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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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Differentiate between isohyetal method and arithmetical average method of rainfall​
prohojiy [21]

Answer:

Explained below

Explanation:

The isohyetal method is one used in estimating Rainfall whereby the mean precipitation across an area is gotten by drawing lines that have equal precipitation. This is done by the use of topographic and other data to yield reliable estimates.

Whereas, the arithmetic method is used to calculate true precipitation by the way of getting the arithmetic mean of all the points or arial measurements that will be considered in the analysis.

7 0
3 years ago
Define waves as it applies to electromagnetic fields
julsineya [31]

Waves in the electric and magnetic fields are known as electromagnetic waves. You must first understand what a field is, which is just a technique of giving each square inch of space a numerical value. You may see that as a temperature field, for instance, when you look at the weather predictions and they mention the temperature in several locations. Every location on Earth has a unique temperature that can be quantified. Everywhere on Earth has its own wind velocity, which is another form of field. This field differs somewhat from the temperature field in that the wind velocity has both a direction and a magnitude, whereas the temperature just has a magnitude (how hot it is). A vector is a quantity that has both magnitude and direction, hence a field that contains vectors at every location is referred to as a vector field. Vector fields include the magnetic and electric fields. We may examine what would happen if we placed a charged particle at any given position in space. If the charged particle were to accelerate, we would state that the electric field there is the direction in which the particle is moving. In general, positively charged particles will move in the electric field's direction, whereas negatively charged particles will move in the opposite way. Because it is a vector field, the magnetic field exhibits comparable behavior. We discovered in the 19th century that the same interaction, electromagnetism, really produces both electric and magnetic fields. Like an electromagnet, a changing electric field will produce a magnetic field, and a changing magnetic field will induce an electric field (like in a generator). If your system is configured properly, you may have an electric field that fluctuates, which in turn produces a magnetic field, which in turn induces another electric field, which in turn generates another magnetic field, and so on indefinitely. At the speed of light, this oscillation between a strong magnetic field and strong electric field spreads out indefinitely. In reality, light is an electromagnetic wave—an oscillation in the electromagnetic fields. An electric or magnetic field may exist without a medium since they exist in a vacuum, which implies that waves in these fields don't require a medium like sound to flow through.

5 0
2 years ago
What is an advantage of a nuclear-fission reactor?.
Kobotan [32]

Answer:

Nuclear fission is almost 8,000 times more efficient than traditional fossil fuels at producing energy. That's a lot of energy packed into a small space. Nuclear energy is more efficient, which means it uses less fuel to power the plant and produces less waste.


advantages:
-produces no polluting gases
-does not contribute to global warming
-very low fuel costs
-Low fuel quantity reduces mining and transportation effects on environment
-High technology research required benefits other industries
-Power station has very long lifetime

Disadvantages:
-Waste is radioactive and safe disposal is very difficult and expensive
-Local thermal pollution from wastewater affects marine life
-Large-scale accidents can be catastrophic
-Public perception of nuclear power is negative
-Costs of building and safely decommissioning are very high
-Cannot react quickly to changes in electricity demand

4 0
2 years ago
Air at 293k and 1atm flow over a flat plate at 5m/s. The plate is 5m wide and 6m long. (a) Determine the boundary layer thicknes
loris [4]

Answer:

a). 8.67 x 10^{-3} m

b).0.3011 m

c).0.0719 m

d).0.2137 N

e).1.792 N

Explanation:

Given :

Temperature of air, T = 293 K

Air Velocity, U = 5 m/s

Length of the plate is L  = 6 m

Width of the plate is b = 5 m

Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X 10^{-5} Pa-s

We know density of air is ρ = 1.21 kg /m^{3}

Now we can find the Reyonld no at x = 1 m from the leading edge

Re = \frac{\rho .U.x}{\mu }

Re = \frac{1.21 \times 5\times 1}{1.822\times 10^{-5} }

Re = 332052.6

Therefore the flow is laminar.

Hence boundary layer thickness is

δ = \frac{5.x}{\sqrt{Re}}

   = \frac{5\times 1}{\sqrt{332052.6}}

   = 8.67 x 10^{-3} m

a). Boundary layer thickness at x = 1 is δ = 8.67 X 10^{-3} m

b). Given Re = 100000

    Therefore the critical distance from the leading edge can be found by,

     Re = \frac{\rho .U.x}{\mu }

     100000 = \frac{1.21\times5\times x}{1.822 \times10^{-5}}

     x = 0.3011 m

c). Given x = 3 m from the leading edge

    The Reyonld no at x = 3 m from the leading edge

     Re = \frac{\rho .U.x}{\mu }

     Re = \frac{1.21 \times 5\times 3}{1.822\times 10^{-5} }

     Re = 996158.06

Therefore the flow is turbulent.

Therefore for a turbulent flow, boundary layer thickness is

    δ = \frac{0.38\times x}{Re^{\frac{1}{5}}}

       = \frac{0.38\times 3}{996158.06^{\frac{1}{5}}}

       = 0.0719 m

d). Distance from the leading edge upto which the flow will be laminar,

  Re = \frac{\rho \times U\times x}{\mu }

5 X 10^{5} = \frac{1.21 \times 5\times x}{1.822\times 10^{-5}}}

 x = 1.505 m

We know that the force acting on the plate is

F_{D} = \frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}

and C_{D} at x= 1.505 for a laminar flow is = \frac{1.328}{\sqrt{Re}}

                                                                         = \frac{1.328}{\sqrt{5\times10 ^{5}}}

                                                                       = 1.878 x 10^{-3}

Therefore, F_{D} =  \frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}

                                          = \frac{1}{2}\times 1.878\times 10^{-3}\times 1.21\times (5\times 1.505)\times 5^{2}

                                         = 0.2137 N

e). The flow is turbulent at the end of the plate.

  Re = \frac{\rho \times U\times x}{\mu }

       = \frac{1.21 \times 5\times 6}{1.822\times 10^{-5} }

       = 1992316

Therefore C_{D} = \frac{0.072}{Re^{\frac{1}{5}}}

                                           = \frac{0.072}{1992316^{\frac{1}{5}}}

                                           = 3.95 x 10^{-3}

Therefore F_{D} = \frac{1}{2}\times C_{D}\times \rho\times A\times U^{2}

                                           = \frac{1}{2}\times 3.95\times 10^{-3}\times 1.21\times (5\times 6)\times 5^{2}

                                          = 1.792 N

3 0
3 years ago
Yo can someone find me an e-boy?
Vera_Pavlovna [14]
I got a friend how old are you and are you ok dating a bi guy
4 0
4 years ago
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