Please hurry need ASAP!!! When a pendulum swings, at which point is kinetic energy highest?
1 answer:
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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:
Heat of vaporization will be 22.59 j
Explanation:
We have given mass m = 10 gram
And heat of vaporization L = 2.259 J/gram
We have to find the heat required to vaporize 10 gram mass
We know that heat of vaporization is given by , here m is mass and L is latent heat of vaporization.
So heat of vaporization Q will be = 10×2.259 = 22.59 J