Answer:

Explanation:
From the law of conservation of energy
Energy lost by the spring, W=Kinetic energy gained, KE+Potential energy gained, PE+Work done by friction, Fr



The required distance from A to B is 
Answer:
distance
Explanation:
it is the distance traveled by light in one year
Answer:
18.7842493212 W
Explanation:
T = Tension = 1871 N
= Linear density = 3.9 g/m
y = Amplitude = 3.1 mm
= Angular frequency = 1203 rad/s
Average rate of energy transfer is given by

The average rate at which energy is transported by the wave to the opposite end of the cord is 18.7842493212 W
Answer:
A. The bird watcher followed the south trail a distance of five kilometers in 45 minutes.