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ExtremeBDS [4]
3 years ago
10

Does anyone have cumulative exams today or tomorrow?(There so boring!)

Engineering
2 answers:
AlekseyPX3 years ago
6 0

Answer:

I feel bad for you

Explanation:

They are so overwhelming and boring

IRINA_888 [86]3 years ago
3 0

Answer:

I do!!

Explanation:

I have to sit for 3 hours lol‍♀️

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A and B connect the gear box to the wheel assemblies of a tractor, and shaft C connects it to the engine. Shafts A and Blie in t
azamat

Answer:

The couples are not all on one axis or plane for that matter but if the A and B connector had to be specified it would go by the yz axis diagonal to the x axis with a magnitude of about 15. The direction of the axis would be pointed up to the second quadrant. Hope this was helpful

Explanation:

8 0
4 years ago
Consider liquid n-hexane in a 50-mm diameter graduated cylinder. Air blows across the top of the cylinder. The distance from the
ra1l [238]

The evaporation rate of the n-Hexane is 7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}

<u>Explanation</u>:

This is a situation regarding diffusing A through non-diffusing B.

A = n-Hexane B=Air

Where the molar flux is provided by,

N_{A}=D_{A B} P_{T}\left(P_{A 1}-P_{A 2}\right) / R T z P_{b m}

\mathrm{D}_{\mathrm{AB}}=8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}

P_{t}=1 a t m=101325 P a\\

\text { so, } P_{A 1}= the vapor pressure at hexane 25 \mathrm{C} =20158.2 \mathrm{Pa}

For wind, assume negligible hexane is present, hence P_{A 2}=0

Now,

\mathrm{P}_{\mathrm{B} 1}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 1}=101325-20158.2 \mathrm{P}_{\mathrm{a}}

\mathrm{P}_{\mathrm{B} 2}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 2}=\mathrm{P}_{\mathrm{T}}=101325 \mathrm{Pa}

P_{B M}=\frac{\left(P_{B 2}-P_{B 1}\right)}{\log _{e}\left(P_{B 2} / P_{B 1}\right)}\\

=\frac{101325-81166.8}{\ln \left(\frac{101325}{81166.8}\right) \mathrm{Pa}}

=90873.57 \mathrm{Pa}

R=8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K}

z=\text { distance }=20 \mathrm{cm}=0.2 \mathrm{m}\\

where T = 298 K

substituting all in the equation, we get

\begin{aligned}&\mathrm{N}_{\mathrm{A}}=\\&\left(8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}\right) \times 101325 \mathrm{Pa} \times(20158.2 \mathrm{Pa}) /(8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K} \times 0.2 \mathrm{m} \times 298 \mathrm{K}\\&\times 90873.57 \mathrm{Pa})\end{aligned}

=0.004 \mathrm{mol} / \mathrm{m}^{2} \mathrm{s}\\

Now,Flux \times area  = Molar rate of evaporation

Evaporation rate = 0.004 \mathrm{mol} / \mathrm{m}^{2}-5 \mathrm{x}\left(\pi \mathrm{d}^{2} / 4 \mathrm{m}^{2}\right)=0.004 \times(3.14 \times 0.05 \times 0.05 / 4)

Evaporation rate =7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}

6 0
3 years ago
A method that uses low temperature heat-treating that imparts toughness without reduction in hardness is called:_______
lbvjy [14]

Answer:

"Tempering Process" seems to be the appropriate choice.

Explanation:

  • Tempering seems to be a method of heat preparation which is mostly used in completely hard materials to increase consistency, strength, durability, and also some decreasing brittleness.
  • The tempering method is used to examine good functionality as well as flexural by reducing stiffness again after the substance has indeed been quenched towards its toughest state.
5 0
3 years ago
The thrust F of a screw propeller is known to depend upon the diameter d,speed of advance \nu ,fluid density p, revolution per s
musickatia [10]

Answer:

<em>screw thrust = ML</em>T^{-2}<em> </em>

Explanation:

thrust of a screw propeller is given by the equation = pV^{2}D^{2} x \frac{ND}{V}Re

where,

D is diameter

V is the fluid velocity

p is the fluid density

N is the angular speed of the screw in revolution per second

Re is the Reynolds number which is equal to  puD/μ

where p is the fluid density

u is the fluid velocity, and

μ is the fluid viscosity = kg/m.s = ML^{-1}T^{-1}

<em>Reynolds number is dimensionless so it cancels out</em>

The dimensions of the variables are shown below in MLT

diameter is m = L

speed is in m/s = LT^{-1}

fluid density is in kg/m^{3} = ML^{-3}

N is in rad/s = LL^{-1}T^{-1} =

If we substitute these dimensions in their respective places in the equation, we get

thrust = ML^{-3}(LT^{-1}) ^{2}L^{2}\frac{T^{-1} L}{LT^{-1} }

= ML^{-3}L^{2}T^{-2}

<em>screw thrust = ML</em>T^{-2}<em> </em>

This is the dimension for a force which indicates that thrust is a type of force

6 0
3 years ago
Compared to arc welding, which of the following statements are true about<br> gas welding?
UNO [17]

Show than arc welding

4 0
3 years ago
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