Let her initial velocity be U.
Let the width of the river be W.
She swims 3/5 the width of the river at U.
Remainder width = (1 - 3/5) = 2/5.
She then swims 2/5 the width with velocity U/2.
Average Speed = (Total Distance Traveled) / ( Total Time Taken).
Distance = Speed * time
time = Distance / Speed.
Time in first trip: = (3/5)W / U = 0.6W/U.
Time in second trip = (2/5)W / (U/2) = 0.4W / 0.5 U = 0.8W/U
Total Distance Traveled = W, width of the river.
Average Speed = W / (0.6W/U + 0.8W/U) = W / (1.4W/U)
= W * U / 1.4W
= U/1.4
= U * 10 / 14
= (5/7) U.
Therefore Average speed is (5/7) of the initial speed.
k = spring constant of the spring = 100 N/m
m = mass hanging from the spring = 0.71 kg
T = Time period of the spring's motion = ?
Time period of the oscillations of the mass hanging is given as
T = (2π) √(m/k)
inserting the values in the above equation
T = (2 x 3.14) √(0.71 kg/100 N/m)
T = (6.28) √(0.0071 sec²)
T = (6.28) (0.084) sec
T = 0.53 sec
hence the correct choice is D) 0.53
Newton's law of universal gravitation<span> states that a particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. hope this helps =)</span>
