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valina [46]
2 years ago
9

The power of a red laser (λ = 630 nm) is 3.25 watts (abbreviated w, where 1 w = 1 j/s). how many photons per second does the las

er emit?
Physics
1 answer:
lisabon 2012 [21]2 years ago
5 0
The energy delivered by the laser in 1 second isE_t=Pt = (3.25 W)(1.0 s)=3.25 J
In order to find how many photons correspond to this energy, we must calculate the energy of a single photon.
Calling h the Planck constant, c the speed of light and \lambda=630 nm=630 \cdot 10^{-9}m the wavelength of the light, the energy of a single photon is given by
E=h \frac{c}{\lambda}=(6.6 \cdot 10^{-34} Js) \frac{3 \cdot 10^8 m/s}{630 \cdot 10^{-9}m}= 3.1 \cdot 10^{-19} J

So, the number of photons emitted by the laser in 1 second is equal to the total energy delivered by the laser divided by the energy of a single photon:
N= \frac{E_t}{E}= \frac{3.25 J}{3.1 \cdot 10^{-19} J} =1.0 \cdot 10^{19} photons
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Calculate the amount of energy produced in a nuclear reaction in which the mass defect is 0.187456 amu.
luda_lava [24]
For nuclear reactions, we determine the energy dissipated from the process from the Theory of relativity wherein energy is equal to the mass defect times the speed of light. We calculate as follows:

E = mc^2 = 0.187456 (3x10^8)^2 = 1.687x10^16 J

Hope this answers the question.
8 0
2 years ago
Read 2 more answers
Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin
Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

\mu_o=4\pi \times 10^{-7}\ T-m/A

Resistivity, \rho=1.72\times 10^{-8}\ \Omega-m

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

R=\rho \dfrac{l}{A}

R=\rho \dfrac{l}{\pi r^2}

r=\sqrt{\dfrac{\rho l}{R\pi}}

r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}

r = 0.00199 m

or

r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m

The magnetic field on a current carrying wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

I=\dfrac{2\pi rB}{\mu_o}

I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

4 0
1 year ago
Two satellites revolve around the Earth. Satellite A has mass m and has an orbit of radius r. Satellite B has mass 6m and an orb
melomori [17]

Answer:

aaaaa

Explanation:

M = Mass of the Earth

m = Mass of satellite

r = Radius of satellite

G = Gravitational constant

F=G\frac{Mm}{r^2}

F=G\frac{M6m}{r_b^2}

G\frac{Mm}{r^2}=G\frac{M6m}{r_b^2}\\\Rightarrow \frac{1}{r^2}=\frac{6}{r_b^2}\\\Rightarrow \frac{r_b^2}{r^2}=6\\\Rightarrow \frac{r_b}{r}=\sqrt{6}\\\Rightarrow r_b=2.44948r

r_b=2.44948r

8 0
2 years ago
evaluate the numerical value of the vertical velocity of the car at time t=0.25 s using the expression from part d, where y0=0.7
likoan [24]

Given :

Displacement , y = 0.75 m .

Angular acceleration , \alpha=0.95\ s^{-2} .

Initial angular velocity , \omega_o=6.3\ s^{-1} .

To Find :

The value of vertical velocity after time t = 0.25 s .

Solution :

By equation of circular motion is given by :

\omega=\omega_o+\alpha t

Putting all given values we get :

\omega=6.3+0.95\times 0.25\\\\\omega= $$6.5375\ s^{-1}

Now , vertical velocity is given by :

v=y\omega\\\\v=0.75\times 6.5375\ m/s\\\\v=4.90\ m/s

Therefore , the numerical value of the vertical velocity of the car at time t=0.25 s is 4.90 m/s .

Hence , this is the required solution .

8 0
2 years ago
A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound
NemiM [27]

Answer:

f3 = 102 Hz

Explanation:

To find the frequency of the sound produced by the pipe you use the following formula:

f_n=\frac{nv_s}{4L}

n: number of the harmonic = 3

vs: speed of sound = 340 m/s

L: length of the pipe = 2.5 m

You replace the values of n, L and vs in order to calculate the frequency:

f_{3}=\frac{(3)(340m/s)}{4(2.5m)}=102\ Hz

hence, the frequency of the third overtone is 102 Hz

8 0
2 years ago
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