Answer:
The airliner travels 1.65 km along the runway before coming to a halt.
Explanation:
Given
Resistive forces = (2.90 × 10⁵) N = 290000 N
Mass of the airliner = (1.70 × 10⁵) kg = 170000 kg
Velocity of airliner = 75 m/s
Let the distance over moved by the airliner be equal to d
According to the work-energy theorem, the work done by the resistive forces in stopping the airliner is equal to the travelling kinetic energy of the airliner.
Work done by the resistive forces = (290000) × d = (290,000d) J
Kinetic energy of the airliner = (1/2)(170000)(75²) = 478,125,000 J
290000d = 478,125,000
d = (478,125,000/290,000)
d = 1648.7 m = 1.65 km
Hope this helps!!!
No
Explanation:Brooo it does not contribute to the eco system
no thanks your a fat h/o/e
The only equation you need is the Δy = v₀t + 1/2at², and in this case, v₀t = 0 because the swimmers are starting from rest. Therefore, Δy = 1/2at², and that is 1/2 x (10m/s²) x (1.5)² = 11.25m.
With sigfigs it is 11m.
Answer:
The angle is 18.3 degree.
Explanation:
A uniformly charged infinite plane, density σ = 4 x 10^-9 C/cm^2, is placed vertically in air. A small ball of mass 8 g, with charge q = 10^-8 C, hangs close to the plane, so that the string is initially parallel to the plane. Take g = 9.8m/s2. When in equilibrium, by what angle is the string hanging the ball to the plane?
surface charge density, σ = 4 x 10^-5 C/m^2
Charge, q = 10^-8 C
mass, m = 0.008 kg
Let the angle is A and the tension in the string is T.
The electric field due to a plane is
Now equate the forces,
