Answer:
Width of the slit will be equal to 1.47 mm
Explanation:
We have given wavelength of the light ![\lambda =550nm=550\times 10^{-9}m](https://tex.z-dn.net/?f=%5Clambda%20%3D550nm%3D550%5Ctimes%2010%5E%7B-9%7Dm)
Distance D = 8 m
Distance between first minimum dark fringe and the central maximum is 2 mm
So ![x=3\times 10^{-3}m](https://tex.z-dn.net/?f=x%3D3%5Ctimes%2010%5E%7B-3%7Dm)
We have to find the width of the slit
For the first order wavelength is equal to
, here a width of slit
So ![a=\frac{\lambda D}{x}=\frac{550\times 10^{-9}\times 8}{3\times 10^{-3}}=1466.666\times 10^{-6}=1.47mm](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B%5Clambda%20D%7D%7Bx%7D%3D%5Cfrac%7B550%5Ctimes%2010%5E%7B-9%7D%5Ctimes%208%7D%7B3%5Ctimes%2010%5E%7B-3%7D%7D%3D1466.666%5Ctimes%2010%5E%7B-6%7D%3D1.47mm)
So width of the slit will be equal to 1.47 mm
Gravitational field exists in
the space surrounding a charged particle and exerts a force on other charged
particles. Gravitational waves are ripples of waves travelling outward from the
source. The more massive the orbit of two bodies, the more it emits
gravitational wave. And everything around it that is near within the wave
experiences a ‘pull’ toward the orbiting bodies.
Complete Question
A person throws a pumpkin at a horizontal speed of 4.0 m/s off a cliff. The pumpkin travels 9.5m horizontally before it hits the ground. We can ignore air resistance.What is the pumpkin's vertical displacement during the throw? What is the pumpkin's vertical velocity when it hits the ground?
Answer:
The pumpkin's vertical displacement is ![H = 27 .67 \ m](https://tex.z-dn.net/?f=H%20%3D%2027%20.67%20%5C%20m)
The pumpkin's vertical velocity when it hits the ground is ![v_v__{f}} = 23.298 \ m/s](https://tex.z-dn.net/?f=v_v__%7Bf%7D%7D%20%3D%2023.298%20%5C%20%20m%2Fs)
Explanation:
From the question we are told that
The horizontal speed is ![v_h = 4 m/s](https://tex.z-dn.net/?f=v_h%20%20%3D%20%204%20m%2Fs)
The horizontal distance traveled is ![d = 9.5 \ m](https://tex.z-dn.net/?f=d%20%3D%20%209.5%20%5C%20m)
The horizontal distance traveled is mathematically represented as
![S = v_h * t](https://tex.z-dn.net/?f=S%20%3D%20%20v_h%20%2A%20t)
Where t is the time taken
substituting values
![9.5 = 4 * t](https://tex.z-dn.net/?f=9.5%20%3D%20%204%20%2A%20t)
=> ![t = \frac{9.5}{4}](https://tex.z-dn.net/?f=t%20%3D%20%20%5Cfrac%7B9.5%7D%7B4%7D)
![t = 2.38 \ sec](https://tex.z-dn.net/?f=t%20%3D%202.38%20%5C%20sec)
Now the vertical displacement is mathematically represented as
![H = v_v t + \frac{1}{2} a_v t^2](https://tex.z-dn.net/?f=H%20%20%3D%20%20v_v%20t%20%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20a_v%20t%5E2)
now the vertical velocity before the throw is zero
So
![H = 0 + \frac{1}{2} (9.8) * (2.375)^2](https://tex.z-dn.net/?f=H%20%3D%20%200%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20%289.8%29%20%2A%20%282.375%29%5E2)
![H = 27 .67 \ m](https://tex.z-dn.net/?f=H%20%3D%2027%20.67%20%5C%20m)
Now the final vertical velocity is mathematically represented as
![v_v__{f}} = v_v + at](https://tex.z-dn.net/?f=v_v__%7Bf%7D%7D%20%3D%20%20v_v%20%2B%20at)
substituting values
![v_v__{f}} = 0 + (9.8)* (2.375)](https://tex.z-dn.net/?f=v_v__%7Bf%7D%7D%20%3D%20%200%20%2B%20%289.8%29%2A%20%282.375%29)
![v_v__{f}} = 23.298 \ m/s](https://tex.z-dn.net/?f=v_v__%7Bf%7D%7D%20%3D%2023.298%20%5C%20%20m%2Fs)
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Answer:
Water falls from a reservoir through a channel to a turbine. The water turns the turbines, which allows the generator to make electricity.
It is falling instead of flowing, because elavation is an important part of the hydroeletric power plant, since gravity is a thing, and there was elevation, than it would be falling and not flowing.