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jonny [76]
3 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]3 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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marishachu [46]

Answer:

As follows,

Explanation:

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In 1st question,

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In 2nd question,

KE=1/2mv^2

6.8=1/2*0.046*v^2

v=sqrt(6.8/0.023)

v=17.19

In 3rd question,

KE=1/2mv^2

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m=0.16kg=160g

For 4th,

a.

1st case,

KE=1/2mv^2=1/2*28*2.4^2=80.64

2nd case,

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6 0
3 years ago
A magnet in the form of a cylindrical rod has a length of 7.30 cm and a diameter of 1.5 cm. It has a uniform magnetization of 5.
yulyashka [42]

Answer:

Magnetic dipole moment is 0.0683 J/T.

Explanation:

It is given that,

Length of the rod, l = 7.3 cm = 0.073 m

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The dipole moment per unit volume is called the magnetization of a magnet. Mathematically, it is given by :

M=\dfrac{\mu}{V}

\mu=M\times \pi r^2\times l

Where

r is the radius of rod, r = 0.0075 m

\mu=5.3\times 10^3\ A/m\times \pi (0.0075)^2\times 0.073\ m

\mu=0.0683\ J/T

So, its magnetic dipole moment is 0.0683 J/T. Hence, this is the required solution.

6 0
3 years ago
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vladimir2022 [97]

Answer:

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Explanation:

<h2>Hope it helps you</h2>
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GREYUIT [131]

Answer: 225N to the right

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The pedaling force is to the right and of 325N.

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The greater force is towards the right proving the direction of the Net force is to the right.

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