Answer: precision
Explanation: Because accuracy is where you keep on getting it right but precision is where you get closer and closer
Answer:
Given that;
Jello there, see explanstion for step by step solving.
A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp/dx1. Find the velocity profiles of two fluids if the height of the flat interface is ha.
Explanation:
A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp/dx1. Find the velocity profiles of two fluids if the height of the flat interface is ha.
See attachment for more clearity
Answer:
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Explanation:
Answer:
The temperature T= 648.07k
Explanation:
T1=input temperature of the first heat engine =1400k
T=output temperature of the first heat engine and input temperature of the second heat engine= unknown
T3=output temperature of the second heat engine=300k
but carnot efficiency of heat engine =![1 - \frac{Tl}{Th} \\](https://tex.z-dn.net/?f=1%20-%20%5Cfrac%7BTl%7D%7BTh%7D%20%5C%5C)
where Th =temperature at which the heat enters the engine
Tl is the temperature of the environment
since both engines have the same thermal capacities <em>
</em> therefore ![n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\](https://tex.z-dn.net/?f=n_%7Bth%7D%20%3Dn_%7Bth1%7D%20%3Dn_%7Bth2%7D%5C%5Cn_%7Bth%20%7D%3D1-%5Cfrac%7BT1%7D%7BT%7D%3D1-%5Cfrac%7BT%7D%7BT3%7D%5C%5C%20%5C%5C%3D%201-%5Cfrac%7B1400%7D%7BT%7D%3D1-%5Cfrac%7BT%7D%7B300%7D%5C%5C)
We have now that
![\frac{-1400}{T}+\frac{T}{300}=0\\](https://tex.z-dn.net/?f=%5Cfrac%7B-1400%7D%7BT%7D%2B%5Cfrac%7BT%7D%7B300%7D%3D0%5C%5C)
multiplying through by T
![-1400 + \frac{T^{2} }{300}=0\\](https://tex.z-dn.net/?f=-1400%20%2B%20%5Cfrac%7BT%5E%7B2%7D%20%7D%7B300%7D%3D0%5C%5C)
multiplying through by 300
-![420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000} \\T=648.07k](https://tex.z-dn.net/?f=420000%2B%20T%5E%7B2%7D%20%3D0%5C%5CT%5E2%20%3D420000%5C%5C%5Csqrt%7BT2%7D%3D%5Csqrt%7B420000%7D%20%20%5C%5CT%3D648.07k)
The temperature T= 648.07k