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2 years ago

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

Physics

1 answer:

Reptile [31]2 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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2 years ago

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A busy chipmunk runs back and forth along a straight line of acorns that has been set out between his burrow and a nearby tree.

saveliy_v [14]

Answer: a = 1.32m/s2

Therefore, the average acceleration is 1.32m/s2

Explanation:

Acceleration is the rate of change in the velocity per time

a = change in velocity/time

a = ∆v/t

average acceleration a = (v2 -v1)/t. ....1

Given;

Final velocity v2 = 1.63m/s

Initial velocity v1 = -1.15ms

time taken t = 2.11s

Substituting into eqn 1

a = [1.63 - (-1.15)]/2.11

a = (1.63+1.15)/2.11

a = 2.78/2.11

a = 1.32m/s2

Therefore, the average acceleration is 1.32m/s2

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3 years ago

A 2.0-kg mass is projected from the edge of the top of a 20-m tall building with a velocity of 24 m/s at some unknown angle abov

Digiron [165]

Ox:vₓ=v₀

x=v₀t

Oy:y=h-gt²/2

|vy|=gt

tgα=|vy|/vₓ=gt/v₀=>t=v₀tgα/g

y=0=>h=gt²/2=v₀²tg²α/2g=>tgα=√(2gh/v₀²)=√(2*10*20/24²)=√(400/576)=0.83=>α=tg⁻¹0.83=39°

cosα=vₓ/v=v₀/v=>v=v₀/cosα=24/cos39°=24/0,77=31.16 m/s

Ec=mv²/2=2*31.16²/2=971.47 J=>Ec≈0.97 kJ

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2 years ago

A car travels a distance of 540km in 6 hours. What speed did it travel at?

maria [59]

Speed v = distance travelled / time taken

v = d / t

v = 540 / 60h

v = 9 km /h

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2 years ago

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Using equations, determine the temperature, pressure and density of the air for a aircraft flying at 19.5 km. Is this aircraft s

Viefleur [7K]

Answer:

a) - 72.5°c

b) pressure = 3625.13 Pa

c) density = 0.063 kg/m^3

d) it is a subsonic aircraft

Explanation:

a) Determine Temperature

Temperature at 19.5 km ( 19500 m )

T = -131 + ( 0.003 * altitude in meters )

= -131 + ( 0.003 * 19500 ) = - 72.5°c

b) Determine pressure and density at 19.5 km altitude

Given :

Po (atmospheric pressure at sea level ) = 101kpa

R ( gas constant of air ) = 0.287 KJ/Kgk

T = -72.5°c ≈ 200.5 k

pressure = 3625.13 Pa

hence density = 0.063 kg/m^3

attached below is the remaining part of the solution

C) determine if the aircraft is subsonic or super sonic

Velocity ( v ) = $\sqrt{CRT}$ = $\sqrt{1.4*287*200.5 }$ = 283.8 m/s

hence it is a subsonic aircraft

4 0

2 years ago