An inflatable structure has the shape of a half-circular cylinder with hemispherical ends. The structure has a radius of 40 ft when inflated to a pressure of 0.60 psi. A longitudinal seam runs the entire length of the structure. The seam fails in tension when the load is 600 pounds per inch of seam. What is the factor of safety with respect to longitudinal seam failure?

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Answer 1

Answer:

Find the given attachment

Related Questions

8.2.1: Function pass by reference: Transforming coordinates. Define a function CoordTransform() that transforms the function's first two input parameters xVal and yVal into two output parameters xValNew and yValNew. The function returns void. The transformation is new

Answers

Answer:

The output will be (3, 4) becomes (8, 10)

Explanation:

#include <stdio.h>

//If you send a pointer to a int, you are allowing the contents of that int to change.

void CoordTransform(int xVal,int yVal,int* xNew,int* yNew){

*xNew = (xVal+1)*2;

*yNew = (yVal+1)*2;

}

int main(void) {

int xValNew = 0;

int yValNew = 0;

CoordTransform(3, 4, &xValNew, &yValNew);

printf("(3, 4) becomes (%d, %d)\n", xValNew, yValNew);

return 0;

}

A conical funnel of half-angle θ = 30 drains through a small hole of diameter d = 6:25 mm at the vertex. The speed of the liquid leaving the funnel is V= √ 2gy where y is the height of the liquid free surface above the hole. The funnel initially is filled to height y0 = 300 mm. Obtain an expression for the time, t, for the funnel to completely drain, and evaluate. Find the time to drain from 300 mm to 150 mm (a change in depth of 150 mm), and from 150 mm to completely empty (also a change in depth of 150 mm). Can you explain the discrepancy in these times?

Answers

Answer:

$3175.424*y^3 + 0.45153*y + 55.552*y^2 = t + 12.0347055$

t = 12.03

t = 81.473

velocity of the fluid decreases with level of fluid. Hence, in later stages the time taken is greater than in earlier stages

Explanation:

Given:

- The half angle θ = 30°

- The diameter of the small hole d = 6.25 mm

- The flow rate out of the funnel Q = A*√ 2gy

- The volume of frustum of cone is given by:

$V = (\pi )/(12)*y*(D^2 + d^2 + d*D})$

Where,

D: Is the larger diameter of the frustum

d: Is the smaller diameter of the frustum

y: The height of the liquid free surface from small diameter d base.

Find:

- Obtain an expression for the time, t, for the funnel to completely drain, and evaluate.

- Find the time to drain from 300 mm to 150 mm (a change in depth of 150 mm) and from 150 mm to completely empty (also a change in depth of 150 mm)

- Can you explain the discrepancy in these times?

Solution:

- We will use rate of change analysis by considering the rate of change of volume, and then apply the chain rule.

- The Volume of the frustum is a function of d , D , y. V = f ( d , D , y ). In our case the diameter of base d remains constant. Then we are left with:

V = f ( D , y )

- Determine a relationship between y and d. We will use half angle θ to determine the relationship between D and y by applying trigonometric ratios:

tan ( θ ) = D / 2*y

D = 2*y*tan ( θ )

- We developed a relationship between D and y in terms of half angle which remains constant. So the Volume is now only a function of one variable y:

V = f ( y )

- The volume of frustum of the cone can be written as:

$V = (\pi )/(12)*y*(D^2 + d^2 + d*D})$

Substituting the relationship for D in terms of y we have:

$V = (\pi )/(12)*y*(4*y^2*tan(Q) ^2 + d^2 + 2*d*y*tan(Q)})$

- Now by rate of change of Volume analysis we have:

dV / dt = [dV / dy] * [dy / dt]

- Computing dV / dy, where V = f(y) only:

$V = (\pi )/(12)*(4*y^3*tan(Q) ^2 + y*d^2 + 2*d*y^2*tan(Q)})\n\n(dV)/(dy) = (\pi )/(12)*(12*y^2*tan(Q) ^2 + d^2 + 4*d*y*tan(Q)})\n\n$

- Where, dV/dt = Volume flow rate:

$(dV)/(dt) = - Q\n(dV)/(dt) = - A*V\n(dV)/(dt) = - (\pi*d^2 )/(4) *√(2*g*y)$

- Then from Chain rule we have:

[dy / dt] = [dV / dt] / [dV / dy]

$(dy)/(dt) = (- (\pi*d^2 )/(4) *√(2*g*y))/((\pi )/(12)*(12*y^2*tan(Q) ^2 + d^2 + 4*d*y*tan(Q))\n\n} \n(dy)/(dt) = (-d^2 *√(2*g*y))/((4*y^2*tan(Q) ^2 + d^2 + 4*d*y*tan(Q))\n\n}$

- Separate variables:

$\frac {(4*y^2*tan(Q) ^2 + d^2 + 4*d*y*tan(Q)} {√(2*g*y)} .dy = {-d^2} .dt\n\n\frac { 1 } { √(2*g) }*(4*y^1^.^5*tan(Q) ^2 + y^-^0^.^5d^2 + 4*d*y^0^.^5*tan(Q))= {-d^2} .dt\n$

- Integrate both sides:

$\frac { 1 } { √(2*g) }*(4*y^1^.^5*tan(Q) ^2 + y^-^0^.^5d^2 + 4*d*y^0^.^5*tan(Q))= {-d^2} .dt\n\n\frac { 1 } { √(2*g) }*((8)/(5)*y^2^.^5*tan(Q) ^2 + 2*d^2*y^0^.^5 + (8)/(3)*d*y^1^.^5*tan(Q)) = -d^2*t + C\n\n\frac { 1 } { √(2*y*g) }*((8)/(5)*y^3*tan(Q) ^2 + 2*d^2*y + (8)/(3)*d*y^2*tan(Q)) = -d^2*t + C\n\n0.1204*y^3 + 0.000017638*y + 0.00217*y^2 = -0.0000390625*t + C\n\n3175.424*y^3 + 0.45153*y + 55.552*y^2 = -t + C$

- Evaluate @ t = 0 , y = 0.3 m

$3175.424*(0.15)^3 + 0.45153*(0.15) + 55.552*(0.15)^2 = 0 + C\n\nC = 12.0347055\n\n3175.424*y^3 + 0.45153*y + 55.552*y^2 = t + 12.0347055$

- Time taken from y = 300 to 150 mm:

$3175.424*0.15^3 + 0.45153*0.15 + 55.552*0.15^2 = -t + 90.871587\n\nt = 0 - -12.0347055\n\nt = 12.0347055 s\n$

- Time taken from y = 150 to 0 mm:

t = t_(0-300) - ( t_(0-150) = 90.871587 - 12.0347055 = 81.4721 s

- The discrepancy is time can be explained by the velocity of fluid coming out of bottom is function of y. The velocity of the fluid decreases as the level of fluid decreases hence the time taken from y =150 mm to 0 mm is larger than y = 300 mm to y = 150 mm.

If the total energy change of an system during a process is 15.5 kJ, its change in kinetic energy is -3.5 kJ, and its potential energy is unchanged, calculate its change in specificinternal energy if its mass is 5.4 kg. Report your answer in kJ/kg to one decimal place.

Answers

Answer:

The change in specific internal energy is 3.5 kj.

Explanation:

Step1

Given:

Total change in energy is 15.5 kj.

Change in kinetic energy is –3.5 kj.

Change in potential energy is 0 kj.

Mass is 5.4 kg.

Step2

Calculation:

Change in internal energy is calculated as follows:

$\bigtriangleup E=\bigtriangleup KE+\bigtriangleup PE+\bigtriangleup U$ $15.5=-3.5+0+\bigtriangleup U$

$\bigtriangleup U=19$ kj.

Step3

Specific internal energy is calculated as follows:

$\bigtriangleup u=(\bigtriangleup U)/(m)$

$\bigtriangleup u=(19)/(5.4)$

$\bigtriangleup u=3.5$ kj/kg.

Thus, the change in specific internal energy is 3.5 kj/kg.

In contrasting the read-evaluation loop and the notification-based paradigm for inter- active programs, construction of a pre-emptive dialog was discussed. How would a programmer describe a pre-emptive dialog by purely graphical means? (Hint: Refer to the discussion in Sec- tion 8.5 concerning the shift from external and independent dialog management to presentation control of the dialog)

Answers

The way a programmer describe a pre-emptive dialog by purely graphical means is; by producing a window that covers the entire screen to make it the currently selected window.

What is Pre - emptive Dialogue?

In a graphics - based interaction, it is supposed that the user can only interact with parts of the system that are visible. However, In a windowing system, the user can only direct input to a single window that was currently selected and the way to change that selected window is to indicate with some gesture within that window.

Finally, to create a pre-emptive dialog, the system would do so through the production of a window that covers the entire screen to make it the currently selected window. Thereafter, all user input would be directed to that window and the user would have no means of selecting any other window. Then the covering window will now pre-empt any other user action with the exception of that which it is defined to support.

Read more about dialogue at; brainly.com/question/5962406

Answer:

In an illustrations based communication, it is expected that the client can just associate with parts of the framework that are obvious. In a windowing framework, for instance, the client can just direct contribution to a solitary, at present chosen window, and the main methods for changing the chose window would be by demonstrating with some signal inside that window. To make a preemptive exchange, the framework can create a window that covers the whole screen and make it the right now chosen window. All client information would then be coordinated to that window and the client would have no methods for choosing another window. The 'covering' window in this way preempts some other client activity with the exception of that which it is characterized to help

A site is compacted in the field, and the dry unit weight of the compacted soil (in the field) is determined to be 18 kN/m3. Determine the relative compaction if the maximum dry unit weight was measured to be 17 kN/m3. Express your answer as a percentage (but do not write the percentage sign in the answer box).

Answers

Answer:

the relative compaction is 105.88 %

Explanation:

Given;

dry unit weight of field compaction, $W_d_((field))$ = 18 kN/m³

maximum dry unit weight measured, $W_d_((max))$ = 17 kN/m³

Relative compaction (RC) of the site is given as the ratio of dry unit weight of field compaction and maximum dry unit weight measured

Relative compaction (RC) = dry unit weight of field compaction / maximum dry unit weight measured

$RC = (W_d_((field)))/(W_d_((max)))$

substitute the given values;

$RC = (18)/(17) = 1.0588$

RC (%) = 105.88 %

Therefore, the relative compaction is 105.88 %

The Clausius inequality expresses which of the following laws? i. Law of Conservation of Mass ii. Law of Conservation of Energy iii. First Law of Thermodynamics iv. Second Law of Thermodynamics

Answers

Answer:

(iv) second law of thermodynamics

Explanation:

The Clausius inequality expresses the second law of thermodynamics it applies to the real engine cycle.It is defined as the cycle integral of change in entropy of a reversible system is zero. It is nothing but mathematical form of second law of thermodynamics . It also states that for irreversible process the cyclic integral of change in entropy is less than zero

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