Answer:
23.5
Explanation:
Dunno how 2 explain but this is correct 4 sureeeeee.
Answer:
ω' = 0.815 rad/s
Explanation:
Given,
R = 1.20 m
Inertia of merry-go- round= 240 kg.m²
Rotating speed = 9 rpm =
=0.9424 rad/s
mass of the child, m = 26 kg
angular speed of the merry-go-round=?
we know
Angular momentum, L = I ω
Moment of inertia of the child
I' = m r² = 26 x 1.2² = 37.44 kgm²
Conservation of angular momentum
initial angular momentum = Final angular momentum
I ω = (I+I')ω'
240 x 0.9424 = (240+37.44) ω'
226.176= 277.44 ω'
ω' = 0.815 rad/s
new angular speed of the merry-go- round is equal to 0.815 rad/s
To solve this exercise it is necessary to apply the equations related to the magnetic moment, that is, the amount of force that an image can exert on the electric currents and the torque that a magnetic field exerts on them.
The diple moment associated with an iron bar is given by,
Where,
Dipole momento associated with an Atom
N = Number of atoms
y previously given in the problem and its value is 2.8*10^{-23}J/T
The number of the atoms N, can be calculated as,
Where
Density
Molar Mass
A = Area
L = Length
Avogadro number
Then applying the equation about the dipole moment associated with an iron bar we have,
PART B) With the dipole moment we can now calculate the Torque in the system, which is
<em>Note: The angle generated is perpendicular, so it takes 90 ° for the calculation made.</em>