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mel-nik [20]
3 years ago
10

A crude but useful approximation for an incompressible boundary layer is a cubic variation from u = 0 at the surface (y = δ) to

the freestream velocity, U, at the edge of the boundary layer (y = 0). The equation for the profile is u/U = 3/2(y/δ) – 1/2(y/δ)3, where δ = cx1/2 and c is a constant. Derive the simplest expression for v/U, the y component of velocity ratio. Determine the y/δ location for the maximum value of the ratio v/U. Determine v/U for y = 1 mm, δ = 5 mm and x = 0.87 m. Determine vmax/U for δ = 5 mm and x = 0.87 m.

Physics
1 answer:
Vinvika [58]3 years ago
7 0

Answer:

Explanation:

The detailed steps and appropriate differentiation and integration is as shown in the attachment.

Starting from the equation for the profile is u/U = 3/2(y/δ) – 1/2(y/δ)3

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A rotation motion is a motion that takes place around a fixed axis.

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Answer:

P = 1471500 [Pa]

Explanation:

We must remember that pressure is defined as the relationship between Force over the area.

P=F/A

where:

P = pressure [Pa] (units of pascals)

F = force [N] (units of Newtons)

A = area of contact = 4 [cm²]

But first we must convert from cm² to m²

A = 4[cm^{2}]*\frac{1^{2} m^{2} }{100^{2} cm^{2} }

A = 0.0004 [m²]

Also, the weight should be calculated as follows:

w = m*g

where:

m = mass = 60 [kg]

g = gravity acceleration = 9.81 [m/s²]

Now replacing:

w = 60*9.81\\w = 588.6[N]

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8 0
3 years ago
A box is being dragged with a horizontal force of 65 N for 12 meters. If there is a force of friction acting on it
asambeis [7]

Answer:

A. 780 J

B. 120 J

C. 660 J

Explanation:

From the question given above the following data were obtained:

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

A. Determination of the work done by the dragging force.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Workdone (Wd) by dragging force =?

Wd = Fₔ × s

Wd = 65 × 12

Wd = 780 J

Therefore, the work done by the dragging force is 780 J

B. Determination of the work done by friction.

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Workdone (Wd) by friction =?

Wd = Fբ × s

Wd = 10 × 12

Wd = 120 J

Therefore, the work done by friction is 120 J

C. Determination of the net work done on the box.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Net work done (Wd) =?

Next, we shall determine the net force acting on the box. This can be obtained as follow:

Dragging force (Fₔ) = 65 N

Force of friction (Fբ) = 10 N

Net force (Fₙ) =?

Fₙ = Fₔ – Fբ

Fₙ = 65 – 10

Fₙ = 55 N

Thus, the net force acting on the box is 55 N

Finally, we shall determine the net work done on the box as follow:

Distance (s) = 12 m

Net force (Fₙ) = 55 N

Net work done (Wd) =?

Wd = Fₙ × s

Wd = 55 × 12

Wd = 660 J

Therefore, the net work done on the box is 660 J

4 0
3 years ago
Convert the following quantities <br><img src="https://tex.z-dn.net/?f=25m%20%7B%7D%5E%7B2%7D%20%20%5C%3A%20into%20%5C%3A%20cm%2
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Answer:

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Refer to the attachment.

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I am absolutely sure its 1.5
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