Answer:
Given the area A of a flat surface and the magnetic flux through the surface
it is possible to calculate the magnitude
.
Explanation:
The magnetic flux gives an idea of how many magnetic field lines are passing through a surface. The SI unit of the magnetic flux
is the weber (Wb), of the magnetic field B is the tesla (T) and of the area A is (
). So 1 Wb=1 T.m².
For a flat surface S of area A in a uniform magnetic field B, with
being the angle between the vector normal to the surface S and the direction of the magnetic field B, we define the magnetic flux through the surface as:

We are told the values of
and B, then we can calculate the magnitude

(a) 1200 rad/s
The angular acceleration of the rotor is given by:

where we have
is the angular acceleration (negative since the rotor is slowing down)
is the final angular speed
is the initial angular speed
t = 10.0 s is the time interval
Solving for
, we find the final angular speed after 10.0 s:

(b) 25 s
We can calculate the time needed for the rotor to come to rest, by using again the same formula:

If we re-arrange it for t, we get:

where here we have
is the initial angular speed
is the final angular speed
is the angular acceleration
Solving the equation,

Answer:
Tension, T = 87.63 N
Explanation:
Given that,
Mass of the object, m = 6.9 kg
The string is acting in the upward direction, a = 2.9 m/s²
Acceleration due to gravity, g = 9.8 m/s²
As the lift is accelerating upwards, it means the net force acting on it is given by :
T = m(a+g)
= 6.9 (2.9+9.8)
= 6.9(2.7)
= 87.63 N
So, the tension in the string is 87.63 N.
Answer:
A u = 0.36c B u = 0.961c
Explanation:
In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains
u ’= (u-v) / (1- uv / c²)
Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory
The data give is u ’= 0.43c and the initial core velocity v = 0.94c
Let's clear the speed with respect to the observer (u)
u’ (1- u v / c²) = u -v
u + u ’uv / c² = v - u’
u (1 + u ’v / c²) = v - u’
u = (v-u ’) / (1+ u’ v / c²)
Let's calculate
u = (0.94 c - 0.43c) / (1+ 0.43c 0.94 c / c²)
u = 0.51c / (1 + 0.4042)
u = 0.36c
We repeat the calculation for the other piece
In this case u ’= - 0.35c
We calculate
u = (0.94c + 0.35c) / (1 - 0.35c 0.94c / c²)
u = 1.29c / (1- 0.329)
u = 0.961c