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3 years ago

You are currently getting 26 sales opportunities per day and closing 64% of them. How many sale opportunists per day?

Physics

1 answer:

Vikki [24]3 years ago

8 0

Answer: 16

Explanation:

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X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to

lys-0071 [83]

Answer:

a) $\Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

b) $\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

c) $E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

Explanation

Part a

For this case we can use the Compton shift equation given by:

$\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

Part b

For this cas we can calculate the wavelength of the phton with this formula:

$\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

$E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

3 0

3 years ago

Read 2 more answers

What experiment should I make using Gravitational Force? <br><br>PLEASE HELP ME :)

Bogdan [553]

You could try the "Spinning Bucket" or the "Center Of Gravity" experiment. There are plenty more that you could research! Hope this helped :)

8 0

3 years ago

A bomb is dropped from a bomber traveling at the speed of 120 km / h, destroying a military objective located at a distance of 2

schepotkina [342]

Answer:

18 km

Explanation:

Convert km/h to m/s:

120 km/h × (1000 m/km) × (1 h / 3600 s) = 33.3 m/s

The time it takes the bomb to travel the 2000 meters is:

2000 m / (33.3 m/s) = 60 s

So it takes 60 seconds for the bomb to fall. The distance it fell is therefore:

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (0 m/s) (60 s) + ½ (10 m/s²) (60 s)²

Δy = 18,000 m

Δy = 18 km

7 0

3 years ago

Rectangular plate has a voltage of +180V and plate to the first plate) has a 'voltage of -5V. Determie another rectangular plate

3241004551 [841]

For a Rectangular plate has a voltage of +180V and a 'voltage of -5V. , the second plate has the Electric field mathematically given as

E=21.5*10^3v/m

<h3>What is the field strength?</h3>

Generally, the equation for the Electric field is mathematically given as

E=v/d

Where

v={180-(-5)}v

v=185v

Therefore

E=185/8.6*10^{-3}

E=21.5*10^3v/m

In conclusion, Electric field

E=21.5*10^3v/m