Answer:
A. The path of least resistance.
Explanation:
ur welcome again ;)
Answer:
True
Explanation:
i searched it up and well this thing is making me do it up till 20 characters long so yea
Complete question:
A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find
(a) the force constant of the spring and (b) the amplitude of the motion.
Answer:
(a) the force constant of the spring = 47 N/m
(b) the amplitude of the motion = 0.292 m
Explanation:
Given;
mass of the spring, m = 200g = 0.2 kg
period of oscillation, T = 0.410 s
total mechanical energy of the spring, E = 2 J
The angular speed is calculated as follows;

(a) the force constant of the spring

(b) the amplitude of the motion
E = ¹/₂kA²
2E = kA²
A² = 2E/k

Answer:
80 Ω.
Explanation:
In this circuit the resistances are in series.The equivalent resistance of a series circuit is equal to the sum of the resistances. Req= 60 + 20 = 80 Ω.