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Rus_ich [418]
3 years ago
15

An aluminum oxide component must not fail when a tensile stress of 12.5 MPa is applied. Determine the maximum allowable surface

crack length if the surface energy of aluminum oxide is 0.90 J/m2. The modulus of elasticity of this material is 393 GPa.
Engineering
1 answer:
aivan3 [116]3 years ago
4 0

Answer:

1.44 mm

Explanation:

Compute the maximum allowable surface crack length using

C=\frac {2E\gamma}{\pi \sigma_c^{2}} where E is the modulus  of elasticity, \gamma is surface energy and \sigma_c is tensile stress

Substituting the given values

C=\frac {2\times 393\times 10^{9}\times 0.9}{\pi\times (16\times 10^{6})^{2}= 0.001441103 m\approx 1.44mm

The maximum allowable surface crack is 1.44 mm

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a) A total charge Q = 23.6 μC is deposited uniformly on the surface of a hollow sphere with radius R = 26.1 cm. Use ε0 = 8.85419
dusya [7]

Answer:

(a) E = 0 N/C

(b) E = 0 N/C

(c) E = 7.78 x10^5 N/C

Explanation:

We are given a hollow sphere with following parameters:

Q = total charge on its surface = 23.6 μC = 23.6 x 10^-6 C

R = radius of sphere = 26.1 cm = 0.261 m

Permittivity of free space = ε0 = 8.85419 X 10−12 C²/Nm²

The formula for the electric field intensity is:

E = (1/4πεo)(Q/r²)

where, r = the distance from center of sphere where the intensity is to be found.

(a)

At the center of the sphere r = 0. Also, there is no charge inside the sphere to produce an electric field. Thus the electric field at center is zero.

<u>E = 0 N/C</u>

(b)

Since, the distance R/2 from center lies inside the sphere. Therefore, the intensity at that point will be zero, due to absence of charge inside the sphere (q = 0 C).

<u>E = 0 N/C</u>

(c)

Since, the distance of 52.2 cm is outside the circle. So, now we use the formula to calculate the Electric Field:

E = (1/4πεo)[(23.6 x 10^-6 C)/(0.522m)²]

<u>E = 7.78 x10^5 N/C</u>

4 0
2 years ago
I have to find the critical points of this function of two variables <img src="https://tex.z-dn.net/?f=%5C%5Cf%28x%2Cy%29%3Dx%5E
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because theres not

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barxatty [35]

Answer:

The specific weight of unknown liquid is found to be 15 KN/m³

Explanation:

The total pressure in tank is measured to be 65 KPa in the tank. But, the total pressure will be equal to the sum of pressures due to both oil and unknown liquid.

Total Pressure = Pressure of oil + Pressure of unknown liquid

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