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77julia77 [94]
3 years ago
9

The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas

ticity of the metal and oxide are, respectively, 67 GPa and 390 GPa, what is the (a) upper-bound, and (b) lower-bound modulus of elasticity values (in GPa) for a composite that has a composition of 44 vol% of oxide particles.
Engineering
1 answer:
Lilit [14]3 years ago
4 0

Answer:

A) 209.12 GPa

B) 105.41 GPa

Explanation:

We are given;

Modulus of elasticity of the metal; E_m = 67 GPa

Modulus of elasticity of the oxide; E_f = 390 GPa

Composition of oxide particles; V_f = 44% = 0.44

A) Formula for upper bound modulus of elasticity is given as;

E = E_m(1 - V_f) + (E_f × V_f)

Plugging in the relevant values gives;

E = (67(1 - 0.44)) + (390 × 0.44)

E = 209.12 GPa

B) Formula for upper bound modulus of elasticity is given as;

E = 1/[(V_f/E_f) + (1 - V_f)/E_m]

Plugging in the relevant values;

E = 1/((0.44/390) + ((1 - 0.44)/67))

E = 105.41 GPa

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When an object has the same number of positive and negative charges, its electrical charge will become neutral.

What is an electric charge?

When the matter is put in an electromagnetic field, it has an electric charge, which causes it to experience a force. A positive or negative electric charge can exist.

Now when the two equal magnitude charges with opposite natures come together they become neutral.

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7 0
2 years ago
Write a method printShampooInstructions(), with int parameter numCycles, and void return type. If numCycles is less than 1, prin
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Answer:

// The method is defined with a void return type

// It takes a parameter of integer called numCycles

// It is declared static so that it can be called from a static method

public static void printShampooInstructions(int numCycles){

// if numCycles is less than 1, it display "Too few"

   if (numCycles < 1){

       System.out.println("Too few.");

   }

// else if numCycles is less than 1, it display "Too many"

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       System.out.println("Too many.");

   }

// else it uses for loop to print the number of times to display

// Lather and rinse

  else {

       for(int i = 1; i <= numCycles; i++){

           System.out.println(i + ": Lather and rinse.");

       }

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}

Explanation:

The code snippet is written in Java. The method is declared static so that it can be called from another static method. It has a return type of void. It takes an integer as parameter.

It display "Too few" if the passed integer is less than 1. Or it display "Too much" if the passed integer is more than 4. Else it uses for loop to display "Lather and rinse" based on the passed integer.

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3 years ago
There are two identical oil tanks. The level of oil in Tank A is 12 ft and is drained at the rate of 0.5 ft/min. Tank B contains
Luba_88 [7]

Answer:

  16 minutes

Explanation:

This is an example of a class of problems in which two quantities start with different initial values and change at different rates. In such problems, the rates of change are generally ones that cause the values to converge.

The question usually asks when the values will be the same. The generic answer is, "when the difference in rates makes up the difference in initial values."

Here the tanks differ in initial fill height by 12 -8 = 4 ft. The rates of change differ by 0.5 -0.25 = 0.25 ft/min. The more filled tank is draining faster (important), so the fill heights will converge after ...

  (4 ft)/(0.25 ft/min) = 16 min

The level in the two tanks will be the same after 16 minutes.

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<em>Additional comment</em>

The oil levels at that time will be 4 ft.

You can write two equations for height:

  y = 12 -0.5x . . . . . . . height in feet after x minutes (tank A)

  y = 8 -0.25x . . . . . .  height in feet after x minutes (tank B)

These will be equal when ...

  y = y

  12 -0.5x = 8 -0.25x

  4 = 0.25x . . . . . . . . . . add 0.5x -8

  16 = x . . . . . . . . . . . . multiply by 4 . . . . time to equal height

The graph shows when the tanks will have equal heights and when they will be drained.

4 0
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Explanation:

As a general rule of thumb, the large the diameter of a bearing, bushing or pin, the larger the tolerance range,” Brieschke points out. “The inverse is true for smaller-diameter pieces.”

Mike Brieschke, vice president of sales at Aries Engineering, says a 0.25-inch-diameter metal dowel that is press-fit into a mild steel hole usually has an interference of ±0.0015 inch. Parts in noncritical assemblies tend to have looser tolerances

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