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T he area of a circle is pr 2. Define r as 5, then find the area of a circle,using MATLAB®.(b) The surface area of a sphere is 4

aksik [14]

Answer:

Area of Circle = 78.5398

Surface Area of Sphere = 1.2566 x 10^3 = 1256.6 ft

Volume of Sphere = 33.5103 ft

Explanation:

Please find below the written MatLab script used to solve the problem. I had to define r in each case to solve for the Area of the circle, the surface area and the volume of the Sphere.

r=5; % define r as 5

a=pi*r^2;% calculate the area of the circle

AreaOfCircle=a

r=10; % define r and 10 ft

sa=4*pi*r^2; %Calculate the surface area of the sphere

SphereSurfaceArea=sa

r=2;% define r as 2 ft

vs=(4/3)*pi*r^3;% Calculate the volume of the sphere

VolumeShere=vs

3 0

3 years ago

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1. What is resilience engineering and how does it pertain to the current<br>corona virus crisis?

saw5 [17]

Answer:

the ability to absorb or avoid damage without suffering complete failure and is an objective of design, maintenance and restoration for buildings and infrastructure

Explanation:

8 0

3 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

g . 3. For each of the following statements Write the statement as an English sentence that does not use the symbols for quan 2.

lubasha [3.4K]

Answer:

(a)

( ∃x ∈ Q) ( x > √2)

There exists a rational number x such that x > √2.

( ∀x ∈ Q) ( ( x ≤ √2)

For each rational number x, x ≤ √2.

(b)

(∀x ∈ Q)(x² - 2 ≠ 0).

For all rational numbers x, x² - 2 ≠ 0

( ∃x ∈ Q ) ( x² - 2 = 0 )

There exists a rational number x such that x² - 2 = 0

(c)

(∀x ∈ Z)(x is even or x is odd).

For each integer x, x is even or x is odd.

( ∃x ∈ Z ) (x is odd and x is even)

There exists an integer x such that x is odd and x is even.

(d)

( ∃x ∈ Q) ( √2 < x < √3 )

There exists a rational number x such that √2 < x < √3

(∀x ∈ Q) ( x ≤ √2 or x ≥ √3 )

For all rational numbers x, x ≤ √2 or x ≥ √3.

3 0

3 years ago

If 4 squares equal 1 foot, how many inches is one square?

xeze [42]

There is 3 inches in one square

6 0

3 years ago

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