I think is between A&B
I think I would answer B
Density =mass/Volume
0.8=mass/60
Mass=0.8*60
Mass=48g
Answer:
The average induced emf in the coil is 0.0286 V
Explanation:
Given;
diameter of the wire, d = 11.2 cm = 0.112 m
initial magnetic field, B₁ = 0.53 T
final magnetic field, B₂ = 0.24 T
time of change in magnetic field, t = 0.1 s
The induced emf in the coil is calculated as;
E = A(dB)/dt
where;
A is area of the coil = πr²
r is the radius of the wire coil = 0.112m / 2 = 0.056 m
A = π(0.056)²
A = 0.00985 m²
E = -0.00985(B₂-B₁)/t
E = 0.00985(B₁-B₂)/t
E = 0.00985(0.53 - 0.24)/0.1
E = 0.00985 (0.29)/ 0.1
E = 0.0286 V
Therefore, the average induced emf in the coil is 0.0286 V
Given Information:
Mass of sock = 0.23 kg
Stretched length of sock = x = 2.54 cm = 0.0254 m
Required Information:
Spring constant = k = ?
Answer:
Spring constant = k = 88.82 N/m
Explanation:
We know from the Hook's law that
F = kx
Where k is spring constant, F is the applied force and x is length of sock being stretched.
k = F/x
Where F is given by
F = mg
F = 0.23*9.81
F = 2.256 N
So the spring constant is
k = 2.256/0.0254
k = 88.82 N/m
Therefore, the spring constant of the sock is 88.82 N/m
