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3 years ago

Que rol tiene el ecosistema el patos

Engineering

1 answer:

zhenek [66]3 years ago

3 0

Answer:

Investigué un poco y descubrí algunas cosas sorprendentes sobre los patos.Espero que los encuentres interesantes:Las aves acuáticas como los patos pueden mantener la diversidad de otros organismos, controlar las plagas, ser bioindicadores efectivos de las condiciones ecológicas y actuar como centinelas de posibles brotes de enfermedades. También proporcionan importantes aprovisionamientos (carne, plumas, huevos, etc.) y servicios culturales a las sociedades indígenas y occidentalizadas.La ubicuidad a menudo ridiculizada de los patos los convierte en vehículos ideales para transportar semillas de un lugar a otro, y eso significa humedales y biodiversidad más saludables para el beneficio de todas las aves y la vida silvestre.Sabías ? :Los patos ayudan al medio ambiente porque limpian los químicos del agua para que sus amigos y otros animales no mueran y se enfermen gravemente.

Explanation:

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a buyer can purchase 70 screwdrivers ten 4-inch length twelve 6 inch length twenty 8-inch length are needed. how many heavy 24-i

jasenka [17]

Answer:

28 , 24-inch screwdrivers

Explanation:

The total number of screwdrivers that can be purchased is = 70

4 - inch length screwdrivers = 10

6- inch length screwdrivers = 12

8- inch length screwdrivers = 20

Total = 20 +12 +10 = 42

Remaining = 70-42 = 28

So, heavy 24-inch screwdrivers = 28

3 0

3 years ago

Air enters a cmpressor at 20 deg C and 80 kPa and exits at 800 kPa and 200 deg C. The power input is 400 kW. Find the heat trans

aksik [14]

Answer:

The heat is transferred is at the rate of 752.33 kW

Solution:

As per the question:

Temperature at inlet, $T_{i} = 20^{\circ}C$ = 273 + 20 = 293 K

Temperature at the outlet, $T_{o} = 200{\circ}C$ = 273 + 200 = 473 K

Pressure at inlet, $P_{i} = 80 kPa = 80\times 10^{3} Pa$

Pressure at outlet, $P_{o} = 800 kPa = 800\times 10^{3} Pa$

Speed at the outlet, $v_{o} = 20 m/s$

Diameter of the tube, $D = 10 cm = 10\times 10^{- 2} m = 0.1 m$

Input power, $P_{i} = 400 kW = 400\times 10^{3} W$

Now,

To calculate the heat transfer, $Q$ , we make use of the steady flow eqn:

$h_{i} + \frac{v_{i}^{2}}{2} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}$

where

$h_{i}$ = specific enthalpy at inlet

$h_{o}$ = specific enthalpy at outlet

$v_{i}$ = air speed at inlet

$p_{s}$ = specific power input

H and H' = Elevation of inlet and outlet

Now, if

$v_{i} = 0$ and H = H'

Then the above eqn reduces to:

$h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}$

$Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}$ (1)

Also,

$p_{s} = \frac{P_{i}}{ mass, m}$

Area of cross-section, A = $\frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}$

Specific Volume at outlet, $V_{o} = A\times v_{o} = 7.85\times 10^{- 3}\times 20 = 0.157 m^{3}/s$

From the eqn:

$P_{o}V_{o} = mRT_{o}$

$m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s$

Now,

$p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg$

Also,

$\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg$

Now, using these values in eqn (1):

$Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW$

Now, rate of heat transfer, q:

q = mQ = $0.925\times 813.33 = 752.33 kW$

4 0

3 years ago

In DC electrode positive, how much power is at the work clamp?

Korolek [52]

Answer:

1/3 power

Explanation:

I'm just a smart guy

7 0

2 years ago

A world class runner can run long distances at a pace of 15 km/hour. That runner expends 800 kilocalories of energy per hour. a)

maks197457 [2]

Answer: a) 1.05kW b) 3.78MJ c) 5.3 bars

Explanation :

Conversions give 900 kcal as 900000 x 4.2 J/cal {4.2 J/cal is the standard factor}

= 3780kJ

And 1 hour = 3600s

Therefore, Power in watts = 3780/3600 = 1.05kW = 1050W

At 15km/hour a 15km run takes 1 hour.

1 hour is 3600s and the runner burns 1050 joule per second.

Energy used in 1 hour = 3600 x 1050 J/s

= 3780000 J or 3.78MJ

1 mile = 1.61km so 13.1 mile is 13.1 x 1.61 = 21.1km

15km needs 3.78 MJ of energy therefore 21.1km needs 3.78 x 21.1/15 = 5.32MJ =5320 kJ

Finally,

1 Milky Way = 240000 calories = 4.2 x 240000 J = 1008000J or 1008kJ

This means that the runner needs 5320/1008 = 5.3 bars

7 0

3 years ago

explain each of the following kinds of rockets: Solid-Fuel Rocket, Liquid-Fuel Rocket, Ion Rocket and Plasma Rocket.

Rudik [331]

Answer:

ur answer friend

Explanation:

answer

Solid-Fuel Rocket - a solid-propellant rocket or solid rocket is a rocket with a rocket engine that uses solid propellants. The earliest rockets were solid-fuel rockets powered by gunpowder; they were used in warfare by the Chinese, Indians, Mongols and Persians, as early as 13th century.

Liquid-Fuel Rocket - a liquid-propellant rocket or liquid rocket utilizes a rocket engine that use liqiud propellants. An inert gas stored in a tank at a high pressure is sometimes used instead of pumps in simpler small engines to force the propellants into the combustion chamber.

Ion Rocket - an ion thruster or ion drive is a form of electric propulsion used for spacecraft propulsion. It creates thrust by accelerating ions using electricity. The Deep Space 1 spacecraft, powered by an ion thruster, changed velocity by 4.3 km/s ( 9600 mph ) while consuming less than 74 kg ( 163 lb ) of xenon.

Plasma Rocket - in this type of rocket, a combination of electric and magnetic fields are used to break down the atoms and molecules of a propellant gas into a collection of particles that have either a positive charge (ions) or a negative charge (electrons). In other words, the propellant gas becomes a plasma.

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5 0

3 years ago