The magnitude of the magnetic dipole moment of the bar magnet is 1.2 Am²
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Magnetic dipole moment of the bar magnet</h3>
The magnitude of the magnetic dipole moment of the bar magnet at distance from its axis is calculated as follows;
where;
- B is magnetic field
- m is dipole moment
- μ is permeability of free space
m = (4π x 0.1³ x 2.4 x 10⁻⁴)/(2 x 4π x 10⁻⁷)
m = 1.2 Am²
The complete question is below:
What is the magnitude of the magnetic dipole moment of the bar magnet from 0.1 m of its axis and magnetic field strength of 2.4 x 10⁻⁴ T.
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Answer:
13.4cm
Explanation:
According to Rayleigh’s criterion the angular resolution to distinguish two objects is given by:
θ = 50.0*10^-7 rad
λ: wavelength of the light = 550nm
b = diameter of the objective
By doing b the subject of the formula and replacing the values of the angle and wavelength you obtain:
hence, the smallest diameter objective lens is 13.4cm
As Potential energy =mgh
m= 0.95kg
h=3 meter
g = 9.8 m/sec^2. ( acceleration due to gravity)
So P.E =(0.95)(9.8)(3)kgm^2/s^2
P.E =27.93 joules
Answer:
Temperature at the exit = 
Explanation:
For the steady energy flow through a control volume, the power output is given as
Inlet area of the turbine = 
To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.
Assuming Argon behaves as an Ideal gas, we have the specific volume 
as
for Ideal gasses, the enthalpy change can be calculated using the formula
hence we have
<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>
evaluating the above equation, we have 
Hence, the temperature at the exit =
