Using this equation,a1v1=a2v2,explain how jet engines provide a forward thrust for an airplane.?
438959
Try here (the content is very techinical).  http://scienceworld.wolfram.com/physics/ContinuityEquation.html\nhttp://www.efm.leeds.ac.uk/CIVE/CIVE1400/Section3/continuity.htm
944938
Using Equation of continuity:\nAv = constant (A1vA = A2v2)\n\n2.54*48.9 = 2720*v2\nv2=4.57E-2 cm/s
54343
Try wikipedia next time:\n\nThrust vectoring is the ability of an aircraft or other vehicle to direct the thrust from its main engine(s) in a direction other than parallel to the vehicle's longitudinal axis. The technique was originally envisaged to provide upward vertical thrust as a means to give aircraft VTOL or STOL capability. Subsequently it was realised that the use of vectored thrust in combat situations enabled an aircraft to perform various maneuvers not available to conventional-engined planes.
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A turbojet engine is the simplest type of jet engine. It uses the high velocity of the exaust gases to generate thrust\n\nSites regardin Turbojet enhines and their working:\n\nhttp://en.wikipedia.org/wiki/Turbojet\n\nhttp://www.grc.nasa.gov/WWW/K-12//airplane/aturbj.html\n\nA TurboFan engine is the modified form of the turbojet engine. It has a fan in the front which is rotated with the help of a turbine connected to it. The hot exaust gases rotates the turbine which in turn rotates the Fan. The fan in turn pushes the air backwards and generates the major thrust. These type of engines are preffered in high speed passenger aircrafts as they can perform well in high altitudes and are quieter compared to turbojet aircrafts. They are also more fuel efficient than Turbojet aircrafts.\n\nSite regardin Turbofan Engine:\n\nhttp://www.grc.nasa.gov/WWW/K-12/airplane/Animation/turbtyp/etfr.html\n\nThe followin site contains description of diff types of jet engines:\n\nhttp://inventors.about.com/library/inventors/bljetenginetypes.htm\n\nHope this helps
258345
The efficiency of an ideal engine is (1- T2/T1) or (T1- T2)/T1 = (600 - 400) / 600 =0.333.\nBut in the problem it is given that it is 0.5 which is not possible.
1131605
Most turbine engines are "free turbine" design, which means that the N1 turbine and the N2 turbine are not physically connected on the same shaft.  (when I say turbine engine, this can be applicable to a turboshaft, turboprop or some jet engines, the principle is all the same).\n\nN1 is the Gas Producer Turbine, whose main function is to turn the compressor, which supplies the engine with air.  The N2 is the power turbine, which is connected through a drive train to a propeller or rotor and carries the load.  These two turbines are both directly downstream from the hot section of the engine.\n\nIf the engine was not governed, rpm changes that occur to the N1 turbine would occur shortly thereafter to the N2.  This would be accomplished by adding more fuel, which would turn the N1 and compressor faster, producing more combustible air, creating a hotter fire, which would cause the N2 to increase in speed.  Free turbine engines, however, are governed to a constant N2 speed.  N2 rpm appears to remain constant.  Small, imperceptible droops in rpm are actually constantly occurring in response to increased load from the prop or rotor.  The governor automatically increases fuel flow, which shows as an increase of N1 and internal temperature, without any obvious change in N2.  Thus, in actual practice, N1 increases with load, (or altitude) and N2 stays the same.\n\nIn a fixed turbine engine, which is used in some airplanes and some older helicopters, the N1 and N2 turbines are connected by the same shaft and turn as one.  Larger fluctuations in rpm and lag times to spool the engine back up after drooping have made this style less attractive to aircraft designers than the free turbine.
292019
If the unit vector has components "a" and "b" (the unknowns for which you want to solve) then your first equation is\n\n1.  a^2 + b^2 = 1\n\nIf you have f and the directional derivative then you can take the partial derivatives of f with respect to x and y (call these f_x and f_y) and get your second equation\n\n2.  a f_x + b f_y = directional derivative\n\nWith two equations in the two unknowns a and b you then try to solve for a and b.
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a) vcenter = (m1*v1+m2+m2)/(m1+m2)\n\nb) v1 rel = v1 - v center\nv2 rel = v2 - v center\n\nremember velocity is a vector so you must keep track of magnitude and direction
1110236
Airflow over the lifting surfaces creates pressure differences between top and bottom sides, which gives you the lifting force.  You have to shove the lifting surfaces through the air in order for them to generate the lift, though, hence why true aeroplanes must have engines to push them along.   If you don't have an engine then you are a glider, and barring thermal upcurrents you will be perpetually falling.   Until...\n\nI think passenger jets can glide at about a 1:20 downward slope with no engine power.
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The system of equations:\n[a11 a12 a13; a21 a22 a23]\n[1 -3 4; 2 3 1]\n\nPerforming indicated row operation:\n[a11 a12 a13; a21' a22' a23']\n[1 -3 4; -1 -6 3]\n\na21' = a11 - a21 = 1 - 2 = -1\na22' = a12 - a22 = -3 - 3 = -6\na23' = a13 - a23 = 4 - 1 = 3
909410
Strange formula, since it has no consideration of the changes when the engine runs out of fuel.\n\nSuggestion, if you have Excel or something like that...\n\nPlug in a series of times in one column - perhaps seconds, or for more precision, try tenths of seconds.\n\nHave the next column calculate the height based on your formula.\n\nThen you could calculate vertical speed in the next column.\nYou can also have Excel plot graphs of this....\n\nSounds like your homework is the usual kind.  Maybe this will help.  \n\nClearly, if you differentiate the expression, you can get a formula for vertical velocity (dH/dT)  \n\nIf you solve for t when vertical velocity = 0, that will be the time of maximum altitude...\n\nAnd so on.  Good luck.
1154907
since P1.V1 / T1 = P2.V2 / T2 ... u don't actually care about the altitude. so V2 = 8.88 cm^3 (btw those degrees are °C, °F ?? cuz u gotta take them to Kevin before using them)\n\n1*10^5 Pa * 2cm^3 / 270 °K = 2*10^4 Pa * V2 / 240 °K\nV2 = 8.88 cm^3
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Horse power can be used for both it is a value given to the amount of energy needed to perform (x) amount of work per second.\n  I believe you are more interested in thrust, which is the forward propulsion of the jet engine.  \n  That would go more towards ft/lbs.  \n\nOne extra little known tid bit.  The super jet engines like on the C-5 military transport uses some water injection on take off.  This is what causes the dark exhaust you see. \n\nIf I remember correctly when water flashes to steam it has a 1:26 pressure expasion rate.  \nAs you can see this has a significant impact on the amount of thrust for the engine as well as a cooling effect.\n\nAnd of course the lift is all by design of the wing.  If you took the wings off of a plane it would never get off the ground...  \n\n    But it would make one fast car !!!
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F = ( m2.v2 - m1.v1)/ t.\n \nF - Force on the body.\n\nm1.v1- initial momentum\n\nm2.v2 - final momentum\n\nt - time for which force acts.\n\nNewton's second law is, "force equals the rate of change in momentum."
561888
Thrust in avionics terms is the same as horse power for a car or truck. It is how much power a jet engine can produce. Drag is things that can limit velocity. A plane need to be sleek to be efficient. when a plane has its landing gear down there is more air resistance, or drag.
783593
V2 = V1 + 0.015v1 = v1(1.015)\ntherefore r = 1.015, and \nVn = 80*1.015^39
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ASSUME V IS OWN SPEED OF THE PLANE & W IS WIND SPEED\n(V+W)*2=225  - WHEN FLYING WITH WIND\n(V-W)*3.75=225 – WHEN FLYING AGAINST\n2 EQUATIONS!!!