A kayaker needs to paddle north across a
2 answers:
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Answer:
the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm
the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J
Explanation:
The torque is given by :
where ;
m = 0.160 A.m²
B = 0.0800 T
θ = 35°
So the magnitude of the torque N = mBsinθ
N = (0.160)(0.0800)(sin 35°)
N = 0.007341
N = 7.34×10⁻³ Nm
Hence, the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm
b) The potential energy
U = -mBcosθ
U = (- 0.160)(0.0800)(cos 45)
U = -0.010485
U = -1.0485 ×10⁻² J
Thus, the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J
Answer:
a. R1 = 0.162 Ω
b. R2 = 0.340 Ω
Explanation:
Since the resistors R1 and R2 are connected in series, the current flowing through them when the 9 V battery is applied is 9/R1 + R2.
When the current increases by 0.450 A wen only R1 is in the circuit, the current is
9/R1 + R2 + 0.450 A = 9/R1 (1)
When the current increases by 0.225 A when only R2 is in the circuit, the current is
9/R1 + R2 + 0.225 A = 9/R2 (2)
equation (1) - (2) equals
9(1/R1 - 1/R2) = 0.450 A - 0.225
9(1/R1 - 1/R2) = 0.125
(1/R1 - 1/R2) = 0.125 A/9 = 0.0138
1/R1 = 0.0138 + 1/R2
R1 = R2/(1 + 0.0138R2) (3)
From (1)
9/R1 - 9/R1 + R2 = 0.450 A
9R2/[R1(R1 + R2)] = 0.450 A
R2/[R1(R1 + R2)] = 0.450 A/9 = 0.5
R2/[R1(R1 + R2)] = 0.5 (4)
From (3) R2/R1 = (1 + 0.0138R2) and from (4) R2/R1 = 0.5(R1 + R2). So,
(1 + 0.0138R2) = 0.5(R1 + R2)
0.5R1 + 0.5R2 = 1 + 0.0138R2
0.5R1 = 1 + 0.0138R2 - 0.5R2
0.5R1 = 1 - 0.4862R2 (5)
Substituting (3) into (5) we have
0.5R2/(1 + 0.0138R2) = 1 - 0.4862R2
R2 = (1 + 0.0138R2)(1 - 0.4862R2)
R2 = 1 - 0.4724R2 - 0.0067R2²
Collecting like terms, we have
0.0067R2² + 0.4724R2 + R2 - 1 = 0
0.0067R2² + 1.4724R2 - 1 = 0
Using the quadratic formula,
We choose the positive answer.
So R2 = 0.340 Ω
From (5)
R1 = 0.5 - 0.9931R2
= 0.5 - 0.9931 × 0.340
= 0.5 - 0.338
= 0.162 Ω
a. R1 = 0.162 Ω
b. R2 = 0.340 Ω