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3 years ago

9

Which of these passages is an example of cohesive writing?

2 answers:

aksik [14]3 years ago

5 0

Answer:

It should be number 1

Oksana_A [137]3 years ago

3 0

The answer mostly likely is: “1” due to the fact it is well ordered, and follows a similar thought process, best of luck!

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What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 3 × 10-4

Vladimir [108]

Answer:

maximum stress is 2872.28 MPa

Explanation:

given data

radius of curvature = 3 × $10^{-4}$ mm

crack length = 5.5 × $10^{-2}$ mm

tensile stress = 150 MPa

to find out

maximum stress

solution

we know that maximum stress formula that is express as

$\sigma m = 2 ( \sigma o ) \sqrt{\frac{a}{\delta t}}$ ......................1

here σo is applied stress and a is half of internal crack and t is radius of curvature of tip of internal crack

so put here all value in equation 1 we get

$\sigma m = 2 ( \sigma o) \sqrt{\frac{a}{\delta t}}$

$\sigma m = 2(150) \sqrt{ \frac{\frac{5.5*10^{-2}}{2}}{3*10^{-4}}}$

σm = 2872.28 MPa

so maximum stress is 2872.28 MPa

8 0

3 years ago

Urgent please help me...

Pepsi [2]

Scuze , am incercat , dat chiar nu stiu. Prefer sa nu raspund si sa scriu asta decat sa raspund gresit.❤️

7 0

3 years ago

The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 5 MPa and an initial temperature

Anit [1.1K]

Answer:

a). Work transfer = 527.2 kJ

b). Heat Transfer = 197.7 kJ

Explanation:

Given:

$P_{1}$ = 5 Mpa

$T_{1}$ = 1623°C

= 1896 K

$V_{1}$ = 0.05 $m^{3}$

Also given $\frac{V_{2}}{V_{1}} = 20$

Therefore, $V_{2}$ = 1 $m^{3}$

R = 0.27 kJ / kg-K

$C_{V}$ = 0.8 kJ / kg-K

Also given : $P_{1}V_{1}^{1.25}=C$

Therefore, $P_{1}V_{1}^{1.25}$ = $P_{2}V_{2}^{1.25}$

$5\times 0.05^{1.25}=P_{2}\times 1^{1.25}$

$P_{2}$ = 0.1182 MPa

a). Work transfer, δW = $\frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}$

$\left [\frac{5\times 0.05-0.1182\times 1}{1.25-1} \right ]\times 10^{6}$

= 527200 J

= 527.200 kJ

b). From 1st law of thermodynamics,

Heat transfer, δQ = ΔU+δW

= $\frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}$

= $\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W$

= $\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200$

= 197.7 kJ

6 0

3 years ago

⦁ An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 27°C, and

Anon25 [30]

Answer:

<u><em>note:</em></u>

<u><em>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</em></u>

8 0

3 years ago

Refrigerant 134a enters an air conditioner compressor at 4 bar, 20 C, and is compressed at steady state to 12 bar, 80 C. The vol

sleet_krkn [62]

Answer:

$Q=15.7Kw$

Explanation:

From the question we are told that:

Initial Pressure $P_1=4bar$

Initial Temperature $T_1=20 C$

Final Pressure $P_2=12 bar$

Final Temperature $T_2=80C$

Work Output $W= 60 kJ/kg$

Generally Specific Energy from table is

At initial state

$P_1=4bar \& T_1=20 C$

$E_1=262.96KJ/Kg$

With

Specific Volume $V'=0.05397m^3/kg$

At Final state

$P_2=12 bar \& P_2=80C$

$E_1=310.24KJ/Kg$

Generally the equation for The Process is mathematically given by

$m_1E_1+w=m_2E_2+Q$

Assuming Mass to be Equal

$m_1=m_1$

Where

$m=\frac{V}{V'}$

$m=frac{0.06666}{V'=0.05397m^3/kg}$

$m=1.24$

Therefore

$1.24*262.96+60)=1.24*310.24+Q$

$Q=15.7Kw$

4 0

3 years ago