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Answer:
the number of turns in the primary coil is 13000
Explanation:
Given the data in the question;
V₁ = 13000 V
V₂ = 120 V
N₁ = ?
N₂ = 120 turns
the relation between the voltages and the number of turns in the primary and secondary coils can be expressed as;
V₁/V₂ = N₁/N₂
V₁N₂ = V₂N₁
N₁ = V₁N₂ / V₂
so we substitute
N₁ = (13000 V × 120 turns) / 120 V
N₁ = 1560000 V-turns / 120 V
N₁ = 13000 turns
Therefore, the number of turns in the primary coil is 13000
Answer:
the magnitude of the total angular momentum of the blades is <em>743.71 kg·m²</em>
Explanation:
Converting the angular speed into radians per second:
ω = 334 rpm · (2π rad / 1 rev) · (1 min / 60 s)
ω = 34.98 rad/s
The rotational kinetic energy of the blades is given by:
EK = 1/2 I ω²
where
- I is the moment of inertia
- ω is the angular speed
Therefore, rearranging the above equation, we get:
1/2 I ω² = EK
I ω² = 2 EK
I = 2(EK) / ω²
I = 2(4.55 × 10⁵ J) / (34.98 rad/s)²
<em>I = 743.71 kg·m²</em>
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Therefore, the magnitude of the total angular momentum of the blades is <em>743.71 kg·m²</em>.
