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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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A helicopter moves horizontally in the x direction at a speed of 120 mi/h. Knowing that the main blades rotate clockwise when vi
devlian [24]

Answer:

The instantaneous axis of rotation=

x = 0 ; z = 8.4 ft

Explanation:

Given:

Speed of helicopter, Vo= 120 mi/h, converting to ft/sec, we have:

\frac{5280 * 120}{60*60}

= 176 ft/s

Angular velociyy, w = 220 rpm, converting to rad/sec, we have: \frac{200*2*pi}{60} =20.95 rad/s

The helicopter moves horizontally in the x direction at a speed of 120 mi/h, this means that the helicopter moves in the positive x direction at 120mi/h

To find the instantaneous axis of rotation of the main blades, we have:

Where Vc = 20.95 rad/s

Vo = 176 ft/s

z = \frac{V_0}{V_c} = \frac{176ft/s}{20.95rad/s}

= 8.4 ft

Therefore the axis of rotation=

x = 0 ; z = 8.4 ft

4 0
3 years ago
In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of the line has resis
Rufina [12.5K]

Question:

In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of the line has resistance R.

For a power line that supplies power to 10 000 households, we can conclude that

a) IV < I²R

b) I²R = 0

c) IV = I²R

d) IV > I²R

e) I = V/R

Answer:

d) IV > I²R

Explanation:

In a typical transmission line, the current I is very small and the voltage V is very high as to minimize the I²R losses in the transmission line.

The power delivered to households is given by

P = IV

The losses in the transmission line are given by

Ploss = I²R

Therefore, the relation IV > I²R  holds true, the power delivered to the consumers is always greater than the power lost in the transmission line.

Moreover, losses cannot be more than the power delivered. Losses cannot be zero since the transmission line has some resistance. The power delivered to the consumers is always greater than the power lost in the transmission.

6 0
3 years ago
Why is the face of the claw on a claw hammer usually a smooth curve? Why isn't it straight or some other shape?
GarryVolchara [31]

Answer:

The face of the claw on the claw hammer is usually a smooth curve so as to improve the ease with which nails are removed when removing nails because as the nail held between the V shaped split claw is being pulled out from the wood, it slides more and more towards cheek, reducing the distance of the nail from the cheek which is the fulcrum, thereby increasing the mechanical advantage because the location of the hand on the grip remains unchanged

Explanation:

7 0
3 years ago
A specimen has circular cross-sectional area with initial diameter d = 26 mm and a gauge length Lo = 200 mm. A force P = 127 kN
steposvetlana [31]
The ahahte ekiebecwkwlwtfe einshgeh dinzgdwbsine ensue d
4 0
3 years ago
Oil with a kinematic viscosity of 4 10 6 m2 /s fl ows through a smooth pipe 12 cm in diameter at 2.3 m/s. What velocity should w
Setler79 [48]

Answer:

Velocity of 5 cm diameter pipe is 1.38 m/s

Explanation:

Use following equation of Relation between the Reynolds numbers of both pipes

Re_{5} = Re_{12}

\sqrt{\frac{V_{5}XD_{5}  }{v_{5}}}= \sqrt{\frac{V_{12}XD_{12}  }{v_{12}}}

Re_{5} = Reynold number of water pipe

Re_{12} = Reynold number of oil pipe

V_{5} = Velocity of water 5 diameter pipe = ?

V_{12} = Velocity of oil 12 diameter pipe = 2.30

v_{5} = Kinetic Viscosity of water = 1 x 10^{-6} m^{2}/s

v_{12} = Kinetic Viscosity of oil =  4 x 10^{-6} m^{2}/s

D_{5} = Diameter of pipe used for water = 0.05 m

D_{12} = Diameter of pipe used for oil = 0.12 m

Use the formula

\sqrt{\frac{V_{5}XD_{5}  }{v_{5}}}= \sqrt{\frac{V_{12}XD_{12}  }{v_{12}}}

By Removing square rots on both sides

{\frac{V_{5}XD_{5}  }{v_{5}}}= {\frac{V_{12}XD_{12}  }{v_{12}}}

{V_{5}= {\frac{V_{12}XD_{12}  }{v_{12}XD_{5}\\}}xv_{5}

{V_{5}= [ (0.23 x 0.12m ) / (4 x 10^{-6} m^{2}/s) x 0.05 ] 1 x 10^{-6} m^{2}/s

{V_{5} = 1.38 m/s

4 0
3 years ago
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