Answer:
Force exerted, F = 1.5 N
Explanation:
It is given that, a boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s.
i.e. u = 0
v = 30 m/s
Time taken, t = 0.06 s
Mass of the paper, m = 0.003 kg
We need to find the force the boxer exert on it. The force can be calculated using second law of motion as :
F = 1.5 N
So, the force the boxer exert on the paper is 1.5 N. Hence, this is the required solution.
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:
The coefficient of kinetic friction between the sled and the snow is 0.0134
Explanation:
Given that:
M = mass of person = 52 kg
m = mass of sled = 15.2 kg
U = initial velocity of person = 3.63 m/s
u = initial velocity of sled = 0 m/s
After collision, the person and the sled would move with the same velocity V.
a) According to law of momentum conservation:
Total momentum before collision = Total momentum after collision
MU + mu = (M + m)V
Substituting values:
The velocity of the sled and person as they move away is 2.81 m/s
b) acceleration due to gravity (g) = 9.8 m/s²
d = 30 m
Using the formula:
The coefficient of kinetic friction between the sled and the snow is 0.0134