The power of the lamp would be calculated with the equation of ohm laws. P = U x I = 122V x 0.1A = 12.2W
<span>v is perpendicular to both E and B and has a magnitude E/B</span>
Answer:
A) μ = A.m²
B) z = 0.46m
Explanation:
A) Magnetic dipole moment of a coil is given by; μ = NIA
Where;
N is number of turns of coil
I is current in wire
A is area
We are given
N = 300 turns; I = 4A ; d =5cm = 0.05m
Area = πd²/4 = π(0.05)²/4 = 0.001963
So,
μ = 300 x 4 x 0.001963 = 2.36 A.m².
B) The magnetic field at a distance z along the coils perpendicular central axis is parallel to the axis and is given by;
B = (μ_o•μ)/(2π•z³)
Let's make z the subject ;
z = [(μ_o•μ)/(2π•B)] ^(⅓)
Where u_o is vacuum permiability with a value of 4π x 10^(-7) H
Also, B = 5 mT = 5 x 10^(-6) T
Thus,
z = [ (4π x 10^(-7)•2.36)/(2π•5 x 10^(-6))]^(⅓)
Solving this gives; z = 0.46m =
Answer:
Check the explanation
Explanation:
The beat frequency is
df = f2 - f1
the wavelength is
lamda1 = (v/f1)
and lamda2 = (v/f2)
where v = 340 m/s,f1 = 25.0 kHz and f2 = 20.0 kHz
The maximum amount of work performed is
Explanation:
The efficiency of a real heat engine is given by the equation:
(1)
where
is the temperature of the cold reservoir
is the temperature of the hot reservoir
However, the efficiency of a real heat engine can be also written as:
where
is the maximum work done
is the heat absorbed from the hot reservoir
can be written as
where
is the heat released to the cold reservoir
So the previous equation can be also written as
(2)
By combining eq.(1) and (2) we get
And re-arranging the equation and solving for , we find
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