A spring stretches 0.2 cm per newton of applied force. An object is suspended from the spring and a deflection of 3 cm is observ
1 answer:
Answer:
Explanation:
To solve this problem we use the Hooke's Law:
(1)
F is the Force needed to expand or compress the spring by a distance Δx.
The spring stretches 0.2cm per Newton, in other words:
1N=k*0.2cm ⇒ k=1N/0.2cm=5N/cm
The force applied is due to the weight
We replace in (1):
We solve the equation for m:
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Answer:
Explanation:
let the charge is q. velocity, v = 24.7 m/s
magnetic force, F = 2.38 x 10^-4 N
Let the magnetic field is B.
Velocity, v' = 5.64 m/s
angle, θ = 21.2°
The force experienced by a charged particle placed in a magnetic field is given by
F = q x v x B x Sinθ
in first case
2.38 x 10^-4 = q x 24.7 x B x Sin 90 .... (1)
in second case
F = q x 5.64 x B x Sin 21.2° .... (2)
Divide equation (2) by equation (1), we get
F = 1.97 x 10^-5 N