![shartiarahoward](/avatars/24214.jpg)
shartiarahoward
20.04.2021 •
Mathematics
What is 1/6 as a percent.
Solved
Show answers
More tips
- F Food and Cooking Discover How to Properly Prepare Dough for Rasstegai...
- P Philosophy Unidentified Flying Object - What is the Nature of this Phenomenon?...
- F Family and Home Protect Your Home or Apartment from Pesky Ants...
- O Other What is a Disk Emulsifier and How Does it Work?...
- F Family and Home What does a newborn need?...
- F Family and Home Choosing the Right Car Seat for Your Child: Tips and Recommendations...
- F Food and Cooking How to Get Reconfirmation of Registration?...
- C Computers and Internet How to Get Rid of Spam in ICQ?...
- A Art and Culture Who Said The Less We Love a Woman, the More She Likes Us ?...
- F Family and Home How to Get Rid of Your Neighbors?...
Answers on questions: Mathematics
- M Mathematics A designer builders a model of a bicycle the finished model is exactly the same as the original but smaller. The scale factor is 2:15 (A) ratio of height ☑️:☑️ (B)surface...
- C Chemistry O grupo inorgânico dos sais tem grande relevância e impacto no cotidiano como, por exemplo, o fluoreto de sódio nos dentifrícios ou o carbonato de cálcio presente...
- M Mathematics Write the equation fully simplified slope intercept form...
- F French 1. Ils sont étonnés que vous... 2. Il est impossible qu ils... 3. Il est bon que nous... 4. As-tu fini de... 5. Vous souhaitez que jelj 6. Faut-il que tu......
- B Business What would one use webex.com for? a. to distribute e-mails to a mailing list b. to post an announcement about a convention c. for online accounts payable d. to hold...
Ответ:
16.66 %
Step-by-step explanation:
1 / 6 as a percent
= ( 1 / 6 ) x 100
= 100 / 6
= 16.66 %
Ответ:
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