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)
<span>ripple factor can be reduced by increasing the value of the load resistor (which means reducing the load of the circuit)</span>
Ohm's Law tells the relationship between voltage, current, and resistance.
It can be written in three different ways, depending on which ones you know,
and which one you want to find.
Here's the one we need:
Resistance = (voltage) divided by (current)
= (120 V) / (0.5 Amp)
= 240 ohms .
Answer:
1. -8.20 m/s²
2. 73.4 m
3. 19.4 m
Explanation:
1. Apply Newton's second law to the car in the y direction.
∑F = ma
N − mg = 0
N = mg
Apply Newton's second law to the car in the x direction.
∑F = ma
-F = ma
-Nμ = ma
-mgμ = ma
a = -gμ
Given μ = 0.837:
a = -(9.8 m/s²) (0.837)
a = -8.20 m/s²
2. Given:
v₀ = 34.7 m/s
v = 0 m/s
a = -8.20 m/s²
Find: Δx
v² = v₀² + 2aΔx
(0 m/s)² = (34.7 m/s)² + 2 (-8.20 m/s²) Δx
Δx = 73.4 m
3. Since your braking distance is the same as the car in front of you, the minimum safe following distance is the distance you travel during your reaction time.
d = v₀t
d = (34.7 m/s) (0.56 s)
d = 19.4 m
Answer:
- 2.7 x 10^-6 J
Explanation:
q1 = 1 nC at x = 0 cm
q2 = - 1 nC at x = 1 cm
q3 = 4 nC at x = 2 cm
The formula for the potential energy between the two charges is given by
where r be the distance between the two charges
By use of superposition principle, the total energy of the system is given by
U = - 2.7 x 10^-6 J
