Answer:
A
Explanation:
this is because the law of reflection states that the angle of incidence equals to the angle of the reflection.
hope this helps, if not please report it
someone else can try it
Answer:
8) 709.8875 J
9) The object is at 7.24375 m from the ground
10) Kinetic energy increases as the object falls.
Explanation:
We use the expression for the displacement h(t) as a function of time of an object experiencing free fall:
h(t) = hi - (g/2) t^2
hi being the initial position of the object (10m) above ground, g the acceleration of gravity (9.8 m/s^2), and t the time (in our case 0.75 seconds):
h(0.75) = 10 - 4/9 (0.75)^2 = 7.24375 m
This is the position of the 10 kg object after 0.75 seconds (answer for part 9)
Knowing this position we can calculate the potential energy of the object when it is at this height, using the formula:
U = m g h = 10kg * 9.8 (m/s^2) * 7.24375 m = 709.8875 J (answer for part 8)
Part 10)
the kinetic energy of the object increases as it gets closer to ground, since its velocity is increasing in magnitude because is being accelerated in its motion downwards.
Answer:
0.58
Explanation:
Sinẞ = opposite ÷ hypotenuse
Sinẞ = 5 ÷ 8.6
Sinẞ = 0.5814
Sinẞ ≈ 0.58
Explanation:
Basaltic lava
Basaltic lava generally takes two distinct forms known by the Hawaiian terms pahoehoe and aa. Pahoehoe has a smooth wavy surface that resembles twisted rope. It advances by extruding molten toes of lava beneath a thin, flexible crust. As it travels pahoehoe lava often changes to blocky flows called aa.
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)
