okay so you have 1/4 and 5 1/2
make 5 1/2 a improper fraction
so you now have 1/4 and 11/2
now do ![\frac{1}{4} x\frac{11}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%7D%20x%5Cfrac%7B11%7D%7B2%7D)
![\frac{1}{4} x\frac{11}{2} = \frac{11}{8} = 1.375](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%7D%20x%5Cfrac%7B11%7D%7B2%7D%20%3D%20%5Cfrac%7B11%7D%7B8%7D%20%3D%201.375)
I hope this helps you
I feel bad study island is the worst!
3) 1/1 2/0 -2/0 good luck i hope this helped
8/21,13/21 because .16/42=x/100 and then cross multiply and simplify
Answer:
Power generated = 4.315 kW
Step-by-step explanation:
Given,
- speed of wind when enters into the turbine, V = 12 m/s
- speed of wind when exits from the turbine, U = 9 m/s
- mass flow rate of the wind, m = 137 kg/s
According to the law of conservation of energy
Energy generated = change in kinetic energy
Since, the air is exiting at same elevation. So, we will consider only kinetic energy.
Energy generated in one second will be given by,
![E\ =\ \dfrac{1}{2}.mV^2-\dfrac{1}{2}.m.U^2](https://tex.z-dn.net/?f=E%5C%20%3D%5C%20%5Cdfrac%7B1%7D%7B2%7D.mV%5E2-%5Cdfrac%7B1%7D%7B2%7D.m.U%5E2)
![=\ \dfrac{1}{2}\times 137\times (12)^2-\dfrac{1}{2}\times 137\times (9)^2](https://tex.z-dn.net/?f=%3D%5C%20%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20137%5Ctimes%20%2812%29%5E2-%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20137%5Ctimes%20%289%29%5E2)
=4315.5 J
= 4.315 kJ
So, energy is generated in one second = 4.315 kJ
Power generated can be given by,
![P\ =\ \dfrac{\textrm{energy generated}}{time}](https://tex.z-dn.net/?f=P%5C%20%3D%5C%20%5Cdfrac%7B%5Ctextrm%7Benergy%20generated%7D%7D%7Btime%7D)
And the energy is generated is already in per second so power generated will be 4.315 kW.