placing a magnetically hard material in a strong magnetic field
Answer:
805.48N/m
Explanation:
According to Hookes law
F = Ke
F is the force = mg
F = 2.4×9.8 = 23.52N
e is the extension = 2.92cm = 0.0292m
Force constant K = F/e
K = 23.52/0.0292
K = 805.48N/m
Hence the force constant of the spring is 805.48N/m
Centripetal force = (mass) x (speed)² / (radius)
= (20 kg) x (20 m/s)² / (20 m)
= (20 x 20 / 20) (kg-meter/sec²)
= 20 newtons
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