<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:
I believe its a and c but my notes are all kinds of messed up so im sorry if its wrong
Explanation:
<span>We can use Coulomb's law to find the force F acting on the proton that is released.
F = k x Q1 x Q2 / r^2
k = 9 x 10^9
Q1 is the charge on one proton which is 1.6 x 10^{-19} C
Q2 is the same charge on the other proton
r is the distance between the protons
F = (9x10^9) x (1.6 x 10^{-19} C) x (1.6 x 10^{-19} C) / (10^{-3})^2
F = 2.304 x 10^{-22} N
We can use the force to find the acceleration.
F = ma
a = F / m
a = (2.304 x 10^{-22} N) / (1.67 x 10^{-27} kg)
a = 1.38 x 10^5 m/s^2
The initial acceleration of the proton is 1.38 x 10^5 m/s^2</span>
Both the object and earth pulls each other towards itself but since the mass and pulling force of objects are very small the pulling force of objects are negligible.
