Answer:
The maximum emf that can be generated around the perimeter of a cell in this field is 
Explanation:
To solve this problem it is necessary to apply the concepts on maximum electromotive force.
For definition we know that

Where,
N= Number of turns of the coil
B = Magnetic field
Angular velocity
A = Cross-sectional Area
Angular velocity according kinematics equations is:



Replacing at the equation our values given we have that




Therefore the maximum emf that can be generated around the perimeter of a cell in this field is 
Answer:
Minimum diameter=1.714
Explanation:
Angular magnification=fob/fey=42
fob=40*fey
Hence minimum diameter=diameter of objective/angular magnification
minimum diameter=72/42=1.714
Answer:
Wavelength = 1.36 * 10^{-34} meters
Explanation:
Given the following data;
Mass = 0.113 kg
Velocity = 43 m/s
To find the wavelength, we would use the De Broglie's wave equation.
Mathematically, it is given by the formula;

Where;
h represents Planck’s constant.
m represents the mass of the particle.
v represents the velocity of the particle.
We know that Planck’s constant = 6.6262 * 10^{-34} Js
Substituting into the formula, we have;


Wavelength = 1.36 * 10^{-34} meters
30000 btuh /3413 btuh/kw. = 8.8 kw
8.8 kw/.746 kw/hp = 11.8 hp if COP is 1
11.8/3 hp (COP coefficient of performance) = 3.99 COP
>>>So yes a 3.0 hp compressor with a nominal COP of 4 will handle the 30,000 btuh load.
3.2 to 4.5 is expected COP range for an air cooled heat pump or a/c unit.