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)
In Newton's Cradle experiment we know that all cradles of same mass and identical to each other
Now we know that when two identical objects collide elastically then they interchange their velocity
So here we have same illustration
When Newton pulls up a cradle and release it will move hit another cradle which is at rest
Due to elastic collision between them first cradle comes to rest and another cradle will move ahead with same speed this process remains the same and one by one all cradle hit another.
So at the last the cradle at the end will move off with the same speed as the first cradle will hit with the speed.
So in this experiment the cradle at the last end will move off at same distance away from the right end as that of left end we pull the cradle.
So here we can say that in horizontal direction when all cradles are colliding each other there is no external force on the system so momentum is conserved and they all will move off with same speed and hence we observe the above condition.
Answer:
the power that can be generated by the river is 117.6 MW
Explanation:
Given that;
Volume flow rate of river v = 240 m³/s
Height above the lake surface a h = 50 m
Amount of power can be generated from this river water after the dam is filled = ?
Now the collected water in the dam contains potential energy which is used for the power generation,
hence, total mechanical energy is due to potential energy alone.
= m(gh)
first we determine the mass flow rate of the fluid m
m = p×v
where p is density ( 1000 kg/m³
so we substitute
m = 1000kg/m³ × 240 m³/s
m = 240000 kg/s
so we plug in our values into ( = m(gh) kJ/kg )
= 240000 × 9.8 × 50
= 117600000 W
= 117.6 MW
Therefore, the power that can be generated by the river is 117.6 MW
Huh? The answer is A Shadow
It causes the global pattern of wind and precipitation (such as rain, snow, or hail).
