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Mandarinka [93]
3 years ago
14

For a given wing–body combination, the aerodynamic center lies 0.03 chord length ahead of the center of gravity. The moment coef

ficient about the center of gravity is 0.0050, and the lift coefficient is 0.50. Calculate the moment coefficient about the aerodynamic center.
Physics
1 answer:
Aloiza [94]3 years ago
5 0

Answer:

-0.01

Explanation:

<u>given:</u>

C_l=0.5

C_gravity=0.0050

C_chord= 0.50

<u>solution:</u>

C_moment=C_chord+C_l(h-h_ac)

                 =0.0050-0.5(0.03)

                 = -0.01

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By how much would its speed reading increase with each second of fall? ... Ex 3.24 For a freely falling object dropped from rest, what is its acceleration at the end of the 5th second ... Pb 3.3 A ball is thrown straight up with an initial speed of 30 m/s. How high does it go, and how long is it in the air (neglecting air resistance)?.


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3 years ago
Which of these factors will increase the speed of a sound wave in air?
Artemon [7]
<h3>Answer;</h3>

-Temperature

<h3><u>Explanation;</u></h3>
  • Sound is a type of mechanical wave, which means it requires a material medium for transmission. It results from the vibration of particles.
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3 years ago
Calculate the pressure exerted on the floor by the boy standing on both feet if the weight of the boy is 40kg. Assume that the a
Jlenok [28]

Answer:

P = 1333.33 N

Explanation:

The pressure exerted by the boy on the floor can be calculated by the following equation:

P = \frac{F}{A}

where,

P = Pressure exerted by the boy = ?

F = Force Applied = Weight of Boy = 40 kg = 40 N (since 1 kg = 1N)

A = Area of application of force = 2(Area of one show) = 2(6 cm x 25 cm)

A = 2(0.06 m x 0.25 m) = 0.03 m²

Therefore,

P = \frac{40\ N}{0.03\ m^2}\\\\

<u>P = 1333.33 N</u>

4 0
3 years ago
Use the Pythagorean theorem to answer the following question. A ping-pong ball is shot straight north from a popgun at 4.0 m/s.
MaRussiya [10]
Resultant is 5 m/s using the Pythagorean theorem<span />
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3 years ago
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A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path wit
makvit [3.9K]

Answer:

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

Initial velocity of block = 0 m/s

Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

Final velocity of bullet = -103 m/s

We need to find final velocity of the block( u )

Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u

We have

            2.87 = -1.442 + 1.8 u

               u = 2.40 m/s

Final velocity of the block = 2.40 m/s east.

7 0
3 years ago
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