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kykrilka [37]
3 years ago
13

Which of the following explains the main reason to cut a piece of wood on the outside of the measurement mark?

Engineering
1 answer:
maks197457 [2]3 years ago
4 0
I think it’s D ?? I’m not completely sure tho
You might be interested in
A Class A fire extingisher is for use on general combustibles such as:​
Tatiana [17]

Answer:

used for ordinary combustibles, such as wood, paper, some plastics, and textiles. This class of fire requires the heat-absorbing effects of water or the coating effects of certain dry chemicals.

Explanation:

7 0
3 years ago
For methyl chloride at 100°C the second and third virial coefficients are: B = −242.5 cm 3 ·mol −1 C = 25,200 cm 6 ·mol −2 Calcu
bogdanovich [222]

Answer:

a)W=12.62 kJ/mol

b)W=12.59 kJ/mol

Explanation:

At T = 100 °C the second and third virial coefficients are

B = -242.5 cm^3 mol^-1

C = 25200 cm^6  mo1^-2

Now according isothermal work of one mole methyl gas is

W=-\int\limits^a_b {P} \, dV

a=v_2\\

b=v_1

from virial equation  

\frac{PV}{RT}=z=1+\frac{B}{V}+\frac{C}{V^2}\\   \\P=RT(1+\frac{B}{V} +\frac{C}{V^2})\frac{1}{V}\\

And  

W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V}  } \, dV

a=v_2\\

b=v_1

Now calculate V1 and V2 at given condition

\frac{P1V1}{RT} = 1+\frac{B}{v_1} +\frac{C}{v_1^2}

Substitute given values P_1\\ = 1 x 10^5 , T = 373.15 and given values of coefficients we get  

10^5(v_1)/8.314*373.15=1-242.5/v_1+25200/v_1^2

Solve for V1 by iterative or alternative cubic equation solver we get

v_1=30780 cm^3/mol

Similarly solve for state 2 at P2 = 50 bar we get  

v_1=241.33 cm^3/mol

Now  

W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V}  } \, dV

a=241.33

b=30780

After performing integration we get work done on the system is  

W=12.62 kJ/mol

(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get  

         dV=RT(-1/p^2+0+C')dP

Hence work done on the system is  

W=-\int\limits^a_b {P(RT(-1/p^2+0+C')} \, dP

a=v_2\\

b=v_1

by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work  

W=12.59 kJ/mol

The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series  

8 0
3 years ago
Ti-6Al-4V has a fracture toughness of 74.6 MPa-m0.5. How much stress (in MPa) would it take to fail a plate loaded in tension th
Nikitich [7]

Answer:

critical stress  = 595 MPa

Explanation:

given data

fracture toughness =  74.6 MPa-\sqrt{m}

crack length = 10 mm

f = 1

solution

we know crack length = 10 mm  

and crack length = 2a as given in figure attach

so 2a = 10

a = 5 mm

and now we get here with the help of plane strain condition , critical stress is express as

critical stress  = \frac{k}{f\sqrt{\pi a}}    ......................1

put here value and we get

critical stress  = \frac{74.6}{1\sqrt{\pi 5\times 10^{-3}}}

critical stress  = 595 MPa

so here stress is change by plane strain condition because when plate become thinner than condition change by plane strain to plain stress.

plain stress condition occur in thin body where stress through thickness not vary by the thinner section.

6 0
3 years ago
A(n) ______ is used to measure fluid flow in engineering
Arte-miy333 [17]

Answer:

A pitot tube is used to measure fluid flow in engineering

3 0
2 years ago
R-134a vapor enters into a turbine at 250 psia and 175°F. The temperature of R-134a is reduced to 20°F in this turbine while its
attashe74 [19]

Answer:

Δ enthalpy = -23 Btu/Ibm

Explanation:

Given data:

Pressure ( P1 ) = 250 psi

Initial Temperature ( T1 ) = 175°F

Final temperature ( T2 ) = 20°F

<u>Calculate the change in the enthalpy of R-134a </u>

From R-134 table

h1 = 129.85 Btu/Ibm

s1 = 0.23281 Btu/Ibm.R

note : entropy is constant  

hence ; s1 = s2

by interpolation  ; h2 = 106.95

Δ enthalpy = h2 - h1

                  =  ( 106.95 - 129.85 ) = -23 Btu/Ibm

5 0
3 years ago
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