Answers:
a) 30 m/s
b) 480 N
Explanation:
The rest of the question is written below:
a. What is the final speed of the falcon and pigeon?
b. What is the average force on the pigeon during the impact?
<h3>a) Final speed</h3>
This part can be solved by the Conservation of linear momentum principle, which establishes the initial momentum 
before the collision must be equal to the final momentum 
after the collision:
(1)
Being:
Where:
the mas of the peregrine falcon
the initial speed of the falcon
is the mass of the pigeon
the initial speed of the pigeon (at rest)
the final speed of the system falcon-pigeon
Then:
(2)
Finding :
(3)
(4)
(5) This is the final speed
<h3>b) Force on the pigeon</h3>
In this part we will use the following equation:
(6)
Where:
is the force exerted on the pigeon
is the time
is the pigeon's change in momentum
Then:
(7)
(8) Since 
Substituting (8) in (6):
(9)
(10)
Finally:
Answer:
Explanation:
When a body is held against a vertical wall , to keep them in balanced position , normal force is applied on their surface . this force creates normal reaction which acts against the normal force and it is equal to the normal force as per newton's third law . Ultimately friction force is created which is proportional to normal force and it acts in vertically upward direction . It prevents the body from falling down .
Hence normal force = reaction force .
From second law also net force is zero , so if normal force is N and reaction force is R
R - N = mass x acceleration = mass x 0 = 0
R = N .
Ranking normal force from highest to smallest
150 N , 130 N , 120 N
B )
Frictional force is equal to the weight of the body because the body is held at rest .
Ranking of frictional force form largest to smallest
7 kg , 5 kg , 3 kg , 1 kg .
Here frictional force is irrespective of the normal force acting on the body because frictional force adjusts itself so that it becomes equal to weight in all cases here because it always balances the weight of the body .
The distance an object falls from rest through gravity is
D = (1/2) (g) (t²)
Distance = (1/2 acceleration of gravity) x (square of the falling time)
We want to see how the time will be affected
if ' D ' doesn't change but ' g ' does.
So I'm going to start by rearranging the equation
to solve for ' t '.
D = (1/2) (g) (t²)
Multiply each side by 2 : 2 D = g t²
Divide each side by ' g ' : 2 D/g = t²
Square root each side: t = √ (2D/g)
Looking at the equation now, we can see what happens
to ' t ' when only ' g ' changes:
-- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'
and smaller 'g' ==> longer 't' .
-- They don't change by the same factor, because 1/g is inside
the square root. So 't' changes the same amount as √1/g does.
Gravity on the surface of the moon is roughly 1/6 the value
of gravity on the surface of the Earth.
So we expect ' t ' to increase by √6 = 2.45 times.
It would take the same bottle (2.45 x 4.95) = 12.12 seconds
to roll off the same window sill and fall 120 meters down to the
surface of the Moon.
