120n
since the speed is doubled, her force is doubled
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:

Explanation:
<h3><u>Given data:</u></h3>
Frequency = f = 200 Hz
Velocity = v = 400 m/s
<h3><u>Required:</u></h3>
Wavelength = λ = ?
<h3><u>Formula:</u></h3>
v = fλ
<h3><u>Solution:</u></h3>
Put the givens in the formula
400 = (200)λ
Divide 200 to both sides
400/200 = λ
2 m = λ
λ = 2 m
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