Answer:
For water
Flow rate= 0.79128*10^-3 Ns
For Air
Flow rate =1.2717*10^-3 Ns
Explanation:
For the flow rate of water in pipe.
Area of the pipe= πd²/4
Diameter = 30/1000
Diameter= 0.03 m
Area= 3.14*(0.03)²/4
Area= 7.065*10^-4
Flow rate = 7.065*10^-4*1.12E-3
Flow rate= 0.79128*10^-3 Ns
For the flow rate of air in pipe.
Flow rate = 7.065*10^-4*1.8E-5
Flow rate =1.2717*10^-3 Ns
Answer:
1. Buy Quiet – select and purchase low-noise tools and machinery
2. Maintain tools and equipment routinely (such 3. as lubricate gears)
3. Reduce vibration where possible
4. Isolate the noise source in an insulated room or enclosure
5. Place a barrier between the noise source and the employee
6. Isolate the employee from the source in a room or booth (such as sound wall or window
Explanation:
Hope my answer will help u.
Answer:
The surface temperature is 921.95°C .
Explanation:
Given:
a=25 cm ,P=350 hp⇒P=260750 W
Power transmitted 
W and remaining will lost in the form of heat.This heat transmitted to air by the convection.
h=230
Actually heat will be transmit by the convection.
In convection Q=hA
So 
T=921.95°C
So the surface temperature is 921.95°C .
Answer:
First you have to separate real and imaginary parts of Tan(x+iy)=Tan(z)=sin(z)/cos(z)
sinz=sin(x+iy)=sinxcos(iy)+cosxsin(iy)=sinxcoshy-icosx sinhy
cosz=cos(x+iy)=cosxcos(iy)-sinxsin(iy)=cosxcoshy−isinxsinhy
Now if you plug in Tan(z) and simplify (it is easy!) you get
Tan(z)=(sin(2x)+isinh(2y))/(cos(2x)+cosh(2y))= A+iB.
This means that
A=sin(2x)/(cos(2x)+cosh(2y)) and B= sinh(2y)/(cos(2x)+cosh(2y))
Now,
A/B=sin(2x)/sinh(2y)
If any questions, let me know.
Answer:
QPSK: 7.5 MHz
64-QAM:2.5 MHz
64-Walsh-Hadamard: 160 MHz
Explanation:
See attached picture.
