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2 answers:
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Answer:
The angle that the wave would be
Explanation:
From the question we are told that the opening to the harbor acts just like a single-slit so a boat in the harbor that at angle equal to the second diffraction minimum would be safe and the on at angle greater than the diffraction first minimum would be slightly affected
The minimum is as a result of destructive interference
And for single-slit this is mathematically represented as
where D is the slit with
is the angle relative to the original direction of the wave
m is the order of the minimum j
is the wavelength
Now since in the question we are told to obtain the largest angle at which the boat would be safe
And the both is safe at the angle equal to the second minimum then
The the angle is evaluated as
Since for second minimum m= 2
The equation becomes
Answer:
10250 N/C leftwards
Explanation:
QA = 4 micro Coulomb
QB = - 5 micro Coulomb
AP = 6 m
BP = 2 m
A is origin, B is at 4 m and P is at 6 m .
The electric field due to charge QA at P is EA rightwards
The electric field due to charge QB at P is EB leftwards
The resultant electric field at P due the charges is given by
E = EB - EA
E = 11250 - 1000 = 10250 N/C leftwards
