Answer:
350 ft/s²
Explanation:
First, convert mph to ft/s.
58 mi/hr × (5280 ft/mi) × (1 hr / 3600 s) = 85.1 ft/s
Given:
v₀ = 85.1 ft/s
v = 0 ft/s
t = 0.24 s
Find: a
v = at + v₀
a = (v − v₀) / t
a = (0 ft/s − 85.1 ft/s) / 0.24 s
a = -354 ft/s²
Rounded to two significant figures, the magnitude of the acceleration is 350 ft/s².
Answer:
The answer is "Choice E".
Explanation:
In this situation the option e is right because its resistance decreases through time, however, the time is the same for the same reason, whereas the sphere deteriorates, somehow it travels shorter distances however if the air resistance becomes are using the amplitude of movement declines, that's why other choices were wrong.
The period T of a pendulum is given by:
where L is the length of the pendulum while
is the gravitational acceleration.
In the pendulum of the problem, one complete vibration takes exactly 0.200 s, this means its period is
. Using this data, we can solve the previous formula to find L:
You must run at least 53.3333333 meters a minute.