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:81.235N
Explanation:
Work=88J
theta=10°
distance=1.1 meters
work=force x cos(theta) x distance
88=force x cos10 x 1.1 cos10=0.9848
88=force x 0.9848 x 1.1
88=force x 1.08328
Divide both sides by 1.08328
88/1.08328=(force x 1.08328)/1.08328
81.235=force
Force=81.235
Answer:
Explanation:
a )
While breaking initial velocity u = 62.5 mph
= 62.5 x 1760 x 3 / (60 x 60 ) ft /s
= 91.66 ft / s
distance trvelled s = 150 ft
v² = u² - 2as
0 = 91.66² - 2 a x 150
a = - 28 ft / s²
b ) While accelerating initial velocity u = 0
distance travelled s = .24 mi
time = 19.3 s
s = ut + 1/2 at²
s is distance travelled in time t with acceleration a ,
.24 = 0 + 1/2 a x 19.3²
a = .001288 mi/s²
= 2.06 m /s²
c )
If distance travelled s = .25 mi
final velocity v = ? a = .001288 mi / s²
v² = u² + 2as
= 0 + 2 x .001288 x .25
= .000644
v = .025 mi / s
= .0025 x 60 x 60 mi / h
= 91.35 mph .
d ) initial velocity u = 59 mph
= 86.53 ft / s
final velocity = 0
acceleration = - 28 ft /s²
v = u - at
0 = 86.53 - 28 t
t = 3 sec approx .
Answer:
While the use of the type of transformer in a rectifier depends on the voltage requirement or to meet desired operating conditions, a step-down transformer is used mainly to reduce the voltage. It is used to bring the high AC voltage level to a reasonable value or the desired output voltage.
Explanation:
Hope it helps
Correct me if Im wrong
