Answer:
Following are the solution to the given question:
Explanation:
The input linear polarisation was shown at an angle of 
. It's a very popular use of a half-wave plate. In particular, consider the case 
, at which the angle of rotation is 
. HWP thereby provides a great way to turn, for instance, a linear polarised light that swings horizontally to polarise vertically. Illustration of action on event circularly polarized light of the half-wave platform. Customarily it is the slow axis of HWP that corresponds to either the rotation. Note that perhaps the vector of polarization is "double-headed," i.e., the electromagnetic current swinging back and forward in time. Therefore the turning angle could be referred to as the rapid axis to reach the same result. Please find the attached file.
Answer:
0.0025116weber/m²
Explanation:
Magnetic field density (B) is the ratio of the magnetic flux (¶) through the loop to its cross sectional area (A).
Mathematically;
B = ¶/A
¶ = BA
Given B = 0.23Tesla which is the magnitude of the magnetic field
Dimension of the rectangular loop = 7.8 cm by 14 cm
Area of the rectangular loop perpendicular to the field B = 7.8cm×14cm
= 109.2cm²
Converting this value to m²
Area of the loop = 109.2 × 10^-4
Area of the loop = 0.01092m²
Magneto flux = 0.23×0.01092
Magnetic flux = 0.0025116weber/m²
Answer:
Magnetic field experienced = 4.5 × 10⁻⁴ T
Explanation:
The magnetic field around an infinite straight current-carrying wire at a distance r from the wire is given by
B = (μ₀I)/(2πr)
B = ?
I = 20 KA = 20000 A
r = 8.9 m
μ₀ = magnetic permeability = 1.257 × 10⁻⁶ T.m/A
B = (1.257 × 10⁻⁶ × 20000)/(2π×8.9) = 4.5 × 10⁻⁴ T
It stands for Unified Computing System
I hope I helped :3
Answer:
a) Em₀ = 42.96 104 J
, b) 
= -2.49 105 J
, c) vf = 3.75 m / s
Explanation:
The mechanical energy of a body is the sum of its kinetic energy plus the potential energies it has
Em = K + U
a) Let's look for the initial mechanical energy
Em₀ = K + U
Em₀ = ½ m v2 + mg and
Em₀ = ½ 50.0 (1.20 102) 2 + 50 9.8 142
Em₀ = 36 104 + 6.96 104
Em₀ = 42.96 104 J
b) The work of the friction force is equal to the change in the mechanical energy of the body
= Em₂ -Em₀
Em₂ = K + U
Em₂ = ½ m v₂² + m g y₂
Em₂ = ½ 50 85 2 + 50 9.8 427
Em₂ = 180.625 + 2.09 105
Em₂ = 1,806 105 J
= Em₂ -Em₀
= 1,806 105 - 4,296 105
= -2.49 105 J
The negative sign indicates that the work that force and displacement have opposite directions
c) In this case the work of the friction going up is already calculated in part b and the work of the friction going down would be 1.5 that job
We have that the work of friction is equal to the change of mechanical energy
= ΔEm
= Emf - Emo
-1.5 2.49 10⁵ = ½ m vf² - 42.96 10⁴
½ m vf² = -1.5 2.49 10⁵ + 4.296 10⁵
½ 50.0 vf² = 0.561
vf = √ 0.561 25
vf = 3.75 m / s
