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vfiekz [6]
3 years ago
14

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
Physics
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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A particle is moving with velocity v(t) = t2 _ 9t + 18 with distance, s measured in meters, left or right of zero, and t measure
Alex787 [66]
V = t^2 - 9t + 18

position, s
s = t^3 /3 - 4.5t^2 +18t + C

       t = 0, s = 1 => 1=C => s = t^3/3 -4.5t^2 + 18t + 1

Average velocity: distance / time

   distance: t = 8 => s = 8^3 / 3 - 4.5 (8)^2 + 18(8) + 1 = 27.67 m
   Average velocity = 27.67 / 8 = 3.46 m/s

t = 5 s

     v = t^2 - 9t + 18 = 5^2 - 9(5) + 18 = -2 m/s
     speed = |-2| m/s = 2 m/s
 
Moving right
     V > 0 => t^2 - 9t + 18 > 0
     (t - 6)(t - 3) > 0

     => t > 6 and t > 3 => t > 6 s => Interval (6,8)

    => t < 6 and t <3 => t <3 s => interval (0,3)

    

Going faster and slowing dowm

acceleration, a = v' = 2t - 9
     a > 0 => 2t - 9 > 0 => 2t > 9 => t > 4.5 s
     Then, going faster in the interval (4.5 , 8) and slowing down in (0, 4.5)
     


4 0
3 years ago
4. Assume a multiple level queue system with a variable time quantum per queue, and that the incoming job needs 50ms to run to c
Troyanec [42]

Answer:

Explanation:

For the completion of incoming job it will take 50ms

First queue takes 5ms quantum time and the subsequent queue takes double of the previous question

So,

First queue T_1 = 5ms

Second queue T_2 = 2 × T_1 = 2 × 5 = 10ms

Third queue T_3 = 2 × T_2 = 2 × 10 = 20ms

Fourth queue T_4 = 2 × T_3 = 2 × 20 = 40ms

Fifth queue T_5= 2 × T_4 = 2 × 40 = 80ms.

Now, the job will be done after the fifth queue.

So, after the first queue, the job is not completed, so, we have first interruption

After the second queue, the job is not completed, so, we have second interruption

After the third queue, the job is not completed, so we have third interruption.

After the fourth queue, the job is not yet completed, so we have the fourth interruption

And in the fifth queue the job is completed, so we don't have any interruption here.

So, the job will be interrupted 4 times and it will finished on the fifth queue.

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3 years ago
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What is the word for a group of stars that’s not a constellation?.
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7 0
2 years ago
A barometer accidentally contains 6.5 inches of water on top of the mercury column (so there is also water vapor instead of a va
Alisiya [41]

Answer:

(a). 14.4 lbf/in^2.

(b). 27.8 in, AS THE TEMPERATURE INCREASES, THE LENGTH OF MERCURY DECREASES.

Explanation:

So, from the question above we are given the following parameters which are going to help us in solving this particular Question;

=> The "barometer accidentally contains 6.5 inches of water on top of the mercury column (so there is also water vapor instead of a vacuum at the top of the barometer)"

=> "On a day when the temperature is 70oF, the mercury column height is 28.35 inches (corrected for thermal expansion)."

With these knowledge, let us delve right into the solution;

(a). The barometric pressure = water vapor pressure + acceleration due to gravity (ft/s^2) × water density(slug/ft^3) × {ft/12 in}^3 × [ height of mercury column + specific gravity of mercury × height of water column].

The barometric pressure= 0.363 + {(62.146) ÷ (12^3) × 390.6425}. = 14.4 lbf/in^2.

(b). { (13.55 × length of mercury) + 6.5 } × (62.15÷ 12^3) = 14.4 - 0.603.

Length of mercury = 27.8 in.

AS THE TEMPERATURE INCREASES, THE LENGTH OF MERCURY DECREASES.

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