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3 years ago

The ______ is the time it takes for a wave to complete one cycle.

Physics

2 answers:

EastWind [94]3 years ago

5 0

It is called the period

kykrilka [37]3 years ago

4 0

Answer:

time period

Explanation:

The time period is defined as the time taken by a wave to complete one cycle.

Its SI unit is second.

It is the reciprocal of frequency.

Frequency is defined as the number of cycles completed in one second.

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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!

4 0

4 years ago

Which parameter of the present universe, more than any other, is considered to be critical in determining the ultimate fate of t

pochemuha

The slope and curvature of space time which is being derived from the Einstein's law of gravitation which was modified later it gives three slopes value (-,0,+ )

8 0

3 years ago

A car’s tire rotates 5.25 times in 3 seconds. What is the tangential velocity of the tire?

Lena [83]

The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire which is 11(r) m/s.

<h3>Angular velocity of the tire</h3>

The angular velocity of the tire is the rate of change of angular displacement of the tire with time.

The magnitude of the angular velocity of the tire is calculated as follows;

ω = 2πN

where;

N is the number of revolutions per second

ω = 2π x (5.25 / 3)

ω = 11 rad/s

<h3>Tangential velocity of the tire</h3>

The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire.

The magnitude of the tangential velocity is caculated as follows;

v = ωr

where;

r is the radius of the car's tire

v = 11r m/s

Learn more about tangential velocity here: brainly.com/question/25780931

4 0

3 years ago

A student performs an experiment in measuring the period of a simple pendulum of known length 49.0 cm.He performed five trials a

Levart [38]

Correct Answer is Bb

4 0

3 years ago

If I start driving 80 miles per hour down the interstate how far will I go if I drop for 4.5 hours

Arte-miy333 [17]

360 miles. Hope this helps

4 0

3 years ago