For the following exercises, use the model for the period of a pendulum, T , such that T = 2 π L g , where the length of the pendulum is L and the acceleration due to gravity is g . 46. If the acceleration due to gravity is 9.8 m / s 2 and the period equals 1 s, find the length to the nearest cm ( 100 c m = 1 m ) .
For the following exercises, use the model for the period of a pendulum, T , such that T = 2 π L g , where the length of the pendulum is L and the acceleration due to gravity is g . 46. If the acceleration due to gravity is 9.8 m / s 2 and the period equals 1 s, find the length to the nearest cm ( 100 c m = 1 m ) .
For the following exercises, use the model for the period of a pendulum, T, such that
T
=
2
π
L
g
, where the length of the pendulum is L and the acceleration due to gravity is g.
46. If the acceleration due to gravity is
9.8
m
/
s
2
and the period equals 1 s, find the length to the nearest cm
(
100
c
m
=
1
m
)
.
The velocity of a particle along the s-axis is given by v = 659/8 where s is in millimeters and vis in millimeters per second. Determine the
acceleration when s is 4.1 millimeters.
Answer: a = i
eTextbook and Media
mm/s²
Answer: 48 Newtons
Suppose a stone is thrown vertically upward with an initial velocity of 48 ft/s from a bridge 64 ft
above a river. By Newton's laws of motion, the position of the stone (measured as the height above
the river) after t seconds is s(t) = -16t² + 48t + 64, where s = 0 is the level of the river.
%3D
a). Find the velocity and acceleration functions.
b). What is the highest point above the river reached by the stone?
c). With what velocity will the stone strike the river?
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, trigonometry and related others by exploring similar questions and additional content below.