Answer:a. Magnetic dipole moment is 0.3412Am²
b. Torque is zero(0)N.m
Explanation: The magnetic dipole moment U is given as the product of the number of turns n times the current I times the area A
That is,
U = n*I*A
But Area A is given as pi*radius² since it is a circular coil
Radius given is 5cm converting to meter we divide by 100 so we have our radius to be 0.05m. So area A is
A = 3.142*(0.05)² =7.86*EXP {-3} m²
Current I is 2 A
Number of turns is 20
So magnetic dipole moment U is
U = 20*2*7.86*EXP {-3}=0.3142A.m²
b. Torque is given as the cross product of the magnetic field B and magnetic dipole moment U
Torque = B x U =B*U*Sine(theta)
But since the magnetic field is directed parallel to the plane of the coil from the question, it means that the angle between them is zero and sine zero is equals 0(zero) if you substitute that into the formula for torque you will find out that your torque would equals zero(0)N.m
What you know:
Vi=0m/s
Vf=143.8m/s
A=-9.8m/s
d=???
Use the equation Vf^2=Vi^2+2A(d)
Rearrange to isolate d: d=Vf^2/2A
d=(143.8)^2/2(-9.8)
d=20678.4/-19.6
d=-1055m
The tank was released from a height of 1055m
Answer:
Explanation:
In order to calculate the angular momentum of the particle you use the following formula:
(1)
r is the position vector respect to the point (0 , 5.0), that is:
r = 0m i + 5.0m j (2)
p is the linear momentum vector and it is given by:
(3)
the direction of p comes from the fat that the particle is moving along the i + j direction.
Then, you use the results of (2) and (3) in the equation (1) and solve for L:
The angular momentum is -30 kgm^2/s ^k
Answer:
1–
Explanation:
The fluorine is the element with biggest electronegativity in the periodic table, so it usually always take an electron and gets charge 1–
Answer:
I = 6.42 A
Explanation:
Given that,
Potential difference, V = 0.9 V
Length of the wire, l = 1.5 m
Area of cross section, 
We need to find the current in the wire. Let I is current. We can find it using Ohm's law as follows :
V = IR
Where R is the resistance of the wire
So, the current in the wire is 6.42 A.
