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
As ABC is similar to LMN
AC AB The ratio remain same if we divide numerator and
---- = ------ denominator by same number
LN LM
AC AB/2 AD
---- = ------------- - ---------
LN LM/2 LO
Let us now work on Triangle ADC and LON
Because Ratio of AC to AD is same Ad to LO these two triangle are also similar hence Ratio of third side DC to No would be also same .
Hence A C/ LM = DC/ NO
16 /9 = x/ 5
x= 16 * 5/9
x= 80/9
That is the answer .
The square root of 85 is
Answer: 9.21954
Step-by-step explanation:
given,
Given:
estimate = 140
actual = 152
Percent of error = |(estimate - actual)/actual| * 100%
% of error = |140 - 152| / 152 * 100%
= |-12|/152 * 100%
= 0.0789 x 100%
% of error = 7.89% rounded to the nearest tenth 7.9%