Answer:
T = 2.83701481512 seconds
Explanation:
Hi!
The formula that you will want to use to solve this question is:
T--> period
L --> length of the pendulum
g --> acceleration due to gravity (9.8m/s^2)
since we know that the mass of the bob at the end of the pendulum does not affect the period of the pendulum, we can go ahead and ignore that bit of information (unless, of course, the weight causes the pendulum to stretch)
so now we can plug in our given info into the formula above and solve!
T = 2*pi * sqrt(2/9.8)
T = 2.83701481512 seconds
*Note*
- I used 3.14 to pi, if you need to use a different value for pi (a longer version, etc) your answer will be slightly different
I hope this helped!
Answer:
This is to optimize storage and transport.
Explanation:
<h2>
a) Displacement of penny = 1300 i + 2400 j - 640 k</h2><h2>b) Magnitude of his displacement = 2729.47 m</h2>
Explanation:
a) He walks 1300 m east, 2400 m north, and then drops the penny from a cliff 640 m high.
1300 m east = 1300 i
2400 m north = 2400 j
Drops the penny from a cliff 640 m high = -640 k
Displacement of penny = 1300 i + 2400 j - 640 k
b) Displacement of man for return trip = -1300 i - 2400 j
Magnitude of his displacement = 2729.47 m
The "waning gibbous" begins immediately after the Full Moon and lasts about a week.
At the end of the week, the moon is no longer gibbous, but it's still waning. At the instant it's exactly half-illuminated, it's called "Third Quarter", and then it continuous to wane for another week.
So 3 to 4 days after the END of the waning gibbous phase, it's a Waning Crescent. It still has another 3 to 4 days of waning to go, before it wanes away to nothing and we have the next New Moon.
This exercise is about Reflection and Mirrors. It explores the relationship between convex mirrors, virtual images, and complex images as well as to object distance d₀.
<h3 /><h3>What range of
d₀ values will result in a real image?</h3>
Any d₀ value greater than 10 for a convex lens will produce a true or real image, while any d₀ value greater than 15 for a complex mirror will produce a real image.
<h3>What range of d
₀ values will result in a virtual image?</h3>
Any d₀ number less than 10 for a convex lens will result in a virtual image. Also, any d₀ value less than 15 for a complex mirror will result in a virtual picture.
<h3>What range of do values will result in a smaller image?</h3>
The range of d₀ values that would result in a reduced picture in both lenses and mirrors is:
d₀ > 2 f, where is the Focal length.
Learn more about Reflection and Mirrors at:
brainly.com/question/3574225