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3 years ago

Where did we use rotational and irrotational flow

Physics

1 answer:

WARRIOR [948]3 years ago

3 0

Answer:

The term rotational and irrotational flow is associated withe the flow of particles in fluid.

The common example of irrrotational flow can be seen on the carriages of the Ferris wheel (giant wheel).

Explanation:

If the fluid is rotating along its axis with the streamline flow of its particles,then this type of flow is rotational flow.
Similarly if fluid particles do not rotate along its axis while flowing in a stream line flow then it is considered as the irrotational flow.
In majority, if the flow of fluid is viscid then it is rotational.
Fluid in a rotating cylinder is an example of rotating flow.

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Answer:

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3 years ago

Density of water is 1000kg/m3. Find out the amount of mass of 5m3 water

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Answer:

5000 m³

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6. A 2-kg ball B is traveling around in a circle of radius r1 = 1 m with a speed (vB)1 = 2 m/s. If the attached cord is pulled d

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Answer:

Explanation:

Given that,

Mass of ball m = 2kg

Ball traveling a radius of r1= 1m.

Speed of ball is Vb = 2m/s

Attached cord pulled down at a speed of Vr = 0.5m/s

Final speed V = 4m/s

Let find the transverse component of the final speed using

V² = Vr²+ Vθ²

4² = 0.5² + Vθ²

Vθ² = 4²—0.5²

Vθ² = 15.75

Vθ =√15.75

Vθ = 3.97 m/s.

Using the conservation of angular momentum,

(HA)1 = (HA)2

Mb • Vb • r1 = Mb • Vθ • r2

Mb cancels out

Vb • r1 = Vθ • r2

2 × 1 = 3.97 × r2

r2 = 2/3.97

r2 = 0.504m

The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m

The required time,

Using equation of motion

V = ∆r/t

Then,

t = ∆r/Vr

t = (r1—r2) / Vr

t = (1—0.504) / 0.5

t = 0.496/0.5

t = 0.992 second

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3 years ago

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The answer should be B. a stable isotope to a decaying isotope.

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3 years ago

Read 2 more answers

A 4.3-kg block slides down an inclined plane that makes an angle of 30° with the horizontal. Starting from rest, the block slide

-BARSIC- [3]

Answer:

0.56

Explanation:

Let the coefficient of friction is μ.

m = 4.3 kg, θ = 30 degree, initial velocity, u = 0, s = 2.7 m, t = 5.8 s

By the free body diagram,

Normal reaction, N = mg Cosθ = 4.3 x 9.8 x Cos 30 = 36.49 N

Friction force, f = μ N = 36.49 μ

Net force acting on the block,

Fnet = mg Sinθ - f = 4.3 x 9.8 x Sin 30 - 36.49 μ

Fnet = 21.07 - 36.49μ

Net acceleartion, a = Fnet / m

a = (21.07 - 36.49μ) / 4.3

use second equation of motion

s = ut + 1/2 a t^2

2.7 = 0 + 1/2 x (21.07 - 36.49μ) x 5.8 x 5.8 / 4.3

By solving we get

μ = 0.56

8 0

3 years ago

Where did we use rotational and irrotational flow​

Where did we use rotational and irrotational flow