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just olya [345]
3 years ago
6

How does an airfoil create lift?

Engineering
1 answer:
scoundrel [369]3 years ago
3 0

Answer:

An airfoil creates lift by exerting a downward force on the air as it flows past

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2 Blocks 1 and 2 rest on rough surfaces with coefficient of frictions ¢1 and ¢2 respectively. The blocks
Amiraneli [1.4K]

Answer:

  • 100N
  • 25N

Explanation:

a) On the verge of tipping over, reaction acts at the corner A

When slippage occurs,

Block moves w/ const. velocity  equilibrium

Three-force member: reaction at A must pass through B

tan b/2h, h b/ 2 θ µ = = ∴= k k ( µ )

b) When slippage occurs,

Block moves w/ const. velocity  equilibrium

Three-force member: reaction at C must pass through G

k tanθ µ =

tan x/ H/2 , x H/2

4 0
3 years ago
A non-licensed person may be the SOLE owner of a civil, electrical, or mechanical engineering business under which of the follow
kotegsom [21]

Answer:

(d) None. No provisions exist.

Explanation:

B&P Code § 6738 prohibits a non-licensed person from being the sole proprietor of an engineering business. The non-licensed can be a partner in an engineering business that offers civil, electrical, or mechanical services. It is mandatory that at least one licensed engineer must be a co-owner of the business.

5 0
3 years ago
What information does the api symbol or donut provide?.
kondor19780726 [428]

Answer:

The API "Donut" identifies oils that meet current API engine oil standards.

Explanation:

hope this helps :)

7 0
2 years ago
A cylindrical buoy is 2m in diameter and 2.5m long and weight 22kN . The specific weight of sea water is 10.25kN/m^3 . (I) Show
aleksandrvk [35]

Answer:

GM

So the bouy does not float with its axis vertical

Explanation:

From the question we are told that:

Diameter d=2m

Length l=2.5m

Weight W=22kN

Specific weight of sea water \mu= 10.25kN/m^3

Generally the equation for weight of cylinder is mathematically given by

Weight of cylinder = buoyancy Force

W=(pwg)Vd

Where

V_d=\pi/4(d)^2y

Therefore

22*10^3=10.25*10^3 *\pi/4(2)^2y\\\\\22*10^3=32201.3247y\\\\\y=1.5m

Therefore

Center of Bouyance B

B=\frac{y}{2}=0.26m\\\\B=0.75

Center of Gravity

G=\frac{I.B}{2}=2.6m

Generally the equation for\BM is mathematically given by

BM=\frac{I}{vd}\\\\BM=\frac{3.142/64*2^4}{3.142/4*2^2*0.5215}\\\\BM=0.479m\\\\

Therefore

BG=2.6-0.476\\\\BG=0.64m

Therefore

GM=BM-BG\\\\GM=0.479m-0.64m\\\\GM=-0.161m\\\\

Therefore

GM

So the bouy does not float with its axis vertical

4 0
3 years ago
To use wiring diagrams, an understanding of the symbols, abbreviations, and connector coding used in the diagrams
Komok [63]

Answer:

True

Explanation:

6 0
3 years ago
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