The emf induced in the coil is -5.65 V
<h3>Induced emf in coil</h3>
The induced emf in the coil is given by ε = -NΔΦ/Δt where
- ΔΦ = change in magnetic flux Φ₂ - Φ₁ where
- Φ₁ = initial magnetic flux = -58 Wb and
- Φ₂ = final magnetic flux = 38 Wb and and
- Δt = change in time = t₂ - t₁ where
- t₁ = initial time = 0 s and
- t₂ = final time = 34 sand
- N = number of loops of coil = 2
Since ε = -NΔΦ/Δt
ε = -N(Φ₂ - Φ₁)/(t₂ - t₁)
Substituting the values of the variables into the equation, we have
ε = -N(Φ₂ - Φ₁)/(t₂ - t₁)
ε = -2(38 Wb - (-58 Wb))/(34 s - 0 s)
ε = -2(38 Wb + 58 Wb)/(34 s - 0 s)
ε = -2(96 Wb)/34 s
ε = -192 Wb/34 s
ε = -5.65 Wb/s
ε = -5.65 V
So, the emf induced in the coil is -5.65 V
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Answer:
The magnitude of the force that the 6.3 kg block exerts on the 4.3 kg block is approximately 41.9 N
Explanation:
Forces on block 4.3 kg are:
63N to the right and R21 (contact force from the 6.3 kg block) to the left
Net force on 4.3 kg block is: 63 N - R21
Forces on the 6.3 kg block are:
R12 to the right (contact force from the 4.3 kg block) and 11 N to the left.
So net force on the 6.3 kg block is: R12 - 11 N
According to the action-reaction principle the contact forces R21 and R12 must be equal in magnitude (let's call them simply "R").
Then, since the blocks are moving with the SAME acceleration, we equal their accelerations:
a1 = (63 N - R)/4.3 = (R - 11 N)/6.3 = a2
solve for R by cross multiplication
6.3 (63 - R) = 4.3 (R - 11)
396.9 - 6.3 R = 4.3 R - 47.3
369.9 + 47.3 = 10.6 R
444.2 = 10.6 R
R = 444.2 / 10.6
R = 41.90 N
Answer:
particles or arranged will be the answer depending on the previous line.
