A circuit has a resistance of 10ohmsband a current of 42 amps.caculate the voltage
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
Explanation:
<u>Formula:</u>
<u>d = distance given</u>
<u>t</u><u> </u><u>=</u><u> </u><u>the amount of time </u><u>given</u>
<u>Substitute the given values into the formula for velocity</u><u>:</u>
velocity is shortened for v.
8 (distance) divided by 4 (time) equals the velocity.
<u>Solve:</u>
The velocity of the toy car equals: B. 2 m/s.