Answer:
The magnitude of the magnetic torque on the coil is 1.98 A.m²
Explanation:
Magnitude of magnetic torque in a flat circular coil is given as;
τ = NIASinθ
where;
N is the number of turns of the coil
I is the current in the coil
A is the area of the coil
θ is the angle of inclination of the coil and magnetic field
Given'
Number of turns, N = 200
Current, I = 7.0 A
Angle of inclination, θ = 30°
Diameter, d = 6 cm = 0.06 m
A = πd²/4 = π(0.06)²/4 = 0.002828 m²
τ = NIASinθ
τ = 200 x 7 x 0.002828 x Sin30
τ = 1.98 A.m²
Therefore, the magnitude of the magnetic torque on the coil is 1.98 A.m²
Answers:
B.) 
C.) 
Explanation:
The image attached shows the way the force 
is acting on the block. Now, if we draw a free body diagram of the situation and write the equations for the Net Force in X and Y, we will have the following:
Net Force in X:
(1)
Where:
is the Normal force
is the magnitude of the force exerted on the block
is the angle
Net Force in Y:
(2)
Where:
is the Friction force (it is expresed with the 
sign because this force may be up or down, we cannot know because the block is at rest)
is the gravity force
Rewrittin (1):
(3) This is according to option B
Rewritting (2):
(3) This is according to option C
Answer:
t = 94.91 nm
Explanation:
given,
wavelength of the light = 522 nm
refractive index of the material = 1.375
we know the equation
c = ν λ
where ν is the frequency of the wave
c is the speed of light
ν = 5.75 x 10¹⁴ Hz
the thickness of the coating will be calculated using
t = 94.91 nm
the thickness of the coating will be equal to t = 94.91 nm
1.velocity and acceleration
2.
3.inertia
4.
5.speed
