Answer:
Solution is in explanation
Explanation:
part a)
For normalization we have
![\int_{0}^{\infty }f(x)dx=1\\\\\therefore \int_{0}^{\infty }ae^{-kx}dx=1\\\\\Rightarrow a\int_{0}^{\infty }e^{-kx}dx=1\\\\\frac{a}{-k}[\frac{1}{e^{kx}}]_{0}^{\infty }=1\\\\\frac{a}{-k}[0-1]=1\\\\\therefore a=k](https://tex.z-dn.net/?f=%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7Df%28x%29dx%3D1%5C%5C%5C%5C%5Ctherefore%20%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7Dae%5E%7B-kx%7Ddx%3D1%5C%5C%5C%5C%5CRightarrow%20a%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7De%5E%7B-kx%7Ddx%3D1%5C%5C%5C%5C%5Cfrac%7Ba%7D%7B-k%7D%5B%5Cfrac%7B1%7D%7Be%5E%7Bkx%7D%7D%5D_%7B0%7D%5E%7B%5Cinfty%20%7D%3D1%5C%5C%5C%5C%5Cfrac%7Ba%7D%7B-k%7D%5B0-1%5D%3D1%5C%5C%5C%5C%5Ctherefore%20a%3Dk)
Part b)
![\int_{0}^{L }f(x)dx=1\\\\\therefore Re(\int_{0}^{L }ae^{-ikx}dx)=1\\\\\Rightarrow Re(a\int_{0}^{L }e^{-ikx}dx)=1\\\\\therefore Re(\frac{a}{-ik}[\frac{1}{e^{ikx}}]_{0}^{L})=1\\\\\Rightarrow Re(\frac{a}{-ik}(e^{-ikL}-1))=1\\\\\frac{a}{k}Re(\frac{1}{-i}(cos(-kL)+isin(-kL)-1))=1](https://tex.z-dn.net/?f=%5Cint_%7B0%7D%5E%7BL%20%7Df%28x%29dx%3D1%5C%5C%5C%5C%5Ctherefore%20Re%28%5Cint_%7B0%7D%5E%7BL%20%7Dae%5E%7B-ikx%7Ddx%29%3D1%5C%5C%5C%5C%5CRightarrow%20Re%28a%5Cint_%7B0%7D%5E%7BL%20%7De%5E%7B-ikx%7Ddx%29%3D1%5C%5C%5C%5C%5Ctherefore%20Re%28%5Cfrac%7Ba%7D%7B-ik%7D%5B%5Cfrac%7B1%7D%7Be%5E%7Bikx%7D%7D%5D_%7B0%7D%5E%7BL%7D%29%3D1%5C%5C%5C%5C%5CRightarrow%20Re%28%5Cfrac%7Ba%7D%7B-ik%7D%28e%5E%7B-ikL%7D-1%29%29%3D1%5C%5C%5C%5C%5Cfrac%7Ba%7D%7Bk%7DRe%28%5Cfrac%7B1%7D%7B-i%7D%28cos%28-kL%29%2Bisin%28-kL%29-1%29%29%3D1)
Answer:
I hear points of low volume sound and points of high volume of sound.
Explanation:
This is because, since the two sources of sound have the same frequency and are separated by a distance, d = 10 mm, there would be successive points of constructive and destructive interference.
Since their frequencies are similar, we should have beats of high and low frequency.
So, at points of low frequency, the amplitude of the wave is smallest and there is destructive interference. The frequency at this point is the difference between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, f - f' = 400 Hz - 400 Hz = 0 Hz. So, the volume of the sound is low(zero) at these points.
Also, at points of high frequency, the amplitude of the wave is highest and there is constructive interference. The frequency at this point is the sum between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, (f + f') = 400 Hz + 400 Hz = 800 Hz. So, the volume of the sound is high at these points.
So, as you wander around the room, I should hear points of high and low sound across the room.
fault-block mountains. hope this helps
Answer:
magnitude of force on charge 2Q = 
Direction of force on charge = 61 ⁰
Explanation:
The magnitude on the force on the charge can be evaluated by finding the net force acting on the charge 2Q i.e x-component of the net force and the y-component of the net force
║F║ = 
= after considering the forces coming from Q, 3Q and 4Q AND APPLYING COULOMBS LAW
magnitude of force acting on 2Q =
The direction of the force on charge 2Q is calculated as
tan ∅ = 
= 1.8284
therefore ∅ = 
1.8284
= 61⁰
