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BARSIC [14]
2 years ago
13

What type of cartilage provides support and shock absorption?.

Engineering
1 answer:
Katena32 [7]2 years ago
7 0

Answer:

fibrocartilage

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For a steel alloy it has been determined that a carburizing heat treatment of 15 h duration will raise the carbon concentration
Amiraneli [1.4K]

Answer:

135 hour

Explanation:

It is given that a carburizing heat treatment of 15 hour will raise the carbon concentration by 0.35 wt% at a point of 2 mm from the surface.

We have to find the time necessary to achieve the same concentration at a 6 mm position.

we know that \frac{x_1^2}{Dt}=constant where x is distance and t is time .As the temperature is constant so D will be also constant

So \frac{x_1^2}{t}=constant

then \frac{x_1^2}{t_1}=\frac{x_2^2}{t_2} we have given x_1=2 mm\ ,t_1=15 hour\ ,x_2=6\ mm and we have to find t_2 putting all these value in equation

\frac{2^2}{15}=\frac{6^2}{t_2}

so t_2=135\ hour

5 0
3 years ago
Discoloration on walls, work surfaces, ceilings, walls, and pipes may indicate a leak that is causing you to waste raw materials
suter [353]

Answer:

True :)

Explanation:

If this is a true or false question.

6 0
2 years ago
Help please really fast!!
ra1l [238]

Answer:368 hdhtygtÿ

901 vuiøöńč

Explanation:

6 0
3 years ago
An escalator handles a steady load of 26 people per minute in elevating them from the first to the second floor through a vertic
photoshop1234 [79]

Answer:

\eta = 70.711\,\%

Explanation:

The power needed to make the escalator working is obtained by means of the Work-Energy Theorem:

\dot W  = \dot U_{g}

\dot W = \dot n \cdot m_{p}\cdot g \cdot \Delta y

\dot W = \left(26\,\frac{persons}{min}\right)\cdot (124\,lbm)\cdot \left(32.174\,\frac{ft}{s^{2}}\right)\cdot \left(\frac{1\,lbf}{32.174\,\frac{lbm\cdot ft}{s^{2}} } \right)\cdot (27.5\,ft)

\dot W = 88660\,\frac{lbf\cdot ft}{min}\,\left(2.687\,hp\right)

The mechanical efficiency of the escalator is:

\eta = \frac{2.687\,hp}{3.8\,hp}\times 100\,\%

\eta = 70.711\,\%

3 0
3 years ago
A cylinder of aluminum-magnesium alloy 0.5 m long is subjected to an elastic tensile stress of 10.2 MPa. The measured elastic el
elena55 [62]

Answer:

E= 15 GPa.

Explanation:

Given that

Length ,L = 0.5 m

Tensile stress ,σ = 10.2 MPa

Elongation ,ΔL = 0.34 mm

lets take young modulus  = E

We know that strain ε given as

\varepsilon =\dfrac{\Delta L}{L}

\varepsilon =\dfrac{0.34}{0.5\times 1000}

\varepsilon =0.00068

We know that

\sigma = \varepsilon  E\\\\E=\dfrac{10.2}{0.00068}\\E= 15000\ MPa\\E=15\ GPa

Therefore the young's modulus will be 15 GPa.

8 0
3 years ago
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