Use this formula to find your answer...
Determine the frequency of a clock waveform whose period is 2us or (micro) and 0.75ms
frequency (f)=1/( Time period).
Frequency of 2 us clock =1/2*10^-6 =10^6/2 =500000Hz =500 kHz.
Frequency of 0..75ms clock =1/0.75*10^-3 =10^3/0.75 =1333.33Hz =1.33kHz.
I think F= mv²/r
And F=ma
So, ma = mv²/r
a = v²/r
a = 100/5
a = 20 m/s
Answer:
See the attached image and the explanation below
Explanation:
We must draw a schematic of the described problem, after the sketch it is necessary to make a free body diagram, at the time before and after cutting the cord.
These free body diagrams can be seen in the attached image.
First we perform a sum of forces on the x & y axes before cutting the cord, to be able to find the T tension of the wire. (This analysis can be seen in the attached image).
In this way we get the T-wire tension equation, before cutting.
Now we make another free body diagram, for the moment when the wire is cut (see in the attached diagram).
It is important to clarify that when the cord is cut, the system will no longer be in statically, therefore newton's second law will be used for summation of forces which will be equal to the product of mass by acceleration.
Finally with equations 1 and 2 we can find the K ratio.
The AU ... Astronomical Unit ... used to be defined as the average distance between the Sun and Earth during the year.
Now it's defined as 149,597,870,700 meters exactly.