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3 years ago

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Consider an NACA 2412 airfoil with a 2-m chord in an airstream with a velocity of 50 m/s at standard sea level conditions. If th

e lift per unit span is 1353 N/m, what is the angle of attac

1 answer:

Akimi4 [234]3 years ago

8 0

Answer:

4°

Explanation:

Given,

chord in the airstream, c = 2 m

speed, v = 50 m/s

lift per unit span = 1353 N/m

density of air, ρ = 1.225 kg/m³

Lift, $L = c_L \times \dfrac{1}{2} \rho v^2(c \times s)$

$\dfrac{L}{s}=c_L \times \dfrac{1}{2} \rho v^2\times c$

$50=c_L \times 0.5\times 1.225\times 50^2\times 2$

$c_L = 0.44$

From the plot of c_L and angle of attack

Angle of attack is equal to 4°

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Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 168 cm , but its circumference is decreasi

DedPeter [7]

Answer:

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Explanation:

We are given that

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Circumference,C= $2\pi r$

$r=\frac{C}{2\pi}$

$r=\frac{168}{2\pi}$ cm

$\frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{1}{2\pi}(-15)=-\frac{15}{2\pi} cm/s$

$\int dr=-\int \frac{15}{2\pi}dt$

$r=-\frac{15}{2\pi}t+C$

When t=0

$r=\frac{168}{2\pi}$

$\frac{168}{2\pi}=C$

$r=-\frac{15}{2\pi}t+\frac{168}{2\pi}$

E= $-\frac{d\phi}{dt}=-\frac{d(B\pi r^2)}{dt}=-2\pi rB\frac{dr}{dt}$

$E=-2\pi(-\frac{5}{2\pi}t+\frac{168}{2\pi})B\times -\frac{15}{2\pi}$

t=8 s

B=0.9

$E=2\pi\times \frac{15}{2\pi}\times 0.9(-\frac{15}{2\pi}(8)+\frac{168}{2\pi})$

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For a moon orbiting its planet, rp is the shortest distance between the moon and its planet andra is the longest distance betwee

Natasha2012 [34]

Answer: D. 0.57

Explanation:

The formula to calculate the eccentricity $e$ of an ellipse is (assuming the moon's orbit in the shape of an ellipse):

$e=\frac{r_{a}-r_{p}}{r_{a}+r_{p}}$

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$r_{a}$ is the apoapsis (the longest distance between the moon and its planet)

$r_{p}=0.27 r_{a}$ is the periapsis (the shortest distance between the moon and its planet)

Then:

$e=\frac{r_{a}-0.27 r_{a}}{r_{a}+0.27 r_{a}}$

$e=\frac{0.73 r_{a}}{1.27 r_{a}}$

$e=0.57$ This is the moon's orbital eccentricity

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4 years ago

The air in a room has a pressure of 1 atm, a dry-bulb temperature of 24°C, and a wet-bulb temperature of 17°C. Using the psychro

Licemer1 [7]

Answer:

Given that

Dry-bulb temperature(T) =24°C

Wet-bulb temperature(Tw) = 17°C

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As we know that psychrometric chart are drawn at constant pressure.

From the diagram

ω= specific humidity

Lets take these two lines Dry-bulb temperature(T) line and Wet-bulb temperature(Tw) cut at point P

From chart at point P

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b)

The enthalpy ( h)

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MrRissso [65]

Answer:

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alpitude

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3 years ago

Under ideal conditions (no atmospheric interference of any kind), if I hit a golf ball at an angle of 25 degrees at an initial s

g100num [7]

Answer:

The required angle is (90-25)° = 65°

Explanation:

The given motion is an example of projectile motion.

Let 'v' be the initial velocity and '∅' be the angle of projection.

Let 't' be the time taken for complete motion.

Let 'g' be the acceleration due to gravity

Taking components of velocity in horizontal(x) and vertical(y) direction.

$v_{x}$ = v cos(∅)

$v_{y}$ = v sin(∅)

We know that for a projectile motion,

t = $\frac{2vsin(∅)}{g}$

Since there is no force acting on the golf ball in horizonal direction.

Total distance(d) covered in horizontal direction is -

d = $v_{x}$ ×t = vcos(∅)× $\frac{2vsin(∅)}{g}$ = $\frac{v^{2}sin(2∅) }{g}$ .

If the golf ball has to travel the same distance 'd' for same initital velocity v = 23m/s , then the above equation should have 2 solutions of initial angle 'α' and 'β' such that -

α +β = 90° as-

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∴ For the initial angles 'α' or 'β' , total horizontal distance 'd' travelled remains the same.

∴ If α = 25° , then

β = 90-25 = 65°

∴ The required angle is 65°.

5 0

3 years ago

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