A potential problem is that you are willing to accept a <u>5% </u>chance of being wrong if you reject the null hypothesis.
The significance level is the probability of rejecting the null hypothesis if it is true. For example, a significance level of 0.05 indicates a 5% risk of concluding that there is a difference when there is actually no difference. Rejecting the true null hypothesis results in a Type I error.
The smaller the value of α the more difficult it is to reject the null hypothesis. Therefore, choosing a low value for α can reduce the likelihood of Type I errors. The result here is that if the null hypothesis is false, it may be more difficult to reject using a lower value for α. The alpha value or statistical significance threshold is arbitrary. Which value to use depends on your field of study.
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Answer:
Explanation:
Surface charge density, σ = 9 μC/m² = 9 x 10^-6 C/m²
According to the Gauss theorem,
Electric field due to the sheet is given by
E = 5.08 x 10^5 N/C
#1
As we are increasing the frequency in the simulation the wavelength is decreasing
So if speed remains constant then wavelength and frequency depends inversely on each other
If we are in boat and and moving over very small wavelengths then these small wavelength will be encountered continuously by the boat in short interval of times
#2
As we are changing the amplitude in the simulation there is no change in the speed frequency and wavelength.
So amplitude is independent of all these parameter
Amplitude of wave will decide the energy of wave
So light of greater intensity is the light of larger amplitude
#3
In our daily life we deal with two waves
1 sound waves
2 light waves
Answer:
The minimum value of width for first minima is λ
The minimum value of width for 50 minima is 50λ
The minimum value of width for 1000 minima is 1000λ
Explanation:
Given that,
Wavelength = λ
For D to be small,
We need to calculate the minimum width
Using formula of minimum width
Where, D = width of slit
= wavelength
Put the value into the formula
Here, 
should be maximum.
So. maximum value of is 1
Put the value into the formula
(b). If the minimum number is 50
Then, the width is
(c). If the minimum number is 1000
Then, the width is
Hence, The minimum value of width for first minima is λ
The minimum value of width for 50 minima is 50λ
The minimum value of width for 1000 minima is 1000λ
