in the case of engineering strains, the strains cannot be added to obtain the total strain.
Engineering
1 answer:
ladessa [460]2 years ago
8 0

Answer:

Kindly check the explanation section.

Explanation:

Determine the strain in units in both steps.

For step one, strain = [ final length - Initial length]/ initial length = 200 - 100/ 100 = 1 unit.

For step two, the strain =  final length - Initial length]/ initial length = 400 - 200/ 200 = 1 unit.

Total strain for the two steps = 1 unit + 1 unit = 2 units.

Estimation using single step is given below;

400 - 100/ 400 = 0.75.

It can be shown from above that the Engineering strength is Addictive.

The next thing to do is to determine the true strength. This is how it can be proved that the strength is Addictive or not.

b1 + b2 = (200 - 100/100) + (400-200/ 200)

b1 + b2 = 1 + 1 = 2.

b(total) = 400 - 100/100 = v2 - v1/v1.

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A binary star system consists of two stars of masses m1m1m_1 and m2m2m_2. The stars, which gravitationally attract each other, r
d1i1m1o1n [39]

Answer:

          a_c_2=\dfrac{a_c_1\times m_1}{m_2}

Explanation:

The question is: <em>Find the magnitude of the centripetal acceleration of the star with mass m₂</em>

The <em>centripetal acceleration</em> is the quotient of the centripetal force and the mass.

                a_c=\dfrac{F_c}{m}

Thus, you can write the equations for each star:

     

       a_c_1=\dfrac{F_c_1}{m_1}

       a_c_2=\dfrac{F_c_2}{m_2}

As per Newton's third law, the centripetal forces are equal in magnitude. Then:

       a_c_1\times m_1=a_c_2\times m_2

Now you can clear a_c_2:

          a_c_2=\dfrac{a_c_1\times m_1}{m_2}

6 0
2 years ago
On the reality television show "Survivor," two tribes compete for luxuries such as food and shelter. During such challenges, one
Ivan

Answer:

Realistic Group Conflict Theory

Explanation:

Realistic Group Conflict theory is a socio-psychological model of internal conflicts among groups.

These groups may compete for a scarce resources based on reality or perception.

These conflicts may be over political power, capital, or social status, etc.

According to this theory, in the competition, only one group wins over the scarce resources while the other has to lose.

Thus The reality show on the television showing two tribes competing for food and shelter where the success of one is the failure for the other is based on Realistic Group Conflict Theory.

5 0
2 years ago
1. Add:<br>(i) 5xy, -2xy, -11xy, 8xy<br>(iv) 3a - 2b + c, 5a + 8b -70​
Cerrena [4.2K]

Answer:

(i) 0

(iv) 8a+6b+c-70

Explanation:

Hope this helps you

8 0
1 year ago
Water (cp = 4180 J/kg·°C) enters the 2.5 cm internal diameter tube of a double-pipe counter-flow heat exchanger at 17°C at a rat
aksik [14]

Answer:

Length = 129.55m, 129.55m

Explanation:

Given:

cp of water = 4180 J/kg·°C

Diameter, D = 2.5 cm

Temperature of water in =  17°C

Temperature of water out = 80°C

mass rate of water =1.8 kg/s.

Steam condensing at 120°C

Temperature at saturation = 120°C

hfg of steam at 120°C = 2203 kJ/kg

overall heat transfer coefficient of the heat exchanger = 700 W/m2 ·°C

U = 700 W/m2 ·°C

Since Temperature of steam is at saturation,

temperature of steam going in = temperature of steam out = 120°C

Energy balance:

Heat gained by water = Heat loss by steam

Let specific capacity of steam = 2010kJ/Kg .°C

Find attached the full solution to the question.

3 0
2 years ago
An 80-L vessel contains 4 kg of refrigerant-134a at a pressure of 160kPa. Determine (a) the temperature, (b) the quality, (c) th
makvit [3.9K]

Answer:

temperature -15.6 C, quality x=0.646, enthalpy h=667.20 KJ, volume of vapor phase Vg= 79.8 L

Explanation:

property table for R-134a

https://www.ohio.edu/mechanical/thermo/property_tables/R134a/R134a_PresSat.html

at 160 KPa , temperature = -15.66 C

quality x=mass of vapour/ total mass of liq-vap mixture

alternaternately: x=(v-vf)/(vg-vf)    

v=total volume i.e. volume of container"80L"   80L=0.08 cubic meter

vf=vol of liquid phase  vg=vol of vapor phase vf, vg values at 160Kpa

x=(0.08-0.0007437)/(0.1235-0.0007437)=0.646

enthalpy

h=hf+xhfg          hf, hfg values at 160Kpa

h=hf+xhfg=31.2+0.646(209.9)=166.80 KJ/Kg

for 4Kg R-134a h=m(166.80 KJ/Kg )=667.20 KJ

volume of vapor phase

vg at 160Kpa=0.1235 cubic meter for quality=1.

in this case quality=0.646 , so it will occupy 64.6% space of the vapor phase at quality=1.

vol. of vapor phase=0.646*0.1235=0.0798 cubic meter=79.8 L

7 0
2 years ago
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