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How much acceleration does a 747 jumbo jet of mass 29000 kg experience in takeoff when the thrust for each of four engines is 30000 N?
Wiki User
∙ 13y ago
Use Newton's Second Law, F=ma (force = mass x acceleration). Solving for acceleration: a = F/m. Since you are using SI units, the answer will be in meters/second2.
Wiki User
∙ 13y ago
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When is the acceleration of a planes takeoff the greatest?
i guess when the engines are set to full thrust, and when the plane leaves the ground...
How much acceleration does a 747 jumbo jet of mass 29600 kg experience in takeoff when the thrust for each of four engines is 30000 N?
You are supposed to use the formula for Newton's Second Law: F=ma
How much acceleration does a 747 jumbo jet of mass 30000 kg experience in takeoff when the thrust for each of four engines is 30000n?
F = M x A 120,000/30,000 = 4 so it would be 4m/s2 (doesn't seem a lot but the figures aren't close 747 are more than 30 tons)
Calculate the acceleration of a 2000-kg single-engine airplane just before takeoff when the thrust of its engine is 500 n?
500 N/ 200kg = .25 m/s/s
Calculate the acceleration of a 2000 kg single-engine airplane just before takeoff when the thrust of its engine is 500 N?
acceleration = net force / mass A = 500 N/ 2000 kg a = 0.25 m/s/s or 0.25 m/s2
When is the acceleration of a planes takeoff the greatest?
i guess when the engines are set to full thrust, and when the plane leaves the ground...
How much acceleration does a 747 jumbo jet of mass 29600 kg experience in takeoff when the thrust for each of four engines is 30000 N?
You are supposed to use the formula for Newton's Second Law: F=ma
Calculate the acceleration of a 3000 kg jumbo jet just before takeoff when the thrust for each of its four engines is 30 000N?
40 m/s2
How much acceleration does a 747 jumbo jet of mass 30000 kg experience in takeoff when the thrust for each of four engines is 30000n?
F = M x A 120,000/30,000 = 4 so it would be 4m/s2 (doesn't seem a lot but the figures aren't close 747 are more than 30 tons)
Calculate the acceleration of a 300000-kg jumbo jet just before takeoff when the thrust of it's four engine is 30000n?
Easy; use the acceleration formula: A=F/M. Now, there are four engines, correct? So you multiple the 30,000 by 4. This gives you 120,000. Divided that by the original 300,000. Your answer should be .4 m/s squared.
How much thrust does a jet use for take off?
For jet engines, higher thrust requires higher hot section temperatures. And higher temperatures reduce engine life, so commercial jets usually do not use full thrust on takeoff. Instead, the needed engine pressure ratio is calculated based on weather conditions, load, and runway length. Military fighters and smaller general aviation aircraft use all they have on takeoff. Do military transports use reduced thrust on takeoff, like the C-17?
Calculate the acceleration of a 2000-kg single-engine airplane just before takeoff when the thrust of its engine is 500 n?
500 N/ 200kg = .25 m/s/s
Does an aeroplane have two engines?
It can have any number of engines of as many different types as needed. For example the B-36 bomber was originally designed with six radial 28 cylinder piston engines driving pusher propellers. Later four small jet engines were added, making it the airplane with the most engines of any mass-produced aircraft ever built. For some airplanes optional jet-assisted takeoff or rocket-assisted takeoff drop away engines can be added for increased thrust at takeoff without extra weight of those engines while in the air in normal flight.
Calculate the acceleration of a 2000 kg single-engine airplane just before takeoff when the thrust of its engine is 500 N?
acceleration = net force / mass A = 500 N/ 2000 kg a = 0.25 m/s/s or 0.25 m/s2
What is a thrust for the jet engines?
The forward propulsion of an aircraft is called thrust.The thrust is given by the engines.
How much acceleration does a 747 jumbo jet of mass 30000kg experience when the thrust for each of its four engines is 30000 N?
Ignoring air resistance, (although we know thatthat's definitely not a valid assumption when we're dealing with an airliner! )Total force = 4 x 30,000 = 120,000 newtonsF = M AA = F / M = (120,000) / (30,000) = 4m/sec2 = about 0.41 of a 'G' .____________________________________________________________________________________________________________In actuality a fully loaded B747-200 airliner has a maximum takeoff weight of 990,000 Lbs which is 450,000 Kg mass.Each General Electric CF6 Turbofan jet engine on the B747-200 has a maximum static thrust output of 63,300 Lbs which is equal 281,685 Newtons of thrust output per engine.There are four (4) engines on the B747-200. Therefore at maximum static thrust output power there will be 1,126,740 Newtons of maximum static thrust on the Boeing 747-200 airliner.The Drag Coefficient (Cd), Aircraft Projected Frontal Area (A), Air Density in Slugs per Cubic Foot (P) and Airspeed (V) are factors which when all combined give the Aerodynamic Drag which resists forward acceleration and counteracts thrust. Tire rolling resistance also resists forward acceleration as well when the airliner is rolling on takeoff during the ground.But let's ignore Aerodynamic Drag & Tire Rolling Resistance in this case and focus purely on the acceleration that is incurred on this airliner at the given thrust output during runway roll acceleration only.Let say a Boeing 747-200 airliner's acceleration is to be calculated during it takeoff roll with the given mass & static thrust information from rest (0 MPH) to 170 MPH within 30 seconds of runway roll time.mass = 450,000 kgforce (thrust) = 1,126,740 NAcceleration = [(Force) / (Mass)]Acceleration = [(Distance) / (Time x Time)]Acceleration = [(Speed / Time)]1 MPH = 1.467 feet per second1 meter = 3.28 feet170 MPH x 1.467 = 249.39 feet per second249.39 feet per second / 3.28 = 76 meters/secAcceleration = [(76 meters per second) / (30 seconds)] = 2.5 meters per second squared.Acceleration = [(1,126,740 Newtons) / (450,000 Kg)] = 2.5 meters per second squared.The acceleration is calculated at: 2.5 m/s^2The horizontal G-Force is a ratio between the horizontal acceleration & gravitational acceleration plus a reference Horizontal G Force of 1.0.The gravitational acceleration on Earth is: 9.8 m/s^2.Horizontal G-Force during runway roll acceleration: [(1) + (2.5 / 9.8)] = +1.25 Horizontal Gs'The Horizontal G-Force is calculated at: +1.25 Horizontal Gs'During cruise flight the airliner's speed is held at a relatively constant rate so therefore the acceleration is zero.Acceleration is a substantial factor on an airliner during runway takeoff roll more than any other time. Even during landing procedures, there is a gradual deceleration but not the extent of the takeoff acceleration where the airliner moves from 0 MPH to 170 MPH (0 m/s to 76 m/s) in a time period of 30 seconds!
What provides thrust to a real airplane?
The thrust is an airplane is provided by the engines.
