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3 years ago

Which permission do you need to shoot on the owner’s property?

Engineering

1 answer:

Elena L [17]3 years ago

4 0

Answer:

filming permit,

( MARK ME BRAINLIEST!!)

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In plumbing what is a video snake used for

aleksley [76]

Answer:

How to stop toilets

Explanation:

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2 years ago

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The statement that is NOT true about the concept of a boundary layer on an object is: a. the Reynolds number is greater than uni

Ilya [14]

Answer:

Option E

Explanation:

All the given statements are true except the velocity gradients normal to the flow direction are small since these are not normally small. It's true that viscous effects are present only inside the boundary layer and the fluid velocity equals the free stream velocity at the edge of the boundary layer. Moreover, Reynolds number is greater than unity and the fluid velocity is zero at the surface of the object.

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3 years ago

Six forces act on a beam that forms part of a building's

IrinaVladis [17]

Answer:

Explanation:

write each force in terms of magnitude and directions

Fx = F sin Ф

Fy = F cos Ф

where Ф is to be measured from x axis.

∑F at y = o

FAy + FBy + FCy + FDy + FEy + FGy = 0

∑F at x = o

FAx + FBx + FCx + FDx + FEx + FGx = 0

Let

FA = FA sin (110) + FA cos (110)

FB = 20 sin (270) + 20 cos (270)

FC = 16 sin (140) + 16 cos (140)

FD = 9 sin (40) + 9 cos (40)

FE = 20 sin (270) + 20 cos (270)

FG = FG sin (50) + FG cos (50)

add x and y forces:

FAx + FBx + FCx + FDx + FEx + FGx = 0

FAy + FBy + FCy + FDy + FEy + FGy = 0

FA sin (110) + 0 + 16 sin (140) + 9 sin (40) + 0 + FG sin (50) = 0

FA cos (110) - 20 + 16 cos (140) + 9 cos (40) - 20 + FG cos (50 = 0

FA sin (110) + 0 + 10.285 + 5.785 + 0 + FG sin (50) = 0

FA cos (110) - 20 - 12.257 + 6.894 - 20 + FG cos (50) = 0

FA sin (110) + 16.070 + FG sin (50) = 0

FA cos (110) - 45.363 + FG cos (50) = 0

solving for FA, and FG

FA = 13 kN

FG = 15.3 kN

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3 years ago

What is definition of<br>computational fluid Dynamics <br>unstructured grid <br>domain<br>geometry

Keith_Richards [23]

Najsjjsjhshehdhdhdhdhhdhdhdhdhhd

6 0

3 years ago

At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho

victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

$\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2} }{2})^{2}+\tau _{xy}^{2} }$

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

$32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2} }$

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

$\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}$

Where

σp1 = 20 ksi

σp2 = -30 ksi

$\sigma _{p1} +\sigma _{p2}=-10 ksi$ (equation 1)

$\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi$ equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

4 0

3 years ago