Answer:
a. 0.9 Hz b. 0.37 Hz
Step-by-step explanation:
The frequency of the simple pendulum f = (1/2π)√g/l where g = acceleration due to gravity and l = length of pendulum
a. Find the frequency of a pendulum whose length is 1 foot and where the gravitational field is approximately 32 ft/s2
To find f on Earth, g = 32 ft/s² and l = 1 ft
So, f = (1/2π)√(g/l)
f = (1/2π)√(32 ft/s²/1 ft)
f = (1/2π)√(32/s²)
f = (1/2π)(5.66 Hz)
f = 0.9 Hz
b. The strength of the gravitational field on the moon is about 1/6 as strong as on Earth.. Find the frequency of the same pendulum on the moon.
On the moon when acceleration due to gravity g' = g/6,
f = (1/2π)√(g'/l)
f = (1/2π)√(g/6l)
f = (1/2π)√[32 ft/s²/(6 × 1 ft)]
f = (1/2π)√(32/s²)/√6
f = (1/2π)(5.66 Hz)/√6
f = 0.9/√6 Hz
f = 0.37 Hz
Answer:
retirement savings plan sponsored by an employer
Step-by-step explanation:
Answer:
It would be best if Peter's report card percentage was based on<em> </em><em>the mean</em><em> </em>and if he could drop a percent, he should drop the <em>44%</em><em>.</em>
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Step-by-step explanation:
The median (without dropping a score) of this set would be 75%
( 44%, 64%, 71%, 75%, 84%, 90%, 98% )
The mean (without dropping a score) of this set would be 75.1428571%
526 ÷ 7 = 75.14 ( 7<em>5.1428571 </em>)
Are you trying to find point slope form bc y2-y1/x2-x1
