To solve this problem it is necessary to apply the concepts related to the energy density in the magnetic fields. Mathematically the expression that determines the relationship between the magnetic field, the permeability constant and the energy density is given by

Where,
B = Magnetic Field
u = Energy density in magnetic field
= Permeability constant
At the same time the energy of a given volume of space is given as
E = uV
Where
E = Magnetic field energy
V = Volume
Our values are given as



Replacing in the first equation we can find the energy density



Now the net energy would be given by



Therefore the Magnetic Field energy is 91.67J
Answer:
m = 28.7[kg]
Explanation:
To solve this problem we must use the definition of kinetic energy, which can be calculated by means of the following equation.

where:
Ek = kinetic energy = 1800 [J]
m = mass [kg]
v = 11.2 [m/s]
![1800=\frac{1}{2}*m*(11.2)^{2}\\m = 28.7[kg]](https://tex.z-dn.net/?f=1800%3D%5Cfrac%7B1%7D%7B2%7D%2Am%2A%2811.2%29%5E%7B2%7D%5C%5Cm%20%3D%2028.7%5Bkg%5D)
You use the equation Velocity = Acceleration X Time. 4x4=16m/s.
The car travels 18m in 3 seconds.
Answer:
a) A´= A
b) θ₁´ = 29º, θ₂´ = - 169º
, θ₃´ = -49º
Explanation:
In this exercise you are asked to give the magnitudes and angles of the vectors from another system of
reference
a) The magnitudes
The magnitude of a vector, the size of which is a scalar, this does not depend on the reference system, since it is obtained by subtracting the coordinates of the end point minus the coordinate of the origin of the vector
A =
- x₀
if the vectors are measured in another reference frame
x_{f}´ = xx_{f}- U
x₀´ = x₀ -U
where U is the distance between the two reference frames
A´ = x_{f}´ - x₀´
we substitute
A´ = (x_{f} - U) - (x₀-U) = x_{f} - x₀
A´ = A
it does not change
b) Angles
The given angles are measured from the positive part of the x axis in a counterclockwise direction, it is asked to give these angles from the x axis
θ₁ = 29º
does not change
θ₁´ = 29º
θ₂ = 191º
we measure clockwise
θ₂´ = θ₂ - 360
θ₂´ = 191 - 360
θ₂´ = - 169º
θ₃ =311º
we measure clockwise
θ₃´ = 311 -360
θ₃´ = -49º