Technician A uses three prong electrical cords when possible.

you have a permanent magnet whit remanent induction. than you cut a piec of that magnet. if you put that piece back into permane

ELEN [110]

Yes, it will. The induction will be the same as long as it’s put back together.

5 0

3 years ago

The following statements are about the laminar boundary layer over a flat plate. For each statement, answer whether the statemen

Nikolay [14]

Answer:

1. B. False

2. B. False

3. A. True

4. B. False

5. A. True

6. A. True

7. A. True

Explanation:

1. B. False

The relation of Reynolds' number, Reₓ to boundary layer thickness δ at a point x is given by the relation

$\delta = \dfrac{x \times C}{\sqrt{Re_x} }$

That is the boundary layer thickness is inversely proportional to the square root of the Reynolds' number so that if the Reynolds' number were to increase, the boundary layer thickness would decrease

Therefore, the correct option is B. False

2. B. False

From the relation

$Re_x = \dfrac{U_o \times x}{v}$

As the outer flow velocity increases, the boundary layer thickness diminishes

3. A. True

As the viscous force is increased the boundary layer thickness increases

4. B. False

Boundary layer thickness is inversely proportional to velocity

5. A. True

The boundary layer model developed by Ludwig Prandtl is a special case of the Navier-Stokes equation

6. A. True

Given a definite boundary layer thickness, the curve representing the boundary layer thickness is a streamline

7. A. True

The boundary layer approximation by Prandtl Euler bridges the gap between the Euler (slip boundary conditions) and Navier-Stokes (no slip boundary conditions) equations.

6 0

3 years ago

Determine the critical load if the bottom is fixed and the top is pinned. ewew = 1. 6 ×(10)3ksi×(10)3ksi ,σyσy = 5 ksiksi

Katen [24]

<h3>What is a Critical Load?</h3>

Critical load Fcr or buckling load is the value of load that causes the phenomenon of change from stable to unstable equilibrium state.

With that beign said, first it is neessary to calculate the moment of inercia about the x-axis:

$Ix= \frac{db^3}{12}\\ Ix = \frac{2.(4)^3}{12} = 10.667in$

Then it is necessary to calculate the moment of inercia about the y-axis:

$Iy = \frac{db^3}{12}\\ Iy = \frac{4.(2)^3}{12} = 2.662in$

Comparing both moments of inercia it is possible to assume that the minimun moment of inercia is the y-axis, so the minimun moment of inercia is 2662in.

And so, it is possible to calculate the critical load:

$Pc\gamma = \frac{2046\pi ^2E.I}{L^2} \\Pc\gamma= \frac{2046.\pi ^2.(1,6.10^3.10^3).2662}{(10.12)^2} \\Pc\gamma= 5983,9db$