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Ksivusya [100]
3 years ago
11

The intensity of electromagnetic radiation from the sun reaching the earth's upper atmosphere is 1.37kW/m2kW/m2. Part A Assuming

an average wavelength of 680 nmnm for this radiation, find the number of photons per second that strike a 1.50 m2m2 solar panel directly facing the sun on an orbiting satellite.
Physics
1 answer:
Juli2301 [7.4K]3 years ago
8 0

Answer:

#_photon = 7  10²¹ photons

Explanation:

Let's look for the power that affects the panel of area of ​​1.5 m2

           I = P / A

           P = I A

           P = 1.37 10³  1.5

           P = 2,055 10³ W

           P = E / t

       

If we use t = 1 s

           E = P t

           E = 2,055 10³ J

This is the power that the panel receives, let's look for the energy of a photon

            E = h f

            c = λ f

            f = c /λ

            E = h c /λ

Let's calculate

            E₀ = 6.63 10⁻³⁴  3 10⁸/680 10⁻⁹

            E₀ = 2.925 10⁻¹⁹ J

In one second the total energy is the number of photons for the energy of each one

             E = #_photon  E₀

             #_photon = E / E₀

             #_photon = 2,055 10³ / 2,925 10⁻¹⁹

            #_photon = 7  10²¹ photons

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vovikov84 [41]
C- Stratus clouds are the rainy clouds.
Hope this helps!
8 0
3 years ago
Un movil viaja a 40km/h y comienza a reducir su velocidad a partir del instante t=0. Al cabo de 6 segundo se detiene completamen
aleksklad [387]

Answer:

1,85 m / s²

Explanation:

De la pregunta anterior, se obtuvieron los siguientes datos:

Velocidad inicial (u) = 40 km / h

Hora inicial (t₁) = 0

Tiempo final (t₂) = 6 s

Velocidad final (v) = 0

Aceleración (a) =?

A continuación, convertiremos 40 km / ha m / s. Esto se puede obtener de la siguiente manera:

1 km / h = 0,2778 m / s

Por lo tanto,

40 km / h = 40 km / h × 0,2778 m / s / 1 km / h

40 km / h = 11,11 m / s

Por tanto, 40 km / h equivalen a 11,11 m / s.

Finalmente, determinaremos la aceleración del móvil durante el período en el que desaceleró. Esto se puede obtener de la siguiente manera:

Velocidad inicial (u) = 11,11 m / s

Hora inicial (t₁) = 0

Tiempo final (t₂) = 6 s

Velocidad final (v) = 0

Aceleración (a) =?

a = (v - u) / (t₂ - t₁)

a = (0 - 11,11) / (6 - 0)

a = - 11,11 / 6

a = –1,85 m / s²

Por tanto, la aceleración del móvil durante el período en el que se ralentizó es de –1,85 m / s²

6 0
3 years ago
A negative charge of - 8.0 x 10^-6 C exerts an attractive force of 12 N on a second charge that is
Umnica [9.8K]

Answer:

<h2>Magnitude of the second charge is -4.17*10^{-7}C</h2>

Explanation:

According to columbs law;

F = kq1q2/r^{2}

F is the attractive or repulsive force between the charges = 12N

q1 and q2 are the charges

let q1 = - 8.0 x 10^-6 C

q2=?

r is the distance between the charges = 0.050m

k is the coulumbs constant =9*10⁹ kg⋅m³⋅s⁻⁴⋅A⁻²

On substituting the given values

12 = 9*10⁹*( - 8.0 x 10^-6)q2/0.050²

Cross multiplying

0.03=9*10^{9}*  -8.0*10^{-6} q2\\0.03 = -72*10^{3} q2\\q2 = \frac{0.03}{ -72*10^{3}} \\q2 = -4.17*10^{-7}C

6 0
4 years ago
Read 2 more answers
Feng and Isaac are riding on a merry-ground. Feng rides on a horse at the outer rim of the circular platform, twice as far from
kogti [31]

Answer: The question is incomplete or missing details. here is the remaining part of the question ;

1. impossible to determine

2. half of Isaac’s

3. the same as Isaac’s

4. twice Isaac’s

The angular speed of feng will be the same as that of Isaac. Hence the answer is option 3

Explanation:

Since we have been told that both feng and isaac are riding on a merry go round i.e in a circular motion, irrespective of how fast one ride above the other, the angular speed will be constant since they are riding on a merry go round, as such both feng and isaac will maintain equal angular speed, hence the angular speed of feng will be the same as that of Isaac.

4 0
4 years ago
Can you answer the question
Dominik [7]
Can u show the whole question plz
3 0
3 years ago
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