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IceJOKER [234]
3 years ago
11

What is Class Night in Puerto Rico?

Engineering
1 answer:
valentina_108 [34]3 years ago
8 0

Answer:

b

Explanation:

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In an orthogonal cutting operation, the tool has a rake angle = 12°. The chip thickness before the cut = 0.32 mm and the cut yie
Snezhnost [94]

Answer:

The shear plane angle and shear strain are 28.21° and 2.155 respectively.

Explanation:

(a)

Orthogonal cutting is the cutting process in which cutting direction or cutting velocity is perpendicular to the cutting edge of the part surface.  

Given:  

Rake angle is 12°.  

Chip thickness before cut is 0.32 mm.

Chip thickness is 0.65 mm.  

Calculation:  

Step1  

Chip reduction ratio is calculated as follows:  

r=\frac{t}{t_{c}}

r=\frac{0.32}{0.65}

r = 0.4923

Step2  

Shear angle is calculated as follows:  

tan\phi=\frac{rcos\alpha}{1-rsin\alpha}

Here, \phi is shear plane angle, r is chip reduction ratio and \alpha is rake angle.  

Substitute all the values in the above equation as follows:  

tan\phi=\frac{rcos\alpha}{1-rsin\alpha}

tan\phi=\frac{0.4923cos12^{\circ}}{1-0.4923sin12^{\circ}}

tan\phi=\frac{0.48155}{0.8976}

\phi=28.21^{\circ}

Thus, the shear plane angle is 28.21°.

(b)

Step3

Shears train is calculated as follows:

\gamma=cot\phi+tan(\phi-\alpha)

\gamma=cot28.21^{\circ}+tan(28.21^{\circ}-12^{\circ})\gamma = 2.155.

Thus, the shear strain rate is 2.155.

6 0
3 years ago
How do u charge ur phone? :)
Pepsi [2]
With a phone charger.
7 0
2 years ago
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What does this middle button on the middle of my Chevrolet equinox steering wheel mean?
almond37 [142]

Answer:

Depending on how new your vehicle is, to me it looks like some sort of turn on button for forward collision safety feature, but i'm not an expert with this particular vehicle. If you want a better answer, I strongly suggest looking at the owner's manual under which should be located in the front dash compartment on the passengers side. Once you have the manual, look in the appendix until you find controls (or something similar) then go to that page and read about your vehicles control buttons. Your answer should be in the manual.

Something that all automobile owners should do right after purchasing a new vehicle, is reading the owners manual. As boring as it may seem, reading the owner's manual will help you get used to your new car quicker and give you instructions on how to take care of your car so that it lasts.

Have a great day, and I wish you safe traveling for now and forever! :)

 

7 0
3 years ago
Read 2 more answers
For a cylindrical annulus whose inner and outer surfaces are maintained at 30 ºC and 40 ºC, respectively, a heat flux sensor mea
miskamm [114]

Answer:

k=0.12\ln(r_2/r_1)\frac {W}{ m^{\circ} C}

where r_1 and r_2 be the inner radius, outer radius of the annalus.

Explanation:

Let r_1, r_2 and L be the inner radius, outer radius and length of the given annulus.

Temperatures at the inner surface, T_1=30^{\circ}C\\ and at the outer surface, T_2=40^{\circ}C.

Let q be the rate of heat transfer at the steady-state.

Given that, the heat flux at r=3cm=0.03m is

40 W/m^2.

\Rightarrow \frac{q}{(2\pi\times0.03\times L)}=40

\Rightarrow q=2.4\pi L \;W

This heat transfer is same for any radial position in the annalus.

Here, heat transfer is taking placfenly in radial direction, so this is case of one dimentional conduction, hence Fourier's law of conduction is applicable.

Now, according to Fourier's law:

q=-kA\frac{dT}{dr}\;\cdots(i)

where,

K=Thermal conductivity of the material.

T= temperature at any radial distance r.

A=Area through which heat transfer is taking place.

Here, A=2\pi rL\;\cdots(ii)

Variation of temperature w.r.t the radius of the annalus is

\frac {T-T_1}{T_2-T_1}=\frac{\ln(r/r_1)}{\ln(r_2/r_1)}

\Rightarrow \frac{dT}{dr}=\frac{T_2-T_1}{\ln(r_2/r_1)}\times \frac{1}{r}\;\cdots(iii)

Putting the values from the equations (ii) and (iii) in the equation (i), we have

q=\frac{2\pi kL(T_1-T_2)}{\LN(R_2/2_1)}

\Rightarrow k= \frac{q\ln(r_2/r_1)}{2\pi L(T_2-T_1)}

\Rightarrow k=\frac{(2.4\pi L)\ln(r_2/r_1)}{2\pi L(10)} [as q=2.4\pi L, and T_2-T_1=10 ^{\circ}C]

\Rightarrow k=0.12\ln(r_2/r_1)\frac {W}{ m^{\circ} C}

This is the required expression of k. By putting the value of inner and outer radii, the thermal conductivity of the material can be determined.

7 0
3 years ago
3<br> Current is measured in units called
bixtya [17]

Answer:

current is measured in Ampere (A)

6 0
2 years ago
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