An AC generator consists of 6 turns of wire. Each turn has an area of 0.040 m2. The loop rotates in a uniform field (B = 0.20 T)
1 answer:
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Answer:
The moment of inertia about the rotation axis is 117.45 kg-m²
Explanation:
Given that,
Mass of one child = 16 kg
Mass of second child = 24 kg
Suppose a playground toy has two seats, each 6.1 kg, attached to very light rods of length r = 1.5 m.
We need to calculate the moment of inertia
Using formula of moment of inertia
m = mass of seat
m₁ =mass of one child
m₂ = mass of second child
r = radius of rod
Put the value into the formula
Hence, The moment of inertia about the rotation axis is 117.45 kg-m²
Answer:
the required minimum length of the attenuator is 3.71 cm
Explanation:
Given the data in the question;
we know that;
= c / 2a
where f is frequency, c is the speed of light in air and a is the plate separation distance.
we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s
plate separation distance a = 7.11 mm = 0.0711 cm
so we substitute
= 3 × 10¹⁰ / 2( 0.0711 )
= 3 × 10¹⁰ cm/s / 0.1422 cm
= 21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }
Now, we determine the minimum wavelength required
Each non propagating mode is attenuated by at least 100 dB at 15 GHz
so
Attenuation constant TE₁ and TM₁ expression is;
∝₁ = 2πf/c × √( ( / f)² - 1 )
so we substitute
∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )
∝₁ = 3.1079 × 10⁻⁷
∝₁ = 310.79 np/m
Now, To find the minimum wavelength, lets consider the design constraint;
20log₁₀ = -100dB
we substitute
20log₁₀ = -100dB
= 3.71 cm
Therefore, the required minimum length of the attenuator is 3.71 cm