<h2>
Answer:442758.96N</h2>
Explanation:
This problem is solved using Bernoulli's equation.
Let
be the pressure at a point.
Let
be the density fluid at a point.
Let
be the velocity of fluid at a point.
Bernoulli's equation states that
for all points.
Lets apply the equation of a point just above the wing and to point just below the wing.
Let
be the pressure of a point just above the wing.
Let
be the pressure of a point just below the wing.
Since the aeroplane wing is flat,the heights of both the points are same.

So,
Force is given by the product of pressure difference and area.
Given that area is
.
So,lifting force is 
Answer:
sun, earth, capitol building, human, atom
Explanation:
Gravity is directly proportional to the size of an object. Therefore, the object with the most mass will have the greatest force of gravity. We know this from the equation. 
Answer: Both cannonballs will hit the ground at the same time.
Explanation:
Suppose that a given object is on the air. The only force acting on the object (if we ignore air friction and such) will be the gravitational force.
then the acceleration equation is only on the vertical axis, and can be written as:
a(t) = -(9.8 m/s^2)
Now, to get the vertical velocity equation, we need to integrate over time.
v(t) = -(9.8 m/s^2)*t + v0
Where v0 is the initial velocity of the object in the vertical axis.
if the object is dropped (or it only has initial velocity on the horizontal axis) then v0 = 0m/s
and:
v(t) = -(9.8 m/s^2)*t
Now, if two objects are initially at the same height (both cannonballs start 1 m above the ground)
And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.
You can notice that the fact that one ball is fired horizontally and the other is only dropped does not affect this, because we only analyze the vertical problem, not the horizontal one. (This is something useful to remember, we can separate the vertical and horizontal movement in these type of problems)
Electric field due to a point charge is given as

here we know that

also the distance is given as

now we will have

so we will have

so above is the electric field due to proton