first what does the N stand for  Max takeoff thrust = 229,329N ?
474431
All you need to know to convert N*m to lb*ft is to know that they are both units of torque. Sure, you could say they are both units of length multiplied by force, but that's besides the point. To do a conversion you will need to convert between units, that's just a fact of the matter. Why convert more units then necessary? If you know they are both units of torque, you are already done. Look up the conversion factor and there you go.\n\n1 N*m = 0.737561 lb*ft
649863
ok..lets start\n\njet. power to lift off, cruise, and land with power.\n     highest alt is aprox 42000 ft.\n\nshuttle. no power to lift off, orbit or cruise, and land\n        without power (glide, no power)\n        it uses rockets for lift off, and burns off the rest\n        of its fuel for landing(deorbit), the rest of the \n       operation is glidding.\n        max altitue. aprox 200 miles.\n\n     hope it helps
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87\n\nProduction  \nIncludes 33 Merlin IIAs, 87 Merlin IIBs, 92 Merlin III and IIIAs and 10 Merlin 300s.
949041
This problem relies on Newton's second law: F=ma\n\na. This question is not completely specified.  The mass of the rocket is then (F/a=m or 2.0E7/a).\n\nb. We first find the net force on the rocket and use the mass from the previous question. Now we compute the acceleration at liftoff (F/m=a or (25E6 - 2.0E7)/(2.0E7/a)).\n\nc. We derive a new formula by double integration on acceleration. \nintegrate(a dt) = at, integrate(at dt) = .5at^2. The sequence gives us acceleration -> speed -> displacement at time t. The rocket will be at velocity: +(25E6 - 2.0E7)/(2.0E7/a)(7*60) m/s and at displacement of: +.5((25E6 - 2.0E7)/(2.0E7/a))(7*60)^2 m.
800850
68 and -224. You are right. Good job.
178900
I can't figure it out, my brain is slow tonight.\nHave x=air plane speed\nhave y=wind speed\nx+0=225\nso x=225\n225+y=300\n225-y=210\nFrom there I'm lost
862142
229/232.87 + fb/1400 and fb = 25.68  substitute\n229/232.87+25.68/1400\n0.98338128569588182247605960407094=0.018342857142857142857142857142857=1.0017241428387389653332024612138
971158
Go to the siye in english and then click fro the german version on the left, scoll down to the technical data\n\nUnder technical data it shows 311 -363 kN of thrust, depending on the type and the engines used. And it has 4 not 6 engines , soory Dr.Z
698763
Vy=sin(36)(30)=17.63m/s\nVx=cos(36)30= 24.27m/s\n\nwhen it reaches the ground, the speed will be the same as Vy\nVf=at+Vi\n-17.63m/s=(-9.8m/s^2)t+17.63\n-35.26m/s=(-9.8m/s^2)t\nt= 3.597 s\n\nwhen it reaches the highest point, its velocity will be 0m/s\nVf^2=2ad+Vi^2\n(0m/s)^2=2(-9.8m/s^2)d+(17.63m/s)2\n-310.81m^2/s^2=(-19.6m/s^2)d\nd= 15.85m\n\nx=vt\nx=(24.27m/s)(3.597s)\nx=87.29m
56899
In the dictionary\n\n(also at Nasa -- counting down to takeoff)
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centripetal force is mv^2/r.\nso it comes around 1,777,777.778 N\n\nI need one more parameter that is coeff of friction.\nif that is available u can use\n\nv(max)= sqrt [(rg (u + tan o)/(1 - u * tan o) ]\n\nassuming frictionless\n\nV= sqrt (rg tan o)\nso \no = 66.2 degrees if g=9.8\nand 65.7 degrees if g= 10.
48198
General characteristics\nWingspan: between 93.0 ft and 112.6 ft (28.3 m and 34.3 m) (36 m for winglet -700, -800, -900) \nLength: \n102.5 ft (31.2 m) (600) \n110.3 ft (33.6 m) (700) \n129.5 ft (39.5 m) (800) \n138.2 ft (42.1 m) (900, 900ER) \nTail height: \n41.3 ft (12.6 m) (600) \n41.2 ft (12.5 m) (700, 800, 900, 900ER) \nEngines: Two high-bypass turbofan engines, rated at between 64.4 kN and 117.3 kN each \nPratt & Whitney JT8D (100, 200) \nCFMI CFM56-3 (300, 400, 500) \nCFMI CFM56-7 (600, 700, 800, 900, 900ER) \nMaximum takeoff weight: \n143,500 lb (65,090 kg) (600) \n174,200 lb (79,010 kg) (700, 800, 900) \nCapacity: 85 to 215 passengers \nCost: USD $45.5 million to $77 million list price in 2005 [2] \nAutoflight, Displays, Navigation and Sensors by Honeywell \nSection 41, fuselage, and most other components produced in Wichita, Kansas. Final assembly is in Seattle-Renton, Washington.
869482
Reynold's number is (speed * length) / kinematic viscosity\nor 6372(Speed* length)
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"Former Lockheed Raptor chief test pilot Paul Metz says the Raptor has a fixed inlet. Paul Metz has also stated that the F-22 has a top speed greater than 1600 mph (Mach 2.42) and its climb rate is faster than the F-15 Eagle."\n\nHmmm, interesting.  He says that the top speed is greater than Mach 2.42.  I wonder how much greater...
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Juno\nhttp://account.juno.com/s/landing;jsessionid=D7B60AED5B991F707302D8F581CA150E.LAX-AC05-0?action=viewProduct&productId=free&zsessionid=25586319|86400|1142577406\nNet Zero\nhttp://account.netzero.net/s/landing?action=viewProduct&productId=free&group=free_2Plat&cf=wwwb1
2741
http://www.northstaraerospace.com/
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Hi. N stands for Newtons. Does that help?