When t=2, the ball has fallen d(2) = 16 (2²) = 64 feet .
When t=5, the ball has fallen d(5) = 16 (5²) = 400 feet .
Distance fallen from t=2 until t=5 is (400 - 64) = 336 feet.
Time period between t=2 until t=5 is (5 - 2) = 3 seconds.
Average speed of the ball from t=2 until t=5 is
(distance covered) / (time to cover the distance)
= 336 feet / 3 seconds = 112 feet per second.
That's what choice-C says.
Work = Force x Distance
Assuming that this work is being done parallel to the displacement that is, but under that assumption:
W = (50)(10)
W = 500 J
We could tell a force is acting on an object if the object is being pushed, pulled, or moved in any way
Answer:drift velocity is 1.7*10^-4m/s and number of charge is 2.7*10^-24
Explanation:
|Acceleration| = (change in speed) / (time for the change).
Change in speed = (6 mi/hr - 25 mi/hr) = -19 mi/hr
Time for the change = 10 sec
|Acceleration| = (-19 mi/hr) / (10 sec) = -1.9 mile per hour per second
Admittedly, that's a rather weird unit.
Other units, perhaps more comfortable ones, are:
-6,840 mi/hr²
-2.79 feet/sec²
