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3 years ago

When Ontario

Engineering

1 answer:

Marat540 [252]3 years ago

4 0

Answer:

Explanation:

This will be possible when setting them up in summer with a certain quantity of sag, they have already know that the cables won't be able to sag any further because of the heat. During winter, when the cables contract because of the cold weather, the sag will therefore be reduced, but much tension will not be put on the cables.

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Write cout statements with stream manipulators that perform the following:

Semenov [28]

Answer:

A)cout<<setw(9)<<fixed<<setprecision(2)<<34.789;

B)cout<<setw(5)<<fixed<<setprecision(3)<<7.0;

C)cout<<fixed<<5.789E12;

D)cout<<left<<setw(7)<<67;

Explanation:

Stream Manipulators are functions specifically designed to be used in conjunction with the insertion (<<) and extraction (>>) operators on stream objects in C++ programming while the 'cout' statement is used to display the output of a C++to the standard output device.

setw: used to specify the minimum number of character positions on the output field

setprecision: Sets the decimal precision to be used to format floating-point values on output operations.

fixed: is used to set the floatfield format flag for the specified str stream.

left: adjust output to the left.

A) To display the number 34.789 in a field of eight spaces with two decimal places of precision. cout<<setw(9)<<fixed<<setprecision(2)<<34.789;

B) To display the number 7.0 in a field of six spaces with three decimal places of precision. cout<<setw(5)<<fixed<<setprecision(3)<<7.0;

C) To print out the number 5.789e+12 in fixed-point notation. cout<<fixed<<5.789E12;

(D) To display the number 67 left-justified in a field of six spaces. cout<<left<<setw(7)<<67;

7 0

4 years ago

Construction lines are thick lines true false

ANTONII [103]

Answer:

true

Explanation:

4 0

3 years ago

Read 2 more answers

How to tie shoe please

Genrish500 [490]

Step 1: Unknot Shoelaces. Make sure that the two ends of the shoelace are completely untangled and free of knots past the tongue. ...
Step 2: Create Overhand Knot. ...
Step 3: Finish the Basic Shoe Knot. ...
Step 4: Create Double Knot. ...
Step 5: Finished!

4 0

2 years ago

An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 278C, and 75

Inessa [10]

Answer:

(a). The value of temperature at the end of heat addition process $T_{3}$ = 2042.56 K

(b). The value of pressure at the end of heat addition process $P_{3}$ = 1555.46 k pa

(c). The thermal efficiency of an Otto cycle $E_{otto}$ = 0.4478

(d). The value of mean effective pressure of the cycle $P_{m}$ = 1506.41 $\frac{k pa}{kg}$

Explanation:

Compression ratio $r_{p}$ = 8

Initial pressure $P_{1}$ = 95 k pa

Initial temperature $T_{1}$ = 278 °c = 551 K

Final pressure $P_{2}$ = 8 × $P_{1}$ = 8 × 95 = 760 k pa

Final temperature $T_{2}$ = $T_{1}$ × $r_{p} ^{\frac{\gamma - 1}{\gamma} }$

Final temperature $T_{2}$ = 551 × $8 ^{\frac{1.4 - 1}{1.4} }$

Final temperature $T_{2}$ = 998 K

Heat transferred at constant volume Q = 750 $\frac{KJ}{kg}$

(a). We know that Heat transferred at constant volume $Q_{S}$ = $m C_{v} ( T_{3} - T_{2} )$

⇒ 1 × 0.718 × ( $T_{3}$ - 998 ) = 750

⇒ $T_{3}$ = 2042.56 K

This is the value of temperature at the end of heat addition process.

Since heat addition is constant volume process. so for that process pressure is directly proportional to the temperature.

⇒ P ∝ T

⇒ $\frac{P_{3} }{P_{2} }$ = $\frac{T_{3} }{T_{2} }$

⇒ $P_{3}$ = $\frac{2042.56}{998}$ × 760

⇒ $P_{3}$ = 1555.46 k pa

This is the value of pressure at the end of heat addition process.

(b). Heat rejected from the cycle $Q_{R} = m C_{v} ( T_{4} - T_{1} )$

For the compression and expansion process,

⇒ $\frac{T_{3} }{T_{2} } = \frac{T_{4} }{T_{1} }$

⇒ $\frac{2042.56}{998} = \frac{T_{4} }{551}$

⇒ $T_{4}$ = 1127.7 K

Heat rejected $Q_{R}$ = 1 × 0.718 × ( 1127.7 - 551)

⇒ $Q_{R}$ = 414.07 $\frac{KJ}{kg}$

Net heat interaction from the cycle $Q_{net}$ = $Q_{S}$ - $Q_{R}$

Put the values of $Q_{S}$ & $Q_{R}$ we get,

⇒ $Q_{net}$ = 750 - 414.07

⇒ $Q_{net}$ = 335.93 $\frac{KJ}{kg}$

We know that for a cyclic process net heat interaction is equal to net work transfer.

⇒ $Q_{net}$ = $W_{net}$

⇒ $W_{net}$ = 335.93 $\frac{KJ}{kg}$

This is the net work output from the cycle.

(c). Thermal efficiency of an Otto cycle is given by

$E_{otto} = 1- \frac{T_{1} }{T_{2} }$

Put the values of $T_{1}$ & $T_{2}$ in the above formula we get,

$E_{otto} = 1- \frac{551 }{998 }$

⇒ $E_{otto}$ = 0.4478

This is the thermal efficiency of an Otto cycle.

(d). Mean effective pressure $P_{m}$ :-

We know that mean effective pressure of the Otto cycle is given by

$P_{m}$ = $\frac{W_{net} }{V_{s} }$ ---------- (1)

where $V_{s}$ is the swept volume.

$V_{s}$ = $V_{1} - V_{2}$ ---------- ( 2 )

From ideal gas equation $P_{1}$ $V_{1}$ = m × R × $T_{1}$

Put all the values in above formula we get,

⇒ 95 × $V_{1}$ = 1 × 0.287 × 551

⇒ $V_{1}$ = 0.6 $m^{3}$

From the same ideal gas equation

$P_{2}$ $V_{2}$ = m × R × $T_{2}$

⇒ 760 × $V_{2}$ = 1 × 0.287 × 998

⇒ $V_{2}$ = 0.377 $m^{3}$

Thus swept volume $V_{s}$ = 0.6 - 0.377

⇒ $V_{s}$ = 0.223 $m^{3}$

Thus from equation 1 the mean effective pressure

⇒ $P_{m}$ = $\frac{335.93}{0.223}$

⇒ $P_{m}$ = 1506.41 $\frac{k pa}{kg}$

This is the value of mean effective pressure of the cycle.

4 0

3 years ago

A good rule of thumb is to design the horizontal stabilizer so that its area is about 1/6 to 1/8 of the area of the wing. If the

erik [133]

Answer:

Stabilizer with area = 14 cm2

Explanation:

i dont know why but its right

4 0

3 years ago