Ask question

Ask question

All categories

3 years ago

13

Which has more pressure, FIRST one gets all the points.

1 answer:

SSSSS [86.1K]3 years ago

8 0

<u>Answer:</u>

area of point has more pressure.

<u>Explanation:</u>

formula's : pressure = force ÷ area

1st pressure,

200 ÷ 0.2 = 1000 Pa

2nd pressure,

200 ÷ 0.004 = 50000 Pa

<h3>Therefore the pressure on the area of the point is more.</h3>

You might be interested in

Helpppppppp<br> ill give brainliest

Nina [5.8K]

Answer:

I believe the answer is

$B. \: 3 \: red, \: 8 \: yellow, \: 1 \: blue$

Explanation:

I may be incorrect because I used to do this a long time ago but I believe I am correct

HOPE THIS HELPS!

(:

4 0

3 years ago

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

Explain how air pollution impacts each of Earth's systems.

dexar [7]

Answer:

1) Atmosphere

Explanation: because if you think about it the only real explaination is atmosphere.

4 0

3 years ago

Olenka [21]

C - I’m pretty sure!

7 0

3 years ago

Using the conversion factor from eV (electron volts) to joules, determine which energy line for an electron dropping from one en

Tom [10]

Explanation:

Wavelength in an emission spectrum, $\lambda=435\ nm=435\times 10^{-9}\ m$

The energy of an electron is given by :

$E=\dfrac{hc}{\lambda}$

Where

h is the Planck's constant

c is the speed of light

For 435 nm, the energy of the electron will be :

$E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8\ m/s}{435\times 10^{-9}}$

$E=4.57\times 10^{-19}\ J$

We know that $1\ eV=1.6\times 10^{-19}\ J$

So, $E=\dfrac{4.57\times 10^{-19}}{1.6\times 10^{-19}}$

So, E = 2.86 eV

The energy of the electron dropping from one energy level is 2.86 eV. We know that,

$\dfrac{hc}{\lambda}=E_{n_2}-E_{n_1}$

From the given energy levels :

$E_5-E_2=-0.544-(-3.403)$

$E_5-E_2=2.859\ eV$

So, the transition must be from E₅ to E₂. Hence, this is the required solution.

5 0

4 years ago

Read 2 more answers