Answer:
Probability of tunneling is ![10^{- 1.17\times 10^{32}}](https://tex.z-dn.net/?f=10%5E%7B-%201.17%5Ctimes%2010%5E%7B32%7D%7D)
Solution:
As per the question:
Velocity of the tennis ball, v = 120 mph = 54 m/s
Mass of the tennis ball, m = 100 g = 0.1 kg
Thickness of the tennis ball, t = 2.0 mm = ![2.0\times 10^{- 3}\ m](https://tex.z-dn.net/?f=2.0%5Ctimes%2010%5E%7B-%203%7D%5C%20m)
Max velocity of the tennis ball,
= 89 m/s
Now,
The maximum kinetic energy of the tennis ball is given by:
![KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv_%7Bm%7D%5E%7B2%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%200.1%5Ctimes%2089%5E%7B2%7D%20%3D%20396.05%5C%20J)
Kinetic energy of the tennis ball, KE' = ![\frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E%7B2%7D%20%3D%200.5%5Ctimes%200.1%5Ctimes%2054%5E%7B2%7D%20%3D%20154.8%5C%20m%2Fs)
Now, the distance the ball can penetrate to is given by:
![\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}](https://tex.z-dn.net/?f=%5Ceta%20%3D%20%5Cfrac%7B%5Cbar%7Bh%7D%7D%7B%5Csqrt%7B2m%28KE%20-%20KE%27%29%7D%7D)
![\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js](https://tex.z-dn.net/?f=%5Cbar%7Bh%7D%20%3D%20%5Cfrac%7Bh%7D%7B2%5Cpi%7D%20%3D%20%5Cfrac%7B6.626%5Ctimes%2010%5E%7B-%2034%7D%7D%7B2%5Cpi%7D%20%3D%201.0545%5Ctimes%2010%5E%7B-%2034%7D%5C%20Js)
Thus
![\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}](https://tex.z-dn.net/?f=%5Ceta%20%3D%20%5Cfrac%7B1.0545%5Ctimes%2010%5E%7B-%2034%7D%7D%7B%5Csqrt%7B2%5Ctimes%200.1%28396.05%20-%20154.8%29%7D%7D)
![\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}](https://tex.z-dn.net/?f=%5Ceta%20%3D%20%5Cfrac%7B1.0545%5Ctimes%2010%5E%7B-%2034%7D%7D%7B%5Csqrt%7B2%5Ctimes%200.1%28396.05%20-%20154.8%29%7D%7D)
![\eta = 1.52\times 10^{-35}\ m](https://tex.z-dn.net/?f=%5Ceta%20%3D%201.52%5Ctimes%2010%5E%7B-35%7D%5C%20m)
Now,
We can calculate the tunneling probability as:
![P(t) = e^{\frac{- 2t}{\eta}}](https://tex.z-dn.net/?f=P%28t%29%20%3D%20e%5E%7B%5Cfrac%7B-%202t%7D%7B%5Ceta%7D%7D)
![P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}](https://tex.z-dn.net/?f=P%28t%29%20%3D%20e%5E%7B%5Cfrac%7B-%202%5Ctimes%202.0%5Ctimes%2010%5E%7B-%203%7D%7D%7B1.52%5Ctimes%2010%5E%7B-35%7D%7D%7D%20%3D%20e%5E%7B-2.63%5Ctimes%2010%5E%7B32%7D%7D)
![P(t) = e^{-2.63\times 10^{32}}](https://tex.z-dn.net/?f=P%28t%29%20%3D%20e%5E%7B-2.63%5Ctimes%2010%5E%7B32%7D%7D)
Taking log on both the sides:
![logP(t) = -2.63\times 10^{32} loge](https://tex.z-dn.net/?f=logP%28t%29%20%3D%20-2.63%5Ctimes%2010%5E%7B32%7D%20loge)
![P(t) = 10^{- 1.17\times 10^{32}}](https://tex.z-dn.net/?f=P%28t%29%20%3D%2010%5E%7B-%201.17%5Ctimes%2010%5E%7B32%7D%7D)
Answer:
3A
Explanation:
Rtoal=R1+R2+R3=5+10+15=30
I=V/R 90/30
I=3
Answer:
152,155 J
Explanation:
115,333 + 36,822 = 152,155J
Answer:
factual evidence of customer-service levels.
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better understanding of the capability range of a process.