Answer:
9500 kJ; 9000 Btu
Explanation:
Data:
m = 100 lb
T₁ = 25 °C
T₂ = 75 °C
Calculations:
1. Energy in kilojoules
ΔT = 75 °C - 25 °C = 50 °C = 50 K

2. Energy in British thermal units

Answer:
a) 
b) 
c)
d) 
Explanation:
We know that the kinetic energy can be expressed in terms of momentum (p= mv):

- p is the momentum
- m is the mass of the particle
So, the momentum will be:
(1)
We can use the energy conservation to relate K and the electric potential energy. We assume that all potential energy becomes kinetic energy. Therefore:
(2)
a) <u>If V=10 [V], K will be:</u>
![K=1.6*10^{-19}*10=1.6*10^{-18}[J]](https://tex.z-dn.net/?f=K%3D1.6%2A10%5E%7B-19%7D%2A10%3D1.6%2A10%5E%7B-18%7D%5BJ%5D)
We can find the momentum using the equation (1)
![p=\sqrt{2*9.1 x 10^{-31}*1.6*10^{-18}}=1.71*10^{-24}[kg*m*s^{-1}]](https://tex.z-dn.net/?f=p%3D%5Csqrt%7B2%2A9.1%20x%2010%5E%7B-31%7D%2A1.6%2A10%5E%7B-18%7D%7D%3D1.71%2A10%5E%7B-24%7D%5Bkg%2Am%2As%5E%7B-1%7D%5D)
The De-Broglie wavelength equation is given by:
(3)
- h is the Plank constant (
)
![\lambda=\frac{6.626 x 10^{-34}}{1.71*10^{-24}}=3.87*10^{-10}[m]=3.87 \r A](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B6.626%20x%2010%5E%7B-34%7D%7D%7B1.71%2A10%5E%7B-24%7D%7D%3D3.87%2A10%5E%7B-10%7D%5Bm%5D%3D3.87%20%5Cr%20A)
b) <u>If V=100 [V],</u> using the same analyze, the De-Broglie wavelength will be:
![K=1.6*10^{-19}*100=1.6*10^{-17}[J]](https://tex.z-dn.net/?f=K%3D1.6%2A10%5E%7B-19%7D%2A100%3D1.6%2A10%5E%7B-17%7D%5BJ%5D)
![p=\sqrt{2*9.1 x 10^{-31}*1.6*10^{-17}}=5.39*10^{-24}[kg*m*s^{-1}]](https://tex.z-dn.net/?f=p%3D%5Csqrt%7B2%2A9.1%20x%2010%5E%7B-31%7D%2A1.6%2A10%5E%7B-17%7D%7D%3D5.39%2A10%5E%7B-24%7D%5Bkg%2Am%2As%5E%7B-1%7D%5D)
![\lambda=\frac{6.626 x 10^{-34}}{5.39*10^{-24}}=1.23*10^{-10}[m]=1.23 \r A](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B6.626%20x%2010%5E%7B-34%7D%7D%7B5.39%2A10%5E%7B-24%7D%7D%3D1.23%2A10%5E%7B-10%7D%5Bm%5D%3D1.23%20%5Cr%20A)
c) <u>V=1000 [V]</u>
![K=1.6*10^{-19}*100=1.6*10^{-16}[J]](https://tex.z-dn.net/?f=K%3D1.6%2A10%5E%7B-19%7D%2A100%3D1.6%2A10%5E%7B-16%7D%5BJ%5D)
![p=\sqrt{2*9.1 x 10^{-31}*1.6*10^{-16}}=1.71*10^{-23}[kg*m*s^{-1}]](https://tex.z-dn.net/?f=p%3D%5Csqrt%7B2%2A9.1%20x%2010%5E%7B-31%7D%2A1.6%2A10%5E%7B-16%7D%7D%3D1.71%2A10%5E%7B-23%7D%5Bkg%2Am%2As%5E%7B-1%7D%5D)
d) <u>V=10000 [V]</u>
![K=1.6*10^{-19}*100=1.6*10^{-15}[J]](https://tex.z-dn.net/?f=K%3D1.6%2A10%5E%7B-19%7D%2A100%3D1.6%2A10%5E%7B-15%7D%5BJ%5D)
![p=\sqrt{2*9.1 x 10^{-31}*1.6*10^{-15}}=5.40*10^{-23}[kg*m*s^{-1}]](https://tex.z-dn.net/?f=p%3D%5Csqrt%7B2%2A9.1%20x%2010%5E%7B-31%7D%2A1.6%2A10%5E%7B-15%7D%7D%3D5.40%2A10%5E%7B-23%7D%5Bkg%2Am%2As%5E%7B-1%7D%5D)
![\lambda=\frac{6.626 x 10^{-34}}{5.40*10^{-23}}=1.23*10^{-11}[m]=0.123 \r A](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B6.626%20x%2010%5E%7B-34%7D%7D%7B5.40%2A10%5E%7B-23%7D%7D%3D1.23%2A10%5E%7B-11%7D%5Bm%5D%3D0.123%20%5Cr%20A)
The fringe spacing interference is proportional to the wavelength, so in our case, the larger fringe spacing occurs when voltage is 10 V, here λ = 3.87 angstroms.
I hope it helps you!
Answer:
The barrel with the sharp edge will drain faster.
Explanation:
The barrel with filed down entrance will gradually allow water to flow out due to the soft round corner.
However, the barrel with sharp edge will create turbulent eddies because of the sudden contraction phenomenon at the exit point. This turbulent eddies will occur in the inside portion of the sharp edge. First, let us understand what Eddies are.
An eddy, is fluid current whose flow direction differs from that of the general flow. Eddies can transfer much more energy and dissolved matter within the fluid and they actually mix together large masses of fluid. They generally become more numerous as the fluid flow velocity increases.
Based on the above definition, we can see that due to the sharp edge created, the flow direction differs from that of the general flow thereby creating eddies that are usually turbulent which causes more outflow of water than that with the soft edge.