A professor drops one lead sinker every 2.0 s from a very high window.
1) How far has the first sinker gone when the second one is dropped?
2) Does the distance between the first and second sinker remain constant? (yes or no)
1. d = Vo*t + 0.5at^2,
d = 0 + 0.5*9.8*2^2 = 19.6m.
To answer these questions, we need to understand the concept of free fall and the equations of motion.
1) How far has the first sinker gone when the second one is dropped?
In free fall, an object accelerates due to the force of gravity alone. The distance traveled by an object in free fall can be calculated using the equation:
d = (1/2) * g * t^2
where:
d is the distance traveled,
g is the acceleration due to gravity (approximately 9.8 m/s^2), and
t is the time.
Since the professor drops one lead sinker every 2.0 seconds, we can calculate the distance traveled by the first sinker at 2.0 seconds:
d1 = (1/2) * (9.8 m/s^2) * (2.0 s)^2
= 19.6 m
Therefore, the first sinker has gone 19.6 meters when the second one is dropped.
2) Does the distance between the first and second sinker remain constant? (yes or no)
No, the distance between the first and second sinker does not remain constant. The reason is that both sinkers are in free fall, and the distance traveled by an object in free fall increases with time according to the equation:
d = (1/2) * g * t^2
As time passes, the distance traveled by each sinker increases, so the distance between the two sinkers will also increase.