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3 years ago

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A 193nm-wavelength UV laser for eye surgery emits a 0.500mJ pulse. (a) How many photons does the light pulse contain?

2 answers:

Law Incorporation [45]3 years ago

6 0

Answer:

The number of photons is 4.8x10^14

Explanation:

The frequency of wave is equal to:

$f=\frac{c}{l}$

where c is the speed of light, l is the wavelength of wave. Replacing values:

$f=\frac{3x10^{8} }{193x10^{9} } =1.5x10^{15} Hz$

The energy of the proton is:

$E=hf$

where h is the Planck´s constant. Replacing

$E=6.626x10^{-34}*1.5x10^{5}=1.03x10^{-18} J$

The number of photons is:

$n=\frac{E1}{E}$

where E1 is the energy of photon. Replacing:

$n=\frac{0.5x10^{-3} }{1.03x10^{-18} }=4.8x10^{14}$

Marrrta [24]3 years ago

4 0

Using the equation E = hc/λ we can find out how much energy a single photon of wavelength 193 nm has. E = Planck Constant * Speed of Light/193 nm

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The angle of incidence of another red ray is 65º. The refractive index of the glass of block

Mamont248 [21]

Answer:

34º

Explanation:

The computation of the angle of refraction in the glass for this ray is shown below:

Given that

the angle of incidence of another red ray is 65º

And, n_1 = 1

n_2 = 1.62

According to shell law

n_1 sin $\theta$ _1 = n_2 sin $\theta$ _r

sin $\theta$ _r = 1 sin 65÷ 1.62

= 0.559

$\theta$ _r = sin^-1 (0.559)

= 34º

6 0

2 years ago

A double-slit interference pattern is created by two narrow slits spaced 0.21 mm apart. The distance between the first and the f

olga nikolaevna [1]

Answer:

$\lambda =533.6 nm$

Explanation:

the slits spacing, d = 0.21 mm

distance of screen, D = 61 cm

The condition for minima is given as

$dsin(\theta) = \left ( n+\frac{1}{2} \right )\lambda$

So, first minima, n = 0

$dsin(\theta_1) = \frac{1}{2}\lambda$

fifth minima, n = 4

$dsin(\theta_5) = \frac{9}{2}\lambda$

$d(sin(\theta_5) -sin(\theta_1))= 4\lambda$

For small angle

$d(tan(\theta_5) -tan(\theta_1))= 4\lambda$

From the figure:

$d(\frac{y_5}{D}-\frac{y_1}{D})= 4\lambda$

$\frac{d}{D} (y_5-y_1) = 4\lambda$

$\lambda = \frac{d}{4D} (y_5-y_1)$

$\lambda = \frac{0.021}{4(61)} (6.2 \times 10^{-3})$

$\lambda =533.6 nm$

4 0

2 years ago

If a proton and an electron are released when they are 4.00 x 10^-10 m apart, find the initial acceleration of each of them.

SVEN [57.7K]

Answer:

For proton

$a=8.8\times 10^{17}\ m/s^2$

For electron

$a=1.5\times 10^{21}\ m/s^2$

Explanation:

We know that

The mass of electron

$m=9.1\times 10^{-31}\ kg$

The mass of proton

$m=1.67\times 10^{-27}\ kg$

Charge on electron and proton

q₁=q₂=q

$q=1.6\times 10^{-19}\ C$

Electrostatics force

$F=K\dfrac{q_1q_2}{r^2}$

Now by putting the values

$F=9\times 10^9\times \dfrac{1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(4\times 10^{-10})^2}$

$F=1.44\times 10^{-9}\ N$

For proton

F = m a

a =F/m

$a=\dfrac{1.4\times 10^{-9}}{1.67\times 10^{-27}}\ m/s^2$

$a=8.8\times 10^{17}\ m/s^2$

For electron

$a=\dfrac{1.4\times 10^{-9}}{9.1\times 10^{-31}}\ m/s^2$

$a=1.5\times 10^{21}\ m/s^2$

5 0

2 years ago

Read 2 more answers

Three small spheres, having masses m1 = 1 kg, m2 = 3 kg, and m3 = 4 kg, are held fixed on the x axis in deep space where the eff

Vlad [161]

Answer: $F_{net} = 0.7411 N$ towards the mass $m_{3}$

Explanation: We know that the gravitational force is a long range force which is always attractive in nature.

Given that:

mass $m_{1} = 1 kg$

mass $m_{2} = 3kg$

mass $m_{3} = 4kg$

The masses are positioned on X-axis at the following points:

Position of mass $m_{1}$ $x_{1} = 0$

Position of mass $m_{2}$ $x_{2} = 3$

Position of mass $m_{3}$ $x_{3} = 6$

Mathematically:

<em>Gravitational force on mass </em> $m_{2}$ <em> due to mass </em> $m_{1}$ <em> is given by </em>

$F_{21} = G \frac{m_{1}.m_{2}}{(r_{21})^2}$ ...................(1)

where: $(r_{21})^2$ = the radial distance between masses $m_{2}$ & $m_{1}$ =3

Similarly, g<em>ravitational force on mass </em> $m_{2}$ <em> due to mass </em> $m_{3}$ <em> is given by </em>

$F_{23} = G \frac{m_{3}.m_{2}}{(r_{23})^2}$ ............................(2)

where: $(r_{23})^2$ = the radial distance between masses $m_{2}$ & $m_{3}$ =3

Now, put the respective values in the above equations.

$F_{21} = 6.67 \times 10^{-11 }\times \frac{1\times 3}{3^2}$

$F_{21} = 2.2233\times 10^{-11} N$

Again,

$F_{23} = 6.67 \times 10^{-11 }\times \frac{1\times 4}{3^2}$

$F_{23} = 2.9644\times 10^{-11} N$

∵Mass $m_{2}$ is in the middle of the masses $m_{3}$ & $m_{1}$ therefore the forces $F_{23}$ & $F_{21}$ will attract them in radially opposite direction.

∴ $F_{net} = F_{23} -F_{21} \\\\F_{net} = 2.9644-2.2233\\\\F_{net} = 0.7411 N$ towards the mass $m_{3}$

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2 years ago

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What is the new kinetic energy of the 1900 kg ship on the right moving at 4 m/s?

Elis [28]

Answer:

soi nuevo

Explanation:me regalan puntos

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2 years ago

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