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andrew-mc [135]

2 years ago

12

The mechanical properties of some metals may be improved by incorporating fine particles of their oxides. If the moduli of elast

icity of a hypothetical metal and its oxide are, respectively, 55 GPa and 430 GPa, what is the upper-bound modulus of elasticity value for a composite that has a composition of 31 vol% of oxide particles

1 answer:

mezya [45]2 years ago

3 0

Answer:

171.2 GPa

Explanation:

I explained every step in the attached file, please kindly go through it.

But the answer is 171.2 GPa

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Water is flowing in a metal pipe. The pipe OD (outside diameter) is 61 cm. The pipe length is 120 m. The pipe wall thickness is

Yuki888 [10]

Answer:

1113kN

Explanation:

The ouside diameter OD of the pipe is 61cm and the thickness T is 0.9cm, so the inside diameter ID will be:

Inside Diameter = Outside Diameter - Thickness

Inside Diameter = 61cm - 0.9cm = 60.1cm

Converting this diameter to meters, we have:

$60.1cm*\frac{1m}{100cm}=0.601m$

This inside diameter is useful to calculate the volume V of water inside the pipe, that is the volume of a cylinder:

$V_{water}=\pi r^{2}h$

$V_{water}=\pi (\frac{0.601m}{2})^{2}*120m$

$V_{water}=113.28m^{3}$

The problem gives you the water density d as 1.0kg/L, but we need to convert it to proper units, so:

$d_{water}=1.0\frac{Kg}{L}*\frac{1L}{1000cm^{3}}*(\frac{100cm}{1m})^{3}$

$d_{water}=1000\frac{Kg}{m^{3}}$

Now, water density is given by the equation $d=\frac{m}{V}$ , where m is the water mass and V is the water volume. Solving the equation for water mass and replacing the values we have:

$m_{water}=d_{water}.V_{water}$

$m_{water}=1000\frac{Kg}{mx^{3}}*113.28m^{3}$

$m_{water}=113280Kg$

With the water mass we can find the weight of water:

$w_{water}=m_{water} *g$

$w_{water}=113280kg*9.8\frac{m}{s^{2}}$

$w_{water}=1110144N$

Finally we find the total weight add up the weight of the water and the weight of the pipe,so:

$w_{total}=w_{water}+w_{pipe}$

$w_{total}=1110144N+2500N$

$w_{total}=1112644N$

Converting this total weight to kN, we have:

$1112644N*\frac{0.001kN}{1N}=1113kN$

7 0

2 years ago

A cylindrical rod of copper (E = 110 GPa, 16 × 106 psi) having a yield strength of 240 MPa (35,000 psi) is to be subjected to a

77julia77 [94]

Answer:

7.65 mm

Explanation:

Stress, $\sigma=\frac {F}{A}$ where F is the force and A is the area

Also, $\sigma=E\times \frac {\triangle L}{L}$

Where E is Young’s modulus, L is the length and $\triangle L$ is the elongation

Therefore,

$\frac {F}{A}= E\times \frac {\triangle L}{L}$

Making A the subject of the formula then

$A=\frac {FL}{E\triangle L}=\frac {6660\times 380}{110\times 10^{9}\times 0.5}=4.60145\times 10^{-5} m^{2}$

Since $A=\frac {\pi d^{2}}{4}$ then

$d=\sqrt {\frac {4A}{\pi}}=\sqrt {\frac {4\times 4.60145\times 10^{-5}}{\pi}}= 0.00765425m= 7.654249728 mm\approx 7.65 mm$

4 0

2 years ago

Derive the probability that a receptor is occupied by a ligand using a model that treats the L ligands in solution as distinguis

love history [14]

that is the same thing as you are not going through this week or something

7 0

2 years ago

print('Predictions are hard.") print(Especially about the future.) user_num = 5 print('user_num is:' user_num)

Bas_tet [7]

Answer:

The correct code is given below:-

print("Predictions are hard.")

print("Especially about the future.")

user_num = 5

print("user_num is:", user_num)

7 0

2 years ago

For laminar flow of air over a flat plate that has a uniform surface temperature, the curve that most closely describes the vari

Aliun [14]

This question is incomplete, the missing diagram is uploaded along this answer below;

Answer:

from the diagram, the curve that most closely describes the variation of the local heat transfer coefficient with position along the plate is Option D

Explanation:

Given the data in the question;

We write the expression for the local Nusselt number for Laminar flow over the flat plate;

Nu = $C$ $(Re_x)^{0.5$ $(Pr)^{1/3$

Nu = $C(\frac{Vx}{v})^{0.5}$ $(Pr)^{1/3$

$\frac{h_xx}{k}$ = $C(\frac{V}{v})^{0.5}$ $(Pr)^{1/3$ $(x)^{0.5$

$h_x$ = $\frac{1}{x^{1/2}}$

Next we write down the expression for the local heat flux from the plate with uniform surface temperature;

q = $h_xA($ T $_s$ - T∞ )

q ∝ $h_x$

∴

q ∝ $\frac{1}{x^{1/2}}$

The local heat flux decreases with the position as it is inversely proportional to the square root of the position from the leading edge and it will not be zero at the end of the plate.

Therefore, from the diagram, the curve that most closely describes the variation of the local heat transfer coefficient with position along the plate is Option D

3 0

2 years ago