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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?
To find the acceleration of the 747 jet during takeoff, we need to calculate the total thrust generated by all four engines. The total thrust is 4 times the thrust of each engine, which equals 120,000 N. Then, we can use Newton's second law (F=ma) to calculate the acceleration, which is 120,000 N divided by the mass of the plane (29,000 kg), resulting in an acceleration of approximately 4.14 m/s^2.
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.
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 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!
How many 30kg in g?
There are 30,000 grams in 30 kilograms.
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?
To find the acceleration of the 747 jet during takeoff, we need to calculate the total thrust generated by all four engines. The total thrust is 4 times the thrust of each engine, which equals 120,000 N. Then, we can use Newton's second law (F=ma) to calculate the acceleration, which is 120,000 N divided by the mass of the plane (29,000 kg), resulting in an acceleration of approximately 4.14 m/s^2.
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.
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 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 is 2 percent of 30000?
2% of 30000 = 2% * 30000 = 0.02 * 30000 = 600
What is 29 percent of 30000?
29% of 30000 = 29% * 30000 = 0.29 * 30000 = 8700
What is 7 percent of 30000?
7 % of 30000 = 7/100 * 30000 = 0.07 * 30000 = 2100
what is 10000+20000?
30000
How do you write 30000 as a fraction?
30000 or 30000 over 1 _____ 1
What is 25987 rounded to the nearest ten thousand?
30,000.
What is 1 percent of 300.00?
1 percent of 30000 = 3001% of 30000= 1% * 30000= 0.01 * 30000= 300
What is 15 percent of 30000?
15% of 30,000= 15% * 30000= 0.15 * 30000= 4,500
