Answer:
The magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is 0.4713 J
Explanation:
Given;
radius of the circular loop of wire = 0.5 m
current in circular loop of wire = 2 A
strength of magnetic field in the wire = 0.3 T
τ = μ x Bsinθ
where;
τ is the magnitude of the magnetic torque
μ is the dipole moment of the magnetic field
θ is the inclination angle, for a plane area perpendicular to the magnetic field, θ = 90
μ = IA
where;
I is current in circular loop of wire
A is area of the circular loop = πr² = π(0.5)² = 0.7855 m²
μ = 2 x 0.7885 = 1.571 A.m²
τ = μ x Bsinθ = 1.571 x 0.3 sin(90)
τ = 0.4713 J
Therefore, the magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is 0.4713 J
Answer: 3.
Explanation:
The correct answer is a higher amplitude and lower frequency. Since an opera singer is lowering his pitch it means that he is creating higher amplitude and because he is raising his voice for a song with that higher amplitude he is creating lower frequency.
The sketch of the system is: two strings, 1 and 2, are attached to the ceiling and to a third string, 3.The third string holds the bag of cement.
The free body diagram of the weight with the string 3, drives to the tension T3 = weihgt => T3 = 325 N
The other free body diagram is around the joint of the three strings.
In this case, you can do the horizontal forces equilibrium equation as:
T1* cos(60) - T2*cos(40) = 0
And the vertical forces equilibrium equation:
Ti sin(60) + T2 sin(40) = T3 = 325 N
Then you have two equations with two unknown variables, T1 and T2
0.5 T1 - 0.766 T2 = 0
0.866 T1 + 0.643T2 = 325
When you solve it you get, T1 = 252.8 N and T2 = 165 N
Answer: T1 = 252.8 N, T2 = 165N, and T3 = 325N
A force over distance is work the unite is joules
