According to Newton's Second Law of Motion, the net force experienced by the system is equal to the mass of the system in question times the acceleration in motion. In this case, the net force is the difference of gravitational force and the force experience by the motion of the airplane. This difference is already given to be 210 N.
Net force = ma
210 N = (73 kg)(a)
a = +2.92 m/s²
Thus, the acceleration of the airplane's motion is 2.92 m/s² to the positive direction which is upwards.
A). m/s ... meters per second.
That's a speed.
b). m/s² ... (meters per second) per second.
Acceleration.
c). m²/s ... square meters per second.
Rate of change of area.
Meaningless unit, unless you happen to be
a painter of houses or a mower of lawns.
d). kg-m/s² ... (mass) x (an acceleration).
This is a unit of force.
In fact, it's the definition of "newton".
We can use the law of conservation of energy to solve the problem.
The total mechanical energy of the system at any moment of the motion is:

where U is the potential energy and K the kinetic energy.
At the beginning of the motion, the ball starts from the ground so its altitude is h=0 and therefore its potential energy U is zero. So, the mechanical energy is just kinetic energy:

When the ball reaches the maximum altitude of its flight, it starts to go down again, so its speed at that moment is zero: v=0. So, its kinetic energy at the top is zero. So the total mechanical energy is just potential energy:

But the mechanical energy must be conserved, Ef=Ei, so we have

and so, the potential energy at the top of the flight is
Answer:

Explanation:
The total charge is distributed over the two objects:

The plate and the rod must have
. So the charge transferred from the plate to the rod is:

Number of electrons:

Answer:
t = 1.05 s
Explanation:
Given,
The distance between your vehicle and car, 100 ft
The constant speed of your vehicle, u = 95 ft/s
Since, the velocity is constant, a =0
If the car stopped suddenly, time left for you to hit the brake, t = ?
Using the second equation of motion,
S = ut + ½ at²
Substituting the given values in the equation
100 = 95 x t
t = 100/95
= 1.05 s
Hence, the time left for you to hit the brakes and stop before rear ending them, t = 1.05 s