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SCORPION-xisa [38]

3 years ago

13

Wave flow of an incompressible fluid into a solid surface follows a sinusoidal pattern. Flow is two-dimensional with the x-axis

normal to the surface and y axis along the wall. The x component of the flow follows the pattern
u = Ax sin (2πt/T)
Determine the y-component of flow (v) and the convective and local components of the acceleration vector.

1 answer:

Artyom0805 [142]3 years ago

4 0

Answer:

sorry , for my point

Explanation:

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What will the following segment of code output? score = 95; if (score > 95) cout << "Congratulations!\n"; cout <<

Anarel [89]

Answer:

That's a high score!

This is a test question!

Explanation:

The reason these two lines are printed and not the first one is simple. After the 'IF' condition has been stated, there is no use of parenthesis such as { and } to enclose the next lines. This means that only the first line after the 'IF' condition may be read or skipped depending on whether the condition (score>95) is met. Since the score is not larger than 95, and the 'IF' condition fails, the line 'Congratulations!' is not printed. The next two lines of the code are read as normal because they do not depend on the 'IF' condition.

5 0

2 years ago

: A pneumatic "cannon" is a device that launches a low mass projectile from a cylindrical tube using pressurized air stored upst

alukav5142 [94]

Answer:

Work done = 125π J

Explanation:

Given:

P = P_i * ( 1 - (x/d)^2 / 25)

d = 5.0 cm

x = 5 * d cm = 25d

Pi = 12 bar

Work done = integral ( F . dx )

F (x) = P(x) * A

F (x) = (πd^2 / 4) * P_i * (1 - (x/d)^2 / 25)

Work done = integral ((πd^2 / 4) * P_i * (1 - (x/d)^2 / 25) ) . dx

For Limits 0 < x < 5d

Work done = (πd^2 / 4) * P_i integral ( (1 - (x/d)^2) / 25)) . dx

Integrate the function wrt x

Work done = (πd^2 / 4) * P_i * ( x - d*(x/d)^3 / 75 )

Evaluate Limits 0 < x < 5d :

Work done = (πd^2 / 4) * P_i * (5d - 5d / 3)

Work done = (πd^2 / 4) * P_i * (10*d / 3)

Work done = (5 π / 6)d^3 * P_i

Input the values:

Work done = (5 π / 6)(0.05)^3 * (1.2*10^6)

Work done = 125π J

5 0

2 years ago

A venturi meter is to be installed in a 63 mm bore section of a piping system to measure the flow rate of water in it. From spac

Alex Ar [27]

Answer:

Throat diameter $d_2$ =28.60 mm

Explanation:

Bore diameter $d_1=63mm$ ⇒ $A_1=3.09\times 10^{-3} m^2$

Manometric deflection x=235 mm

Flow rate Q=240 Lt/min⇒ Q=.004 $\frac{m^3}{s}$

Coefficient of discharge $C_d$ =0.8

We know that discharge through venturi meter

$Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}$

$h=x(\dfrac{S_m}{S_w}-1)$

$S_m$ =13.6 for Hg and $S_w$ =1 for water.

$h=0.235(\dfrac{13.6}{1}-1)$

h=2.961 m

Now by putting the all value in

$Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}$

$0.004=0.8\times \dfrac{3.09\times 10^{-3} A_2\sqrt{2\times 9.81\times 2.961}}{\sqrt{(3.09\times 10^{-3})^2-A_2^2}}$

$A_2=6.42\times 10^{-4} m^2$

⇒ $d_2$ =28.60 mm

So throat diameter $d_2$ =28.60 mm

4 0

3 years ago

Calculate the equivalent capacitance of the three series capacitors in Figure 12-1

GrogVix [38]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

Calculate the equivalent capacitance of the three series capacitors in Figure 12-1

a) 0.01 μF

b) 0.58 μF

c) 0.060 μF

d) 0.8 μF

Answer:

The equivalent capacitance of the three series capacitors in Figure 12-1 is 0.060 μF

Therefore, the correct option is (c)

Explanation:

Please refer to the attached Figure 12-1 where three capacitors are connected in series.

We are asked to find out the equivalent capacitance of this circuit.

Recall that the equivalent capacitance in series is given by

$\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$

Where C₁, C₂, and C₃ are the individual capacitance connected in series.

C₁ = 0.1 μF

C₂ = 0.22 μF

C₃ = 0.47 μF

So the equivalent capacitance is

$\frac{1}{C_{eq}} = \frac{1}{0.1} + \frac{1}{0.22} + \frac{1}{0.47}$

$\frac{1}{C_{eq}} = \frac{8620}{517}$

$C_{eq} = \frac{517}{8620}$

$C_{eq} = 0.0599$

Rounding off yields

$C_{eq} = 0.060 \: \mu F$

The equivalent capacitance of the three series capacitors in Figure 12-1 is 0.060 μF

Therefore, the correct option is (c)

5 0

3 years ago

The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

$\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0$ Equation 1

Similarly, the equation for outer node “o” will be

$\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0$ Equation 2

The conventive thermal resistance in i-node will be

$R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w$ Equation 3

The conventive hermal resistance per unit area is

$R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w$ Equation 4

The conductive thermal resistance per unit area is

$R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w$ Equation 5

Since $q_{rad}$ is given as 100, $T_{o}$ is 40 $T_ \infty$ is 300 $T_{\infty, o}$ is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

$\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0$ Equation 6

$\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0$

$T_{i}-40= \frac {L}{0.05}*150$

$T_{i}-40=3000L$

$T_{i}=3000L+40$ Equation 7

From equation 6 we can substitute wherever there’s $T_{i}$ with 3000L+40 as seen in equation 7 hence we obtain

$\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0$

The above can be simplified to be

$\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0$

$\frac {260-3000L}{0.033}=50$

-3000L=1.665-260

$L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm$

Therefore, insulation thickness is 86mm

8 0

3 years ago