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1 year ago

What is the value of the thickness of the boundary layer at leading edge of the plate?

1 answer:

7 0

Along the flow direction, the boundary layer's thickness varies. For ReT values between 260 and 780, the boundary layer thickness—defined as the depth at which the normalized concentration has a value of 1/e—ranges between 800 and 250 m.

<h3>What is the Boundary layer?</h3>

A boundary layer is the thin layer of fluid that forms immediately around a bounded surface in physics and fluid mechanics as a result of the fluid flowing along the surface.
A no-slip boundary condition is created as a result of the fluid and wall interaction (zero velocity at the wall).
After that, the flow velocity above the surface steadily rises until it reaches the bulk flow velocity again.
The term "velocity boundary layer" refers to the thin layer of fluid whose velocity has not yet recovered to that of the main flow.

<h3>What purpose does a boundary layer serve?</h3>

Because it is on the fluid's boundary, engineers refer to this layer as the boundary layer.
Many aerodynamics issues, such as wing stall, skin friction drag on an object, and the heat transfer that takes place in high-speed flight, depend heavily on the specifics of the flow within the boundary layer.

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Two stationary positive point charges, charge 1 of magnitude 3.90 nC and charge 2 of magnitude 1.80 nC, are separated by a dista

soldi70 [24.7K]

Answer:

$v = 7793150 m/s$

Explanation:

First, we are going to calculate the electrical potential in the point middle between the two charges

Remember that the electrical potential can be calculated as:

$v = \frac{kQ}{r}$

Where $k = 8.9874 x 10^{9} \frac{Nm^{2} }{C^{2} }$

and it is satisfy the superposition principle, thus

$v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.23} + \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.23}$

$v = 222.73v$

The electrical potential at 10 cm from charge 1 is:

$v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.1} + \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.36}$

$v = 395.44 v$

Since the work - energy theorem, we have:

$q\Delta v = \frac{mv^{2} }{2}$

where q is the electron's charge and m is the electron's mass

Therefore:

$v = \sqrt{\frac{2q\Delta v}{m} }$

$v = 7793150 m/s$

6 0

3 years ago

A 3.6kg mass is accelerated at 2.5m/s. Calculate the resultant force<br> acting on it.

AleksandrR [38]

Force = mass*acceleration so
3.6*2.5 =9 Newtons

5 0

3 years ago

Need some help

Keith_Richards [23]

The answer is C) the density of the rock

Density of rock is the dependent variable, because it depends on the temperature. The temperature can’t be the dependent variable because ,the density of a rock does not have magical powers that can change temperature of a room. However changing the temperature of the room ,will change the density of the rock. Hope this helps !

4 0

3 years ago

In what way were augusts<br> comte and immanual Kant<br> alike

vodka [1.7K]

1. they were both educated in france
2. they both felt that the end of the monarchy was inevitable
3. they were both artist
4. they both felt that science would improve the world

5 0

3 years ago

Read 2 more answers

3. A roller coaster car sits at rest at the top of a hill, 23.5 meters tall. What will be the final

Sloan [31]

Answer:

21.5 m/s

Explanation:

KE at bottom = PE at top

½ mv² = mgh

v = √(2gh)

v = √(2 × 9.8 m/s² × 23.5 m)

v = 21.5 m/s

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3 years ago