The answer is D hope it helps
The wire vibrates back and forth between the poles of the magnet.
The frequency of the vibration is the frequency of the AC supply.
Answer:
Workdone = 465766038 Joules.
Explanation:
<u>Given the following data;</u>
Mass = 1167
Initial velocity = 10m/s
Final velocity =28m/s
To find the workdone;
We know that from the workdone theorem, the workdone by an object or a body is directly proportional to the kinetic energy possessed by the object due to its motion.
Mathematically, it is given by the equation;
W = Kf - Ki
W = ½MVf² - ½MVi²
Substituting into the equation
W = ½(1167)*28² - ½(1167)*10²
W = ½ * 1361889* 784 - ½ * 1361889 * 100
W = 533860488 - 68094450
Workdone = 465766038 Joules.
Answer:m1v1 + m2v2 = (m1f + m2f)vf. 3000kg(10.0m/s) + (15000kg)(0.0m/s) = (18000kg)(vf).
Explanation:
