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valina [46]
3 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]3 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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Which biome is most likely to have animals that hibernate, or sleep for long periods during the winter to avoid the cold?
TiliK225 [7]
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4 years ago
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A 286-kg motorcycle is accelerating up along a ramp that is inclined 31.6° above the horizontal. The propulsion force pushing th
kkurt [141]

Answer:

The acceleration motorcycle

a = 5.13 m / s²

Explanation:

Now to determine the acceleration of the motorcycle

Use the force to analysis motion

∑ F = m * a

∑ F = E - D - m*g * sin ( β ) = m * a

E = 3168 N

D = 230 N

β = 31.6 °

3168 N - 230 N - 286 kg * 9.8 m / s² * sin ( 31.6° ) = 286 kg * a

Now solve to a'

a = [ 3168 N - 230 N - 286 kg * 9.8  m / s² * sin ( 31.6° ) ] / (286 kg)

a = 5.13 m / s²

7 0
3 years ago
What is the SI unit used to mesasure force
lukranit [14]

Answer:

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Hope this helps

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6 0
3 years ago
A 525 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satel
Dafna1 [17]

Answer:

1.69\cdot 10^{10}J

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

E=-G\frac{mM}{2r}

where

G is the gravitational constant

m = 525 kg is the mass of the satellite

M=5.98\cdot 10^{24}kg is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m

So the initial total energy is

E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J

When the satellite hits the ground, it is now on Earth's surface, so

r=R=6370 km=6.37\cdot 10^6 m

so its gravitational potential energy is

U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J

And since it hits the ground with speed

v=1.90 km/s = 1900 m/s

it also has kinetic energy:

K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J

So the total energy when the satellite hits the ground is

E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:

\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J

8 0
3 years ago
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