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jonny [76]
2 years ago
11

If the Test Section area has an air speed of 10 m/s, what would be the air speed at the Air Intake area

Physics
1 answer:
FrozenT [24]2 years ago
4 0

The speed of air at the intake area is \frac{10A_2}{A_1} (m/s).

<h3 /><h3>Continuity equation</h3>

The continuity equation is used to determine the flow rate at different sections of a pipe or fluid conduit.

The continuity equation is given as;

A_1v_1 = A_2 v_2

  • Let the intake area = A₁
  • Let the velocity of air in intake area = v₁
  • Let the area of the test section = A₂
  • Velocity of air in test section, v₂ = 10 m/s

The speed of air at the intake area is calculated as follows;

A_1 v_1 = A_2 v_2\\\\&#10;v_1 = \frac{A_2 v_2}{A_1} \\\\&#10;v_1 = \frac{10A_2}{A_1}

Learn more about continuity equation here: brainly.com/question/14619396

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The answer is zero hope I helped
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3 years ago
A jet airplane is in level flight. The mass of the airplane is m=9010kg. The airplane travels at a constant speed around a circu
Lyrx [107]

Answer:

The magnitude of the lift force L = 92.12 kN

The required angle is ≅ 16.35°

Explanation:

From the given information:

mass of the airplane = 9010 kg

radius of the airplane R = 9.77 mi

period T = 0.129 hours = (0.129 × 3600) secs

= 464.4 secs

The angular speed can be determined by using the expression:

ω = 2π / T

ω = 2 π/ 464.4

ω = 0.01353 rad/sec

The direction \theta = tan^{-1} ( \dfrac{\omega ^2 R}{g})

\theta = tan^{-1} ( \dfrac{0.01353 ^2 \times (9.77\times 1609)}{9.81})

θ = 16.35°

The magnitude of the lift force  L = mg ÷ Cos(θ)

L = (9010 × 9.81) ÷ Cos(16.35)

L = 88388.1  ÷ 0.9596

L = 92109.32 N

L = 92.12 kN

3 0
3 years ago
The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that
pav-90 [236]

Answer:

The permittivity of rubber is  \epsilon  = 8.703 *10^{-11}

Explanation:

From the question we are told that

     The  magnitude of the point charge is  q_1 =  70 \ nC  =  70 *10^{-9} \  C

      The diameter of the rubber shell is  d = 32 \ cm  =  0.32 \ m

       The Electric field inside the rubber shell is  E =  2500 \ N/ C

The radius of the rubber is  mathematically evaluated as

              r =  \frac{d}{2} =  \frac{0.32}{2}  =  0.16 \ m

Generally the electric field for a point  is in an insulator(rubber) is mathematically represented as

         E =  \frac{Q}{ \epsilon }  *  \frac{1}{4 *  \pi r^2}

Where \epsilon is the permittivity of rubber

    =>     E  *  \epsilon  *  4 * \pi *  r^2 =  Q

   =>      \epsilon  =  \frac{Q}{E *  4 *  \pi *  r^2}

substituting values

            \epsilon  =  \frac{70 *10^{-9}}{2500 *  4 *  3.142 *  (0.16)^2}

            \epsilon  = 8.703 *10^{-11}

7 0
3 years ago
The potential energy of a negative charge moved from point A to point B will increase.A negative charge moved from point A to po
AysviL [449]

Answer:

<em>The K.E from A to B won't increase...</em>

Explanation:

That's because the P.E from A to B is increasing. The K.E will increase if charge moves from a higher potential to a lower potential i.e., from B to A.

That is the reason there is no effect on net K.E when moving from a potential to same potential over and over (A to C).

4 0
3 years ago
What do the mean by cubic expansivity​
olganol [36]
A real cubic expansivity is an increase in the volume of a liquid per unit volume per degree rise in temperature when heated in an inexpansible vessel.
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2 years ago
Read 2 more answers
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