Answer:
The magnitude of momentum of the airplane is 
.
Explanation:
Given that,
Mass of the airplane, m = 3400 kg
Speed of the airplane, v = 450 miles per hour
Since, 1 mile per hour = 0.44704 m/s
v = 201.16 m/s
We need to find the magnitude of momentum of the airplane. It is given by the product of mas and velocity such that,
or
So, the magnitude of momentum of the airplane is . Hence, this is the required solution.
Normally, the number of electrons is equal to the number of protons, which makes atoms electrically neutral. The number of protons in an atom is the defining feature of an atom. It's what makes one element different from another
HOPE THIS HELPS!!!
This may helpv^2=u^2+2as. v=0 at top of flight. a=acceleration of gravity(vo^2)/2a=s.
Answer:
A) μ = A.m²
B) z = 0.46m
Explanation:
A) Magnetic dipole moment of a coil is given by; μ = NIA
Where;
N is number of turns of coil
I is current in wire
A is area
We are given
N = 300 turns; I = 4A ; d =5cm = 0.05m
Area = πd²/4 = π(0.05)²/4 = 0.001963
So,
μ = 300 x 4 x 0.001963 = 2.36 A.m².
B) The magnetic field at a distance z along the coils perpendicular central axis is parallel to the axis and is given by;
B = (μ_o•μ)/(2π•z³)
Let's make z the subject ;
z = [(μ_o•μ)/(2π•B)] ^(⅓)
Where u_o is vacuum permiability with a value of 4π x 10^(-7) H
Also, B = 5 mT = 5 x 10^(-6) T
Thus,
z = [ (4π x 10^(-7)•2.36)/(2π•5 x 10^(-6))]^(⅓)
Solving this gives; z = 0.46m =
