Answer:
Explanation:
Given a square side loop of length 10cm
L=10cm=0.1m
Then, Area=L²
Area=0.1²
Area=0.01m²
Given that, frequency=60Hz
And magnetic field B=0.8T
a. Flux Φ
Flux is given as
Φ=BA Sin(wt)
w=2πf
Φ=BA Sin(2πft)
Φ=0.8×0.01 Sin(2×π×60t)
Φ=0.008Sin(120πt) Weber
b. EMF in loop
Emf is given as
EMF= -N dΦ/dt
Where N is number of turns
Φ=0.008Sin(120πt)
dΦ/dt= 0.008×120Cos(120πt)
dΦ/dt= 0.96Cos(120πt)
Emf=-NdΦ/dt
Emf=-0.96NCos(120πt). Volts
c. Current induced for a resistance of 1ohms
From ohms law, V=iR
Therefore, Emf=iR
i=EMF/R
i=-0.96NCos(120πt) / 1
i=-0.96NCos(120πt) Ampere
d. Power delivered to the loop
Power is given as
P=IV
P=-0.96NCos(120πt)•-0.96NCos(120πt)
P=0.92N²Cos²(120πt) Watt
e. Torque
Torque is given as
τ=iL²B
τ=-0.96NCos(120πt)•0.1²×0.8
τ=-0.00768NCos(120πt) Nm
Answer:
Capacitance, C = 26.1 picofarad
Explanation:
It is given that,
Side of square, x = 4.3546 cm = 0.043546 m
Distance between electrodes, d = 0.6408 mm = 0.0006408 m
Voltage, V = 73.68 V
Capacitance of parallel plates is given by :
or
C = 26.1 picofarad
So, the capacitance of the capacitor is 26.1 picofarad. Hence, this is the required solution.
Answer:
Seriously I have no idea. I need help with my homework.
Explanation:
I really need help with my homework. Sorry
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1.06 is the <u>maximum</u> refractive index that the liquid may have for the light to be totally reflected.
Only when a light source passes from a denser to a rarer medium can it completely reflect.
When the angle of incidence surpasses a specific critical value, specular reflection occurs in the more highly refractive of the two mediums at their interface, and this reflection is known as total reflection.
sin = μ / μ
From the diagram
Angle of incidence = 60°
sin60° ≥ sin = μ/μ
μ ≤ μ sin60°
μ ≤ √1.5 × √3/2
= 1.06
Hence, the maximum index that the liquid may have for the light to be totally reflected is 1.06
Learn more about refractive index here brainly.com/question/10729741
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