Two long, straight parallel wires 8.2 cm apart carry currents of equal magnitude I. The parallel wires repel each other with a f
1 answer:
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Answer:
m = 170 g
Explanation:
Meter stick is suspended at 40 cm mark
So here the torque due to additional mass and torque due to weight of the spring must be counter balanced
given that
1) 220 g is suspended at x = 5 cm
2) 120 g is suspended at x = 90 cm
3) mass of the scale is acting at its mid point i.e. x = 50 cm
now with respect to the suspension point the torque must be balanced
so we have
by solving above equation we have
m = 170 g
Answer:
The compression of the spring is 24.6 cm
Explanation:
magnitude of the charge on the left, q₁ = 4.6 x 10⁻⁷ C
magnitude of the charge on the right, q₂ = 7.5 x 10⁻⁷ C
distance between the two charges, r = 3 cm = 0.03 m
spring constant, k = 14 N/m
The attractive force between the two charges is calculated using Coulomb's law;
The extension of the spring is calculated as follows;
F = kx
x = F/k
x = 3.45 / 14
x = 0.246 m
x = 24.6 cm
The compression of the spring is 24.6 cm