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Shtirlitz [24]
3 years ago
8

The three components of velocity in a velocity field are given by u = Ax + By + Cz, v = Dx + Ey + Fz, and w = Gx + Hy + Jz. Dete

rmine the relationship among the coefficients A through J that is necessary if this is to be a possible incompressible flow field.
Physics
1 answer:
Alexxandr [17]3 years ago
6 0

Answer:

The relationship is only between the coefficients A, E and J which is:

A + E + J = 0. The remaining coefficients can be anything without any constraints.

Explanation:

Given:

The three components of velocity is a velocity field are given as:

u = Ax + By + Cz\\\\v = Dx + Ey + Fz\\\\w = Gx + Hy + Jz

The fluid is incompressible.

We know that, for an incompressible fluid flow, the sum of the partial derivatives of each component relative to its direction is always 0. Therefore,

\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0

Now, let us find the partial derivative of each component.

\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}(Ax+By+Cz)\\\\\frac{\partial u}{\partial x}=A+0+0=A\\\\\frac{\partial v}{\partial y}=\frac{\partial }{\partial y}(Dx+Ey+Fz)\\\\\frac{\partial v}{\partial y}=0+E+0=E\\\\\frac{\partial w}{\partial z}=\frac{\partial }{\partial z}(Gx+Hy+Jz)\\\\\frac{\partial w}{\partial z}=0+0+J=J

Hence, the relationship between the coefficients is:

A+E+J=0

There is no such constraints on other coefficients. So, we can choose any value for the remaining coefficients B, C, D, F, G and H.

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2 years ago
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A small branch is wedged under a 200 kg rock and rests on a smaller object. The small object that creates a pivot point is calle
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Answer:

F = 326.7 N

Explanation:

given data

mass m = 200 kg

distance d = 2 m

length L = 12 m

solution

we know force exerted by the weight of the rock that is

W = m × g   ..............1

W = 200 × 9.8

W = 1960 N

and

equilibrium the sum  of the moment about that is

∑Mf = F(cos∅) L - W (cos∅) d  = 0

here ∅ is very small so cos∅ L = L and cos∅ d = D

so F × L - W × d = 0       .................2

put here value

F × 12 - 1960 × 2 = 0

solve it we get

F = 326.7 N

8 0
3 years ago
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A bullet is fired from a gun at 45° angle to the horizontal with a velocity of 500 m/s. Find the
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A pendulum of length =1.0 m is pulled to the side and released on the moon. It's period is measured to be 4.82 seconds. What is
Shkiper50 [21]

Answer:

Gravity on the moon, g = 1.69 m/s²

Explanation:

It is given that,

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We have to find the gravity of the moon. The time period of the pendulum is given by :

T=2\pi\sqrt{\dfrac{l}{g}}

g = acceleration due to gravity on moon

g=\dfrac{4\pi^2l}{T^2}

g=\dfrac{4\pi^2\times 1\ m}{(4.82\ s)^2}

g = 1.69 m/s²

Hence, the gravity on the moon is 1.69 m/s².

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3 years ago
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