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:
503
Step-by-step explanation:
The slope would be -5/4 or -1.25 because the y-intercept is (0,5) and x-intercept is (4,0), thus giving y = 5 - (5 x)\/4
Answer:
I think it is 6/1
Step-by-step explanation:
You are wrong because 0.167 into a fraction is 167/1000 and that isn't one of the options but when I did it 1 1/2=3/2=1.5 and 1/4=0.25 so I did 1.5 divide by 0.25 and it gave me 6 so that why I think that my answer is correct and yours is not.
Answer:
c= 18.4
Step-by-step explanation:
Since this is a right triangle, we can use trig functions
cos theta = adj/ hyp
cos theta = a/c
cos 57 = 10/c
Multiply each side by c
c cos 57 =10
Divide each side by cos 57
c = 10/cos 57
c=18.36078459
Rounding to the nearest tenth
c= 18.4
