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3 years ago

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Light from a single laser is directed through two slits that are separated by a small distance. On the other side of the slits i

s an interference pattern of alternating dark(black) and bright bands. That dark bands coincide with areas of?
A.harmonic resonance
B.harmonic dissonance
C.destructive interference
D.destructive interference

1 answer:

nekit [7.7K]3 years ago

3 0

The answer is destructive interference. You have this for both C and D. I suspect one of C or D is supposed to be constructive interference... But destructive interference is the answer

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How much work does the electric field do in moving a proton from a point with a potential of +125 v to a point where it is -55 v

777dan777 [17]

The work W done by the electric field in moving the proton is equal to the difference in electric potential energy of the proton between its initial location and its final location, therefore:
$W= qV_i - qV_f$
where q is the charge of the proton, $q=1 e = 1.6\cdot 10^{-19}C$ , with $e$ being the elementary charge, and $V_i = +125 V$ and $V_f = -55 V$ are the initial and final voltage.

Substituting, we get (in electronvolts):
$W=e(125 V-(-55 V))=180 eV$
and in Joule:
$W=(1.6 \cdot 10^{-19})(125 V-(-55V))=2.88 \cdot 10^{-17}J$

5 0

3 years ago

A regular polygon has angkes of size 150° each.how many side has the polygon

emmasim [6.3K]

Answer: 12

Explanation:

ıf each interior is 150 degrees, then each exterior angle is 180–150 or 30 degrees. Hence the polygon has 360/30 = 12 sides

5 0

2 years ago

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

A passenger train left station A at 6:00 p.m. Moving with the average speed 45 mph, it arrived at station B at 10:00 p.m. A tran

marishachu [46]

<h2>Average speed of transit train is 60 mph</h2>

Explanation:

Average speed of passenger train = 45 mph

Time taken from station A to station B for passenger train = 10:00 - 6:00 = 4 hours

Distance between station A to station B = 45 x 4 = 180 miles.

Time taken from station A to station B for transit train = 4 - 1 = 3 hours

Distance between station A to station B = Average speed of transit train x Time taken from station A to station B for transit train

180 = Average speed of transit train x 3

Average speed of transit train = 60 mph

Average speed of transit train is 60 mph

8 0

3 years ago

A typical wall outlet voltage is 120 volts in the United States. Personal MP3 players require much smaller voltages, typically 2

butalik [34]

Answer:

Number of turns in secondary will be 7

Explanation:

We have given primary voltage $V_p=120volt$

Number of turns in the primary is $N_p=3575$

Secondary voltage is given $V_s=235mV=0.235volt$

We have to find the number of turns in secondary

We know that $\frac{N_p}{N_s}=\frac{V_p}{V_s}$

So $\frac{3575}{N_s}=\frac{120}{0.235}$

$N_s=6.60$

As the number of turns can not be in decimal so number of turns will be 7

6 0

3 years ago