Answer: When a ray of light approaches a smooth polished surface and the light ray bounces back, it is called the reflection of light. The incident light ray which lands upon the surface is said to be reflected off the surface. The ray that bounces back is called the reflected ray.<u>
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</u>Have a great day and stay safe !
I would say it's <span>C. the Moon
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Answer:
Minimum uncertainty in velocity of a proton,
Explanation:
It is given that,
A proton is confined to a space 1 fm wide, ![\Delta x=10^{-15}\ m](https://tex.z-dn.net/?f=%5CDelta%20x%3D10%5E%7B-15%7D%5C%20m)
We need to find the minimum uncertainty in its velocity. We know that the Heisenberg Uncertainty principle gives the uncertainty between position and the momentum such that,
![\Delta p.\Delta x\ge \dfrac{h}{4\pi}](https://tex.z-dn.net/?f=%5CDelta%20p.%5CDelta%20x%5Cge%20%5Cdfrac%7Bh%7D%7B4%5Cpi%7D)
Since, p = mv
![\Delta (mv).\Delta x\ge \dfrac{h}{4\pi}](https://tex.z-dn.net/?f=%5CDelta%20%28mv%29.%5CDelta%20x%5Cge%20%5Cdfrac%7Bh%7D%7B4%5Cpi%7D)
![m \Delta v.\Delta x\ge \dfrac{h}{4\pi}](https://tex.z-dn.net/?f=m%20%5CDelta%20v.%5CDelta%20x%5Cge%20%5Cdfrac%7Bh%7D%7B4%5Cpi%7D)
![\Delta v\ge \dfrac{h}{4\pi m\Delta x}](https://tex.z-dn.net/?f=%5CDelta%20v%5Cge%20%5Cdfrac%7Bh%7D%7B4%5Cpi%20m%5CDelta%20x%7D)
![\Delta v\ge \dfrac{6.63\times 10^{-34}}{4\pi \times 1.67\times 10^{-27}\times 10^{-15}}](https://tex.z-dn.net/?f=%5CDelta%20v%5Cge%20%5Cdfrac%7B6.63%5Ctimes%2010%5E%7B-34%7D%7D%7B4%5Cpi%20%5Ctimes%201.67%5Ctimes%2010%5E%7B-27%7D%5Ctimes%2010%5E%7B-15%7D%7D)
![\Delta v\ge 3.15\times 10^7\ m/s](https://tex.z-dn.net/?f=%5CDelta%20v%5Cge%203.15%5Ctimes%2010%5E7%5C%20m%2Fs)
So, the minimum uncertainty in its velocity is greater than
. Hence, this is the required solution.
Answer:
The discharge rate is ![Q = 0.0192 \ m^3 /s](https://tex.z-dn.net/?f=Q%20%3D%200.0192%20%5C%20%20m%5E3%20%2Fs)
Explanation:
From the question we are told that
The diameter is ![d = 60 \ mm = 0.06 \ m](https://tex.z-dn.net/?f=d%20%3D%20%2060%20%5C%20mm%20%20%20%3D%20%200.06%20%5C%20m)
The head is ![h = 6 \ m](https://tex.z-dn.net/?f=h%20%20%3D%20%206%20%5C%20m)
The coefficient of contraction is ![Cc = 0.68](https://tex.z-dn.net/?f=Cc%20%20%3D%20%200.68)
The coefficient of velocity is ![Cv = 0.92](https://tex.z-dn.net/?f=Cv%20%20%3D%20%200.92)
The radius is mathematically evaluated as
![r = \frac{d}{2}](https://tex.z-dn.net/?f=r%20%3D%20%20%5Cfrac%7Bd%7D%7B2%7D)
substituting values
![r = \frac{ 0.06 }{2}](https://tex.z-dn.net/?f=r%20%3D%20%20%5Cfrac%7B%200.06%20%7D%7B2%7D)
![r = 0.03 \ m](https://tex.z-dn.net/?f=r%20%3D%20%200.03%20%5C%20m)
The area is mathematically represented as
![A = \pi r^2](https://tex.z-dn.net/?f=A%20%3D%20%20%5Cpi%20r%5E2)
substituting values
![A = 3.142 * (0.03)^2](https://tex.z-dn.net/?f=A%20%3D%20%203.142%20%2A%20%20%280.03%29%5E2)
![A = 0.00283 \ m^2](https://tex.z-dn.net/?f=A%20%3D%200.00283%20%5C%20m%5E2)
The discharge rate is mathematically represented as
![Q = Cv *Cc * A * \sqrt{ 2 * g * h}](https://tex.z-dn.net/?f=Q%20%3D%20%20Cv%20%2ACc%20%20%2A%20%20A%20%20%2A%20%20%5Csqrt%7B%202%20%2A%20g%20%2A%20%20h%7D)
substituting values
![Q = 0.68 * 0.92* 0.00283 * \sqrt{ 2 * 9.8 * 6}](https://tex.z-dn.net/?f=Q%20%3D%200.68%20%2A%20%200.92%2A%20%20%200.00283%20%20%2A%20%20%5Csqrt%7B%202%20%2A%209.8%20%2A%20%206%7D)
![Q = 0.0192 \ m^3 /s](https://tex.z-dn.net/?f=Q%20%3D%200.0192%20%5C%20%20m%5E3%20%2Fs)
Answer:
Bounce 1 , pass 3, emb2
Explanation:
(By the way I am also doing that question on College board physics page) For the Bounce arrow, since it bumps into the object and goes back, it means now it has a negative momentum, which means a larger momentum is given to the object. P=mv, so the velocity is larger for the object, and larger velocity means a larger kinetic energy which would result in a larger change in the potential energy. Since K=0.5mv^2=U=mgh, a larger potential energy would have a larger change in height which means it has a larger angle θ with the vertical line. Comparing with the "pass arrow" and the "Embedded arrow", the embedded arrow gives the object a larger momentum, Pi=Pf (mv=(M+m)V), it gives all its original momentum to the two objects right now. (Arrow and the pumpkin), it would have a larger velocity. However for the pass arrow, it only gives partial of its original momentum and keeps some of them for the arrow to move, which means the pumpkin has less momentum, means less velocity, and less kinetic energy transferred into the potential energy, and means less change in height, less θangle. So it is Bounce1, pass3, emb2.