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Reil [10]
4 years ago
7

If red light hits a red filter, what kind of light comes through the filter?

Physics
1 answer:
Roman55 [17]4 years ago
8 0
A. red, is the correct answer. A red filter only lets red light through, whether white or red light passes through it.
Hope this helps, and have a great day!

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The Tevatron acceleator at the Fermi National Accelerator Laboratory (Fermilab) outside Chicago boosts protons to 1 TeV (1000 Ge
Eva8 [605]

Answer:

a) v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s

b) v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s

c) v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s

d) v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s

e) v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s

Explanation:

At that energies, the speed of proton is in the relativistic theory field, so we need to use the relativistic kinetic energy equation.

KE=mc^{2}(\gamma -1) = mc^{2}(\frac{1}{\sqrt{1-\beta^{2}}} -1)           (1)

Here β = v/c, when v is the speed of the particle and c is the speed of light in vacuum.

Let's solve (1) for β.

\beta = \sqrt{1-\frac{1}{\left (\frac{KE}{mc^{2}}+1 \right )^{2}}}

We can write the mass of a proton in MeV/c².

m_{p}=938.28 MeV/c^{2}

Now we can calculate the speed in each stage.

a) Cockcroft-Walton (750 keV)

\beta = \sqrt{1-\frac{1}{\left (\frac{0.75 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.04

v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s

b) Linac (400 MeV)

\beta = \sqrt{1-\frac{1}{\left (\frac{400 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.71

v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s

c) Booster (8 GeV)

\beta = \sqrt{1-\frac{1}{\left (\frac{8000 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.994

v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s

d) Main ring or injector (150 Gev)

\beta = \sqrt{1-\frac{1}{\left (\frac{150000 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.999

v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s

e) Tevatron (1 TeV)

\beta = \sqrt{1-\frac{1}{\left (\frac{1000000 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.9999

v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s

Have a nice day!

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