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anygoal [31]
3 years ago
9

What is the frequency of radiation emitted by a photon of light if the energy released during its transition to ground state is

3.611 × 10-15 joules? (Planck's constant is 6.626 × 10-34 joule seconds) 5.50 × 10-11 Hz 2.39 × 10-50 Hz 5.45 × 1018 Hz 1.633 × 1027 Hz
Physics
1 answer:
miskamm [114]3 years ago
3 0
If i was feeling harsh today, I'd say the answer to your question is impossible to obtain due to the fact that photons do not emit radiation, photons ARE the radiation emitted. Though for the sake of it, here is the method...

<u>The simple method:
</u>

E=hf

therefore f=e/h

f=(3.611x10^-15) / 6.63x10^-34)

Answer: 5.45x10^18
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Explanation:

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6 0
2 years ago
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Describe the phases of the moon along with a picture.​
Alex787 [66]

Answer:

The eight Moon phases:

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The phases of the Moon are the different ways the Moon looks from Earth over about a month. As the Moon orbits around the Earth, the half of the Moon that faces the Sun will be lit up. The different shapes of the lit portion of the Moon that can be seen from Earth are known as phases of the Moon.

<h2>The 8 phases (in order) are:</h2>
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Explanation:

Hope it is helpful....

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3 years ago
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A 23 kg child is riding a 6.7 kg bike with a velocity of 4.4 m/s to the northwest.
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3 years ago
Dos cargas puntuales (q1 y q2 ) se atraen inicialmente entre sí con una fuerza de 550 N , si la separación entre ellas es de 90
antiseptic1488 [7]

Answer:

q2 = 9.02*10^{-4}C

Explanation:

To find the value of the other charge you use the Coulomb's law:

F=k\frac{q_1q_2}{r^2}

k: Coulomb's constant = 8.98*10^{9}Nm^2/C^2

q1: charge 1 = 55*10^{-6}C

r: distance between charges = 90cm = 0.9m

F: electric force = 550N

By doing q2 the subject of the formula and replacing you obtain:

q_2=\frac{r^2F}{kq_1}=\frac{(0.9m)^2(550N)}{(8.98*10^9Nm^2/C^2)(55*10^{-6}C)}=9.02*10^{-4}C

hence, the value of the other charge q2 is 9.02*10^{-4}C

6 0
3 years ago
An armada of spaceships that is 1.60 ly long (in its rest frame) moves with speed 0.800c relative to the ground station in frame
miskamm [114]

Answer:

(a). The trip take a time is 3.61 year.

(b). The trip take a time is 3.96 year.

(c). The trip take a time is 8.73 year.

Explanation:

Given that,

Length = 1.60 ly

Speed of spaceships= 0.800c

Speed of messenger = 0.910 c

(a). We need to calculate the velocity of armada

Using formula of velocity

v'=\dfrac{v-v_{m}}{1-\dfrac{vv_{m}}{c^2}}

Put the value into the formula

v'=\dfrac{0.800c-0.910c}{1-\dfrac{0.800\times0.910 c^2}{c^2}}

v'=\dfrac{-0.11c}{1-0.800\times0.910}

v'=-0.404c

We need to calculate the length

Using formula of length

l=l'\sqrt{1-\dfrac{v'^2}{c^2}

Put the value into the formula

l=1.60\sqrt{1-(-0.404)^2}

l=1.46\ ly

We need to calculate the length of the trip

Using formula of time

t=\dfrac{l}{|v'|}

Put the value into the formula

t=\dfrac{1.46}{0.404}

t=3.61\ year

(b). If the armada's rest frame

We need to calculate the length

Using formula of length

l=l'\sqrt{1-\dfrac{v'^2}{c^2}

Put the value into the formula

l=1.60\sqrt{1-(0)^2}

l=1.60\ ly

Using formula of time

t=\dfrac{l}{|v'|}

Put the value into the formula

t=\dfrac{1.60}{0.404}

t=3.96\ year

(c). If an observer in frame S

We need to calculate the length

Using formula of length

l=l'\sqrt{1-\dfrac{v'^2}{c^2}

Put the value into the formula

l=1.60\sqrt{1-(0.800)^2}

l=0.96\ ly

We need to calculate the time

t=\dfrac{0.96}{0.910-0.800}

t=8.73\ year

Hence, (a). The trip take a time is 3.61 year.

(b). The trip take a time is 3.96 year.

(c). The trip take a time is 8.73 year.

8 0
3 years ago
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