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3 years ago

How many photons strike the cd surface during the playing of a cd 69 minutes in length?

Physics

1 answer:

Reptile [31]3 years ago

8 0

Answer:

Photons strike the CD $n=1.624 x10^{18}$

Explanation:

First of all, it is necessary to calculate the energy produced by a semiconductor laser for 69 minutes:

$69minutes*\frac{60s}{1 minute}=4140s$

$E=0.1x10^{-3}w*4140s=0.414J$

Next, it is possible to calculate the number of photons striking the CD surface during this time:

$E=\frac{n*h*c}{l}$

$l=780x10^{-9}m$ long in CD

$c=3x10^8 m/s$ light velocity

$h=6.626x10^{-34} J*s$

Solve to n'

$n=\frac{E*l}{c*h}=\frac{0.414 J*780x10^{-9}m}{3x10^8m/s*6.626x10^{-34}}$

$n=1.624 x10^{18}$

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vesna_86 [32]

Answer:

Explanation:

by adding all the numbers

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3 years ago

As a bat flies toward a wall at a speed of 6.0 m/s, the bat emits an ultrasonic sound wave with frequency 30.0 kHz. What frequen

Leto [7]

Answer:

$f_b = 29.98 Hz$

Explanation:

speed of bat = 6 m/s

sound wave frequency emitted by bat = 30.0 kHz

as we know,

speed of sound (c)= 343 m/s

$f_w = f(\dfrac{c+v_r}{c+v_s})$

$f_w = 30(\dfrac{343+0}{343+6})$

$f_w = 29.48 Hz$

now frequency received by bat is equal to

$f_b = f(\dfrac{c+v_r}{c+v_s})$

$f_b = 29.48(\dfrac{343+6}{343+0})$

$f_b = 29.98 Hz$

hence the frequency hear by bat will be 29.98 Hz

7 0

3 years ago

Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular alumin

Zarrin [17]

Answer:

$\rm 9.186\times 10^{-7}\ C.$

Explanation:

<u>Given:</u>

Diameter of the plates of the capacitor, D = 21 cm = 0.21 m.
Distance of separation between the plates, d = 1.0 cm = 0.01 m.
Minimum value of electric field that produces spark, $\rm E=3\times 10^6\ N/C.$

When the dimensions of the plate of the capacitor is comparatively much larger than the distance of separation between the plates, then, according to the Gauss' law of electrostatics, the value of the electric field strength in the region between the plates of the capacitor is given by

$\rm E=\dfrac{\sigma}{\epsilon_o}.$

where,

$\rm \sigma$ = surface charge density of the plate of the capacitor = $\dfrac qA$ .

$\rm q$ = magnitude of the charge on each of the plate.
$\rm A$ = surface area of each of the plate = $\rm \pi \times (Radius)^2=\pi \times\left ( \dfrac{D}{2}\right )^2= \pi \times \left ( \dfrac{0.21}{2}\right )^2=3.46\times 10^{-2}\ m^2.$
$\epsilon_o$ = electrical permittivity of free space, having value = $8.85\times 10^{-12}\rm \ C^2N^{-1}m^{-2}.$

For the minimum value of electric field that produces spark,

$\rm E = \dfrac{q}{A\epsilon_o}\\\Rightarrow q = E\ A\epsilon_o\\=3\times 10^6\times 3.46\times 10^{-2}\times 8.85\times 10^{-12}\\=9.186\times 10^{-7}\ C.$