Answer:
B. NET force: 2 resultant motion: left
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C. Net force: 3 Resultant motion: Left
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D. Net Force: 7 Resultant motion: right
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E. Net Force:0 resultant motion: NO MOTION
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F. NET Force: 3 resultant motion: Down
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G. NET FORCE: 10 resultant motion: up
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H. Net force: 3 Resultant motion: left
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I. Net force: 50 Resultant motion: right
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J. NET FORCE: 75 Resultant motion: down
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K. Net force :200 Resultant motion: Right
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L. Net force: 0 resultant motion:No motion
Explanation:
Answer:
For a given spring the extension is directly proportional to the force applied For example if the force is doubled, the extension doubles When an elastic object is stretched beyond its limit of proportionality the object does not return to its original length when the force is removed
Explanation:
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)
The valence electrons of metals are weakly attracted to the parent nuclei, so the electrons break free and float. The moving electrons form a electron <u>negative</u> blanket that binds the atomic <u>positive</u> nuclei together, forming a metallic bond.
So the answers are <u>{ Negative }</u> and <u>{ Positive }.</u>
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