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3 years ago

What does the constant c mean and what is it’s value

Physics

1 answer:

Luba_88 [7]3 years ago

6 0

Answer:

Constants in C are the fixed values that are used in a program, and its value remains the same during the entire execution of the program.

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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

A block is projected up a frictionless inclined plane with initial speed v0 = 1.72 m/s. The angle of incline is θ = 44.8°. (a) H

Snowcat [4.5K]

Explanation:

Given

initial velocity(v_0)=1.72 m/s

$\theta =44.8{\circ}$

using $v^2-u^2=2as$

Where v=final velocity (Here v=0)

u=initial velocity(1.72 m/s)

a=acceleration $(gsin\theta )$

s=distance traveled

$0-(1.72)^2=2(-9.81\times sin(44.8))s$

s=0.214 m

(b)time taken to travel 0.214 m

v=u+at

$0=1.72-gsin(44.8)\times t$

$t=\frac{1.72}{9.8\times sin(44.8)}$

t=0.249 s

(c)Speed of the block at bottom

$v^2-u^2=2as$

Here u=0 as it started coming downward

$v^2=2\times gsin(44.8)\times 0.214$

$v=\sqrt{2.985}$

v=1.72 m/s

3 0

4 years ago

Two parallel slits are illuminated by light composed of two wavelengths. Wavelength A is 564 nm and the other is wavelength B an

anygoal [31]

Answer:

423nm

Explanation:

To find the unknown wavelength you take into account the distance y to the maximum central fringe, for light fringes and dark fringes.

- for light fringes:

$dsin\theta=m\lambda\\\\sin\theta\approx\theta=\frac{y}{D}\\\\y=\frac{m\lambda_1D}{d}$

- for dark fringes:

$y=\frac{m\lambda_2/2 D}{d}$

The third-order bright fringe (m= 3) of wavelength A coincides with the fourth dark fringe (m=4) of the wavelength B. Hence you have that:

$\frac{(3)\lambda_1D}{d}=\frac{(4)\lambda_2D}{d}\\\\\lambda_2=\frac{3}{4}\lambda_1=\frac{3}{4}(564nm)=423nm$

hence, the wavelength B is 423nm

7 0

4 years ago

Read 2 more answers

0.5000 kg of water at 35.00 degrees Celsius is cooled, with the removal of 6.300 E4 J of heat. What is the final temperature of

kogti [31]

Answer:

=30.22°C

Explanation:

The enthalpy change made the water to cool.

Enthalpy change = MC∅ where M is the mass of the water, C is the specific heat capacity of water and ∅ the temperature change.

ΔH=MC∅

∅=ΔH/MC

=(6.3×10⁴J)/(0.5kg×4186J/(kg°C))

=30.1

Final temperature = 35.00°C-30.1°C

=4.9°C

5 0

3 years ago

Read 2 more answers

You are traveling at 14 m/s for 20 seconds. What is your displacement?

enyata [817]

Velocity = 14 m/s

Time = 20 s

Displacement = Velocity×Time = (14×20) m = 280 m

The displacement is 280 m towards the direction of motion.

5 0

3 years ago

Read 2 more answers