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OleMash [197]
2 years ago
4

It is desired to produce and aligned carbon fiber-epoxy matrix composite having a longitudinal tensile strength of 800 MPa. Calc

ulate (a) the critical fiber length (in mm), and (b) the volume fraction of fibers necessary if (1) the average fiber diameter is 0.017 mm, (2) the average fiber length is 2.3 mm, (3) the fiber fracture strength is 5500 MPa, (4) the fiber-matrix bond strength is 18 MPa, (5) the matrix stress at fiber failure is 13.5 MPa.
Engineering
1 answer:
aliya0001 [1]2 years ago
5 0

Answer:

A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

Explanation:

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Michelle is the general manager of a power plant. This morning, she will meet with city officials to discuss environmental issue
Irina-Kira [14]

Answer:

interpersonal.

Explanation:

Out of all the activities performed by Michelle, three activities involves the interpersonal skills.

1. Meeting with city officials

2. Meeting with section managers

3. Handling the complaint filed by an employee

All these activities involves interpersonal skills. Hence, we can say that she had spent her most of the day by using the interpersonal skills.

6 0
2 years ago
Your task is to fill in the missing parts of the C code to get a program equivalent to the generated assembly code. Recall that
Rudik [331]

Answer:

See Explaination

Explanation:

//Function

long loop (long x, long n)

{

//Declare a variable named result and initialize it to zero

long result = 0;

//Declare a variable named mask

long mask;

//For loop

for(mask = 1; mask != 0; mask = mask << (n & 0xFF))

{

//Calculate

result | = (x&mask);

}

//Return result

return result;

}

6 0
2 years ago
A thermal energy storage unit consists of a large rectangular channel, which is well insulated on its outer surface and encloses
yaroslaw [1]

Answer:

the temperature of the aluminum at this time is 456.25° C

Explanation:

Given that:

width w of the aluminium slab = 0.05 m

the initial temperature T_1 = 25° C

T{\infty} =600^0C

h = 100 W/m²

The properties of Aluminium at temperature of 600° C by considering the conditions for which the storage unit is charged; we have ;

density ρ = 2702 kg/m³

thermal conductivity k = 231 W/m.K

Specific heat c = 1033 J/Kg.K

Let's first find the Biot Number Bi which can be expressed by the equation:

Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{h \dfrac{w}{2}}{k}

Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{100 \times \dfrac{0.05}{2}}{231}

Bi = \dfrac{2.5}{231}

Bi = 0.0108

The time constant value \tau_t is :

\tau_t = \dfrac{pL_cc}{h} \\ \\ \tau_t = \dfrac{p \dfrac{w}{2}c}{h}

\tau_t = \dfrac{2702* \dfrac{0.05}{2}*1033}{100}

\tau_t = \dfrac{2702* 0.025*1033}{100}

\tau_t = 697.79

Considering Lumped capacitance analysis since value for Bi is less than 1

Then;

Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]

where;

Q = -\Delta E _{st} which correlates with the change in the internal energy of the solid.

So;

Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]= -\Delta E _{st}

The maximum value for the change in the internal energy of the solid  is :

(pVc)\theta_1 = -\Delta E _{st}max

By equating the two previous equation together ; we have:

\dfrac{-\Delta E _{st}}{\Delta E _{st}{max}}= \dfrac{  (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]} { (pVc)\theta_1}

Similarly; we need to understand that the ratio of the energy storage to the maximum possible energy storage = 0.75

Thus;

0.75=  [1-e^{\dfrac {-t}{ \tau_1}}]}

So;

0.75=  [1-e^{\dfrac {-t}{ 697.79}}]}

1-0.75=  [e^{\dfrac {-t}{ 697.79}}]}

0.25 =  e^{\dfrac {-t}{ 697.79}}

In(0.25) =  {\dfrac {-t}{ 697.79}}

-1.386294361= \dfrac{-t}{697.79}

t = 1.386294361 × 697.79

t = 967.34 s

Finally; the temperature of Aluminium is determined as follows;

\dfrac{T - T _{\infty}}{T_1-T_{\infty}}= e ^ {\dfrac{-t}{\tau_t}}

\dfrac{T - 600}{25-600}= e ^ {\dfrac{-967.34}{697.79}

\dfrac{T - 600}{25-600}= 0.25

\dfrac{T - 600}{-575}= 0.25

T - 600 = -575 × 0.25

T - 600 = -143.75

T = -143.75 + 600

T = 456.25° C

Hence; the temperature of the aluminum at this time is 456.25° C

3 0
2 years ago
A dielectric material, such as Teflon®, is placed between the plates of a parallel-plate capacitor without altering the structur
Lina20 [59]

Answer: The electric field decreases because of the insertion of the Teflon.

Explanation:

If the charge on the capacitor is held fixed, the electric field as a consequence of this charge distribution (directed from the positive charged plate to the negative charged one remains unchanged.

However, as the Teflon is a dielectric material, even though doesn't allow the free movement of the electrons as an answer to an applied electric field, it allows that the electrons be displaced from the equilibrium position, leaving a local negative-charged zone close to the posiitive plate of the capacitor, and an equal but opposite charged layer close to the negative plate.

In this way, a internal electric field is created, that opposes to the external one due to the capacitor, which overall effect is diminishing the total electric field, reducing the voltage between the plates, and  increasing the capacitance proportionally to the dielectric constant of the Teflon.  

8 0
2 years ago
Sharon is designing a house in an area that receives a lot of rainfall all year. Which material should she use to stick the wood
kakasveta [241]

Explanation:

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4 0
2 years ago
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