Answer: v = ![1.19 * 10^{6} m/s](https://tex.z-dn.net/?f=1.19%20%2A%2010%5E%7B6%7D%20m%2Fs)
Explanation: q = magnitude of electronic charge = ![1.609 * 10^{-19} c](https://tex.z-dn.net/?f=1.609%20%2A%2010%5E%7B-19%7D%20c)
mass of an electronic charge =
V= potential difference = 4V
v = velocity of electron
by using the work- energy theorem which states that the kinetic energy of the the electron must equal the work done use in accelerating the electron.
kinetic energy =
, potential energy = qV
hence, ![\frac{mv^{2} }{2} = qV](https://tex.z-dn.net/?f=%5Cfrac%7Bmv%5E%7B2%7D%20%7D%7B2%7D%20%3D%20qV)
![\frac{9.10 *10^{-31} * v^{2} }{2} = 1.609 * 10^{-16} * 4\\\\\\\\9.10*10^{-31} * v^{2} = 2 * 1.609 *10^{-16} * 4\\\\\\9.10 *10^{-31} * v^{2} = 1.287 *10^{-15} \\\\v^{2} = \frac{1.287 *10^{-15} }{9.10 *10^{31} } \\\\v^{2} = 1.414*10^{15} \\\\v = \sqrt{1.414*10^{15} } \\\\v = 1.19 * 10^{6} m/s](https://tex.z-dn.net/?f=%5Cfrac%7B9.10%20%2A10%5E%7B-31%7D%20%2A%20v%5E%7B2%7D%20%20%7D%7B2%7D%20%3D%201.609%20%2A%2010%5E%7B-16%7D%20%2A%204%5C%5C%5C%5C%5C%5C%5C%5C9.10%2A10%5E%7B-31%7D%20%20%2A%20v%5E%7B2%7D%20%3D%202%20%2A%201.609%20%2A10%5E%7B-16%7D%20%2A%204%5C%5C%5C%5C%5C%5C9.10%20%2A10%5E%7B-31%7D%20%2A%20v%5E%7B2%7D%20%3D%201.287%20%2A10%5E%7B-15%7D%20%5C%5C%5C%5Cv%5E%7B2%7D%20%3D%20%5Cfrac%7B1.287%20%2A10%5E%7B-15%7D%20%7D%7B9.10%20%2A10%5E%7B31%7D%20%7D%20%5C%5C%5C%5Cv%5E%7B2%7D%20%3D%201.414%2A10%5E%7B15%7D%20%5C%5C%5C%5Cv%20%3D%20%5Csqrt%7B1.414%2A10%5E%7B15%7D%20%7D%20%5C%5C%5C%5Cv%20%3D%201.19%20%2A%2010%5E%7B6%7D%20m%2Fs)
Answer:
521 nm
Explanation:
Given the values and units we are given, I'm assuming 5.76*10^14 Hz is frequency.
The formula to use here is λ * υ = c, where λ is wavelength, υ is frequency, and c is the speed of light.
λ = ![\frac{3*10^8\frac{m}{s} }{5.76*10^{14}Hz} = {5.20833*10^{-7} m}\approx{521 *10^{-9}m}={521 nm}](https://tex.z-dn.net/?f=%5Cfrac%7B3%2A10%5E8%5Cfrac%7Bm%7D%7Bs%7D%20%7D%7B5.76%2A10%5E%7B14%7DHz%7D%20%3D%20%7B5.20833%2A10%5E%7B-7%7D%20m%7D%5Capprox%7B521%20%2A10%5E%7B-9%7Dm%7D%3D%7B521%20nm%7D)
Answer:
B) 0.3Hz
Explanation:
I just took the test i hope i helped and i hope you pass the test
The moment of inertia of a point mass about an arbitrary point is given by:
I = mr²
I is the moment of inertia
m is the mass
r is the distance between the arbitrary point and the point mass
The center of mass of the system is located halfway between the 2 inner masses, therefore two masses lie ℓ/2 away from the center and the outer two masses lie 3ℓ/2 away from the center.
The total moment of inertia of the system is the sum of the moments of each mass, i.e.
I = ∑mr²
The moment of inertia of each of the two inner masses is
I = m(ℓ/2)² = mℓ²/4
The moment of inertia of each of the two outer masses is
I = m(3ℓ/2)² = 9mℓ²/4
The total moment of inertia of the system is
I = 2[mℓ²/4]+2[9mℓ²/4]
I = mℓ²/2+9mℓ²/2
I = 10mℓ²/2
I = 5mℓ²