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:
The 
electrons are moving through the superconductor per second.
Explanation:
Given :
Current 
A
Charge of electron 
C
Time 
sec
From the formula of current,
Current is the number of charges flowing per unit time.
Where 
number of charges means in our case number of electrons
Therefore, electrons are moving through the superconductor per second.
Answer:
Explanation:
Given that,
Weight of jet
W = 2.25 × 10^6 N
It is at rest on the run way.
Two rear wheels are 16m behind the front wheel
Center of gravity of plane 10.6m behind the front wheel
A. Normal force entered on the ground by front wheel.
Taking moment about the the about the real wheel.
Check attachment for better understanding
So,
Clock wise moment = anti-clockwise moment
W × 5.4 = N × 16
2.25 × 10^6 × 5.4 = 16•N
N = 2.25 × 10^6 × 5.4 / 16
N = 7.594 × 10^5 N
B. Normal force on each of the rear two wheels.
Using the second principle of equilibrium body.
Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces
ΣFy = 0
Nr + Nr + N — W = 0
2•Nr = W—N
2•Nr = 2.25 × 10^6 — 7.594 × 10^5
2•Nr = 1.491 × 10^6
Nr = 1.491 × 10^6 / 2
Nr = 7.453 × 10^5 N
Answer:
Period of brightness variation and luminosity.
Explanation:
The Cepheid variables are used as distance indicators. This requires estimation of periods and (usually) intensity-mean magnitudes in order to establish a period—apparent luminosity relation. It is particularly important for the techniques employed to be as accurate and efficient as possible.
Mechanical advantage is the amount by which a machine multiplies input force. It is a measure of the force amplification using a tool or machine system.
