if input frequency is 50 hz, the frequency of output wave in a half wave diode rectifier circuit is
50 hz
a frequency converter, also known as a frequency changer, is a device that takes incoming power, normally 50 or 60 hz and converts it to 400 hz output power. (a) f = 1/t = 2.11 hz

(b) 2*pi*f = 13.26 radians/s = sqrt(k/m)
m = 0.296 kg
solve for k
(c) (1/2)ka^2 = 0.150 j
solve for amplitude a, in meters the value of head loss due to friction using chezy formula with constant c= 60 for a fluid, flowing through a 30 cm diameter and 70 m long pipe. velocity of fluid inside the pipe is 5 m/s. area of a semi-circle= 77 m2 v = (4/3)πr³ 

dv/dt = 4πr².dr/dt

so, when the diameter is 20 cm we have:

-4π = 4πr².dr/dt

hence, dr/dt = -1/r²

i.e. dr/dt = -1/10² => -1/100

now, d = 2r

so, d(d)/dt = 2.dr/dt

=> d(d)/dt = -1/50 cm/hour

or, -0.2 mm/hour

:)> class intervals
frequency
(fi)
xi
di = xi - a
fi di


0 - 20
17
10
-40
-680


20 - 40
28
30
-20
-560


40 - 60
32
50 = a
0
0


60 - 80
p
70
20
20p


80 - 100
19
90
40
760


total
Î£
fi = 96 + p


Î£ fi di = 20p -
480


![](https://s3mn%20mnimgs%20com/img/shared/discuss_editlive/5130830/2013_07_08_13_01_36/mathmlequation8813644107607516415%20png) from the figure:total resistance of the circuit, r = (rab + 15) ω = 20 ωcurrent in the circuit , \[i = \frac{v}{r} = \frac{8}{20} a\] ∴ voltage across ab, vab = i.rab = 2 vthe emf of the cell connected as above is given by: \[e = \frac{l}{l} v\_0\] here:l = 60 cm (balance point)ab = l = 1 m = 100 cm (total length of the wire) \[\therefore e = \frac{60}{100}\left( 2 \right) = 1 . 2 v\] The code is as shown below 

**clc;**

**close** **all;**

**clear** **all;**

**% define n**

**n = 0:18;**

**N = 19;** **% N tap filter**

**M = (N-1)/2;**

**% define window**

**% define the filter**

**h = 1./((n-M)*pi).*sin((n-M)*pi/5).*blackman(19)'.*((n>=0)-(n>=N));**

**h(M+1) = 2/15;** **% this corresponds to n=1 where the denominator is zero**

**% plot the filter**

**figure;**

**stem(n,h,'filled');grid** **on;title('h[n]');**

**xlabel('n');ylabel('stem height');**

**% plot the frequency response of the filter**

**figure;**

**freqz(h,1);grid** **on;**

  


  


![](/image?k=f31b88da6411d9811f191c8b336bed5e)![](/image?k=d351b5006ccd60c46820c64292935f2f) 0.5 solution: in a given data, the highest frequency is 41, which lies in the interval 1000-15000.   
 here, l=10000, `f\_(m)=41=26, f\_(2)=16 and h=5000`   
 `"mode""=l+(f\_(m)-f\_(1))/(2f\_(m)-f\_(1)-f\_(2))xxh`   
 `=1000+((41-26)/(2xx41-26-16))xx5000`   
 `=1000+((15)/(82-42))xx5000`   
 `=1000+((15)/(40))xx5000`   
 `=1000+15xx125=10000+1875=11875`   
 hence, the modal income is 11875.