Answer:
The displacement of the spring due to weight is 0.043 m
Explanation:
Given :
Mass
Kg
Spring constant 
According to the hooke's law,

Where
force,
displacement
Here,
(
)
N
Now for finding displacement,

Here minus sign only represent the direction so we take magnitude of it.

m
Therefore, the displacement of the spring due to weight is 0.043 m
To solve this problem it is necessary to apply the concepts related to the magnetic dipole moment in terms of the current and the surface area, as well as the current density, as a function of the current over the area.
Part A) By definition we know that magnetic dipole moment is

Where,
I = Current
S = Area

Replacing with our values we have that,

Re-arrange to find I,

Part B) To find the Current density we need to find the cross sectional area of the Wire:

Finally the current density is simply J

PART C) Finally to make the comparison with the given values we have to cross-sectional area would be

Therefore the current density would be

Comparing the two values we can see that the 2mm wire has a higher current density.
Answer:

Explanation:
We are given that
Frequency,f=800KHz=

Distance,d=4.5 km=
1 km=1000 m
Electric field,E=0.63V/m
We have to find the magnetic field amplitude of the signal at that point.

We know that



Calculate for the x and y-components of the velocities involved in this item.
2 mi/h (45° east)
x-component = (2 mi/h)(sin 45°)
= 1.41 mi/h
y-component = (2 mil/h)(cos 45°)
= 1.41 mi/h
4 mi/h (east)
x-component = 4 mi/h
y-component = 0 mi/h
Adding up the corresponding components:
x-component = 5.4142 mi/h
y-component = 1.4142 mi/h
Calculating for the resultant,
R = sqrt ((x²) + (y²))
R = sqrt ((5.4142 mi/h)² + (1.4142 mi/h)²)
R = 5.60 mi/h
Answer: 5.6 mi/h
Heat flows irreversibly from hot to cold