Answer:
The coil radius of other generator is 5.15 cm
Explanation:
Consider the equation for induced emf in a generator coil:
EMF = NBAω Sin(ωt)
where,
N = No. of turns in coil
B = magnetic field
A = Cross-sectional area of coil = π r²
ω = angular velocity
t = time
It is given that for both the coils magnetic field, no. of turn and frequency is same. Since, the frequency is same, therefore, the angular velocity, will also be same. As, ω = 2πft.
Therefore, EMF for both coils or generators will be:
EMF₁ = NBπr₁²ω Sin(ωt)
EMF₂ = NBπr₂²ω Sin(ωt)
dividing both the equations:
EMF₁/EMF₂ = (r₁/r₂)²
r₂ = r₁ √(EMF₂/EMF₁)
where,
EMF₁ = 1.8 V
EMF₂ = 3.9 V
r₁ = 3.5 cm
r₂ = ?
Therefore,
r₂ = (3.5 cm)√(3.9 V/1.8 V)
<u>r₂ = 5.15 cm</u>
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Answer:
0.782 s
Explanation:
The water flows horizontally from the hose, so its initial vertical velocity is 0.
Given:
y₀ = 3 m
y = 0 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
y = y₀ + v₀ t + ½ at²
0 m = 3 m + (0 m/s) t + ½ (-9.8 m/s²) t²
t = 0.782 s
Round as needed.
