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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 30000 kg experience in takeoff when the thrust for each of four engines is 30000n?
The total thrust produced by all four engines is 120,000 N (30,000 N x 4). To calculate acceleration, we use Newton's second law: acceleration = total force / mass. Plugging in the numbers, the acceleration experienced during takeoff is 4 m/s^2 (120,000 N / 30,000 kg).
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?
To find the acceleration, we first need to calculate the total thrust produced by all four engines. Since each engine produces 30,000 N of thrust, the total thrust is 4 * 30,000 = 120,000 N. The acceleration can be calculated using Newton's second law, F = ma, where F is the total thrust (120,000 N) and m is the mass of the jet (29,600 kg). Plugging in the values, we get a = 120,000 N / 29,600 kg ≈ 4.05 m/s^2.
What causes an airplane to accelerate?
An airplane accelerates due to the thrust generated by its engines. As the engines produce forward thrust, the aircraft gains speed. The thrust must overcome drag forces acting on the airplane to achieve acceleration.
Why is thrust so important in aviation?
In aviation, thrust is the force along the axis of the aircraft which moves it through the air. Together with drag, lift and weight, it is one of the four forces required to make an aircraft fly.
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...
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?
The total thrust produced by all four engines is 120,000 N (30,000 N x 4). To calculate acceleration, we use Newton's second law: acceleration = total force / mass. Plugging in the numbers, the acceleration experienced during takeoff is 4 m/s^2 (120,000 N / 30,000 kg).
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?
To find the acceleration, we first need to calculate the total thrust produced by all four engines. Since each engine produces 30,000 N of thrust, the total thrust is 4 * 30,000 = 120,000 N. The acceleration can be calculated using Newton's second law, F = ma, where F is the total thrust (120,000 N) and m is the mass of the jet (29,600 kg). Plugging in the values, we get a = 120,000 N / 29,600 kg ≈ 4.05 m/s^2.
What causes an airplane to accelerate?
An airplane accelerates due to the thrust generated by its engines. As the engines produce forward thrust, the aircraft gains speed. The thrust must overcome drag forces acting on the airplane to achieve acceleration.
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?
How does a rocket take off and what forces are involved?
A rocket takes off by igniting its engines, which produce thrust that propels the rocket upward. The main forces involved in the rocket's takeoff are thrust and gravity. Thrust overcomes gravity, allowing the rocket to lift off the ground and travel into space.
Why is thrust so important in aviation?
In aviation, thrust is the force along the axis of the aircraft which moves it through the air. Together with drag, lift and weight, it is one of the four forces required to make an aircraft fly.
What is the force used in the taking off of the plane?
The main force used in the takeoff of a plane is thrust generated by the engines. This thrust propels the plane forward and allows it to overcome drag, enabling the wings to generate lift for the plane to become airborne. Additional factors like lift from the wings and rotation of the aircraft also play a role in the takeoff process.
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 energy does a jet plane need at the point of takeoff?
A jet plane needs a significant amount of kinetic energy for takeoff, which is generated by the engines providing thrust to overcome drag forces and lift the aircraft off the ground. Additionally, potential energy is required to lift the plane against gravity. The engines must work at full power during takeoff to generate enough speed and lift for the aircraft to become airborne.
