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Three complete orders on each side of the m=0 order can be produced in addition to the m=0 order.
The ruling separation is d=1/(470mm-1)
Diffraction lines occurs at an angle θ such that dsin=mλ,when λ is the wavelength and m is an integer.
Notice that for a given order,the line associated with a long wavelength is produced at a greater angle than the line associated with shorter wavelength.
we take λ to be the longest wavelength in the visible spectrum (538nm) and find the greatest integer value of m such that θ is less than 90°.
That is,find the greater integer value of m for which mλ<d.
since,d/λ
There are three complete orders on each side of the m=0 order.
The second and third orders overlap.
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Answer:
Explanation:
The time taken to complete one revolution is the period and is defined as:
The angular speed is given by:
Replacing (2) in (1) and solving for :
Answer:
So then we will have at least 75% of the data within two deviations from the mean .
So then the interval would be (97.14, 99.5)
Explanation:
Previous concepts
Chebyshev’s rule is appropriate for any distribution. "Chebyshev’s inequality applies to all distributions, regardless of shape". And is useful since provides a "minimum percentage of the observations that lies within k standard deviations of the mean. "
If k = 2, at least 3/4 of the measurements lie within 2 standard deviations to within the mean.
And the general formula is (1-1/k^2) represent the fraction of the data within the mean .
Solution to the problem
For this case we want to find the percentage of data that would be at least within two deviations from the mean so for this case the value of k =2 and if we replace we got:
So then we will have at least 75% of the data within two deviations from the mean .
For the other part we have the mean and deviation provided the interval would be:
So then the interval would be (97.14, 99.5)