Answer:
Work done, W = 1.44 kJ
Explanation:
Given that,
Mass of boy, m = 74 kg
Initial speed of boy, u = 1.6 m/s
The boy then drops through a height of 1.56 m
Final speed of boy, v = 8.5 m/s
To find,
Non-conservative work was done on the boy.
Solution,
The work done by the non conservative forces is equal to the sum of total change in kinetic energy and total change in potential energy.
W = 1447.21 Joules
or
W = 1.44 kJ
Therefore, the non conservative work done on the boy is 1.44 kJ.
The magnetic field strength is 9.47 ×10⁻⁴ T
The energy density at the center of the loop is 0.36 J/m³
<h3>Calculating Magnetic field strength & Energy density </h3>From the question, we are to find the magnetic field strength
The magnetic field strength of a loop can be calculated by using the formula,
Where B is the magnetic field strength
is the permeability of free space
is the current
and R is the radius
From the give information,
and
Putting the parameters into the formula, we get
Hence, the magnetic field strength is 9.47 ×10⁻⁴ T
Now, for the energy density
Energy density can be calculated by using the formula,
Where is the energy density
Then,
Hence, the energy density at the center of the loop is 0.36 J/m³
Learn more on Magnetic field stregth & Energy density here: brainly.com/question/13035557