Answer:
2750
Explanation:
The number of windings and the voltage are proportional.
__
Let n represent the number of windings to produce 110 Vac. Then the proportion is ...
n/110 = 300,000/12,000
n = 110(300/12) = 2750 . . . . multiply by 110
2750 windings would be needed to produce 110 Vac at the output.
Answer:
#See solution for details.
Explanation:
#The commutative property of multiplication tells us that it doesn't matter in what order you multiply numbers. The formula for this property is a * b = b * a:
![40\times32\times10\times 25=320,000\\\\25\times10\times32\times 40=320,000\\\\10\times32\times25\times 40=320,000\\\\32\times25\times40\times 10=320,000](https://tex.z-dn.net/?f=40%5Ctimes32%5Ctimes10%5Ctimes%2025%3D320%2C000%5C%5C%5C%5C25%5Ctimes10%5Ctimes32%5Ctimes%2040%3D320%2C000%5C%5C%5C%5C10%5Ctimes32%5Ctimes25%5Ctimes%2040%3D320%2C000%5C%5C%5C%5C32%5Ctimes25%5Ctimes40%5Ctimes%2010%3D320%2C000)
Hence, the product of the four numbers remains 320,000 irrespective of their order.
Answer:
0.08kg/s
Explanation:
For this problem you must use 2 equations, the first is the continuity equation that indicates that all the mass flows that enter is equal to those that leave the system, there you have the first equation.
The second equation is obtained using the first law of thermodynamics that indicates that all the energies that enter a system are the same that come out, you must take into account the heat flows, work and mass flows of each state, as well as their enthalpies found with the temperature.
finally you use the two previous equations to make a system and find the mass flows
I attached procedure
Answer:
5.6 mm
Explanation:
Given that:
A cylindrical tank is required to contain a:
Gage Pressure P = 560 kPa
Allowable normal stress
= 150 MPa = 150000 Kpa.
The inner diameter of the tank = 3 m
In a closed cylinder there exist both the circumferential stress and the longitudinal stress.
Circumferential stress ![\sigma = \dfrac{pd}{2t}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Cdfrac%7Bpd%7D%7B2t%7D)
Making thickness t the subject; we have
![t = \dfrac{pd}{2* \sigma}](https://tex.z-dn.net/?f=t%20%3D%20%5Cdfrac%7Bpd%7D%7B2%2A%20%5Csigma%7D)
![t = \dfrac{560000*3}{2*150000000}](https://tex.z-dn.net/?f=t%20%3D%20%5Cdfrac%7B560000%2A3%7D%7B2%2A150000000%7D)
t = 0.0056 m
t = 5.6 mm
For longitudinal stress.
![\sigma = \dfrac{pd}{4t}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Cdfrac%7Bpd%7D%7B4t%7D)
![t= \dfrac{pd}{4*\sigma }](https://tex.z-dn.net/?f=t%3D%20%5Cdfrac%7Bpd%7D%7B4%2A%5Csigma%20%7D)
![t = \dfrac{560000*3}{4*150000000}](https://tex.z-dn.net/?f=t%20%3D%20%5Cdfrac%7B560000%2A3%7D%7B4%2A150000000%7D)
t = 0.0028 mm
t = 2.8 mm
From the above circumferential stress and longitudinal stress; the stress with the higher value will be considered ; which is circumferential stress and it's minimum value with the maximum thickness = 5.6 mm