PLEASE HELP FAST The object distance for a convex lens is 15.0 cm, and the image distance is 5.0 cm. The height of the object is 9.0 cm. What is the height of the image?

Answer:

A.   number of neutrons of Magnesium Mg = 13

B.   The average mass of Mg = 22.29 amu

C.   the magnesium composition on Mars is not the same as that on Earth.

Explanation:

Isotopes are atoms with the same atomic number but different mass number. This is due to the difference in mass of the neutrons.

The atomic number of Magnesium Mg = 12

The atomic number of an element is the number of protons present in the atomic nucleus of the element

i.e Atomic number = number of protons = 12

The mass number of an element is the sum of the protons and neutrons in the atomic nucleus of the element.

Mass number = number of protons + number of neutrons

Given that the mass number of Mg = 25

Then;

25 = 12 + number of neutrons

25 - 12 = number of neutrons

13 = number of neutrons

number of neutrons of Magnesium Mg = 13

B. What is the average atomic mass of magnesium in these rocks?

The average atomic mass of an element which exhibit isotopy is the average mass of its various isotopes as they occur naturally in any quantity of the element.

Therefore the average atomic mass of magnesium can be calculated as:

= [tex]\mathtt{\dfrac{(23.9872 \times 79.70) + ( 24.9886 \times 10.13) + (25.9846 \times 10.17) }{79.7 + 10.13 +10.17}}[/tex]

= [tex]\mathtt{\dfrac{(1911.77984) + ( 53.134518) + (264.263382) }{100}}[/tex]

= [tex]\mathtt{\dfrac{2229.17774 }{100}}[/tex]

The average mass of Mg = 22.29 amu

C. Is the magnesium composition on Mars the same as that on Earth? Explain.

The average atomic weight of magnesium on Earth is said to be 24.305 amu while that of Mars is 22.29 amu.

There difference in the average atomic weight result into difference in their composition. Therefore,the magnesium composition on Mars is not the same as that on Earth.

Answer:

p-aminoazobenzene is formed

Explanation:

The reaction of benzene diazonium chloride and aniline takes place in a basic medium and leads to the formation of an azo compound which is also a dye. The terminal diazonium nitrogen of the benzene diazonium ion is coupled to the aniline at the para-position. The product of the reaction, p-aminoazobenzene is a yellow dye.

Benzene diazonium chloride is prepared by diazotization of aniline in the presence of hydrochloric acid. The full reaction of aniline and benzene diazonium chloride​ is shown in the image attached to this answer.

Answer:

The blade undergoes 40 revolutions, so neither of the given options is correct!

Explanation:

The revolutions can be found using the following equation:

[tex]\theta_{f} = \theta_{i} + \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]

Where:

α is the angular acceleration

t is the time = 2.5 s

[tex]\omega_{i}[/tex] is the initial angular velocity = 1500 rev/min                

First, we need to find the angular acceleration:

[tex] \alpha = \frac{\omega_{f} - \omega_{i}}{t} = \frac{400 rev/min*2\pi rad*1 min/60 s - 1500 rev/min *2\pi rad*1 min/60 s}{2.5 s} = -46.08 rad/s^{2} [/tex]

Now, the revolutions that the blade undergo are:

[tex]\theta_{f} - \theta_{i} = \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]

[tex]\Delta \theta = 1500 rev/min *2\pi rad*1 min/60 s*2.5 s - \frac{1}{2}*(46.08 rad/s^{2})*(2.5)^{2} = 248.7 rad = 39.9 rev[/tex]        

Therefore, the blade undergoes 40 revolutions, so neither of the given options is correct!

I hope it helps you!                              

Answer:

C

They allow light to pass only if their directions of polarizations are exactly 90° apart.

Answer:

The percentage is  [tex]k = 12.5 \%[/tex]

Explanation:

From the question we are told that

    The  axis is is  at  [tex]\theta = 45 ^o[/tex]

Generally the of intensity light emerging from the first polarizer is mathematically represented as

                [tex]I_{1} = \frac{I_o}{ 2}[/tex]

Where  [tex]I_o[/tex] is the intensity of unpolarized light

       Now the light emerging from the second polarizer is mathematically represented as

         [tex]I_2 = I_ 1 * cos ^2(\theta )[/tex]

         [tex]I_2 = \frac{I_o}{2} * cos ^2(45 )[/tex]

        [tex]I_2 = \frac{I_o}{2} * \frac{1}{2} = \frac{I_o}{4}[/tex]

  Now the light emerging from the third  polarizer is mathematically represented as      

       [tex]I_3 = I_ 2 * cos ^2(\theta )[/tex]

       [tex]I_3 = \frac{I_o}{4} * cos ^2(45 )[/tex]

      [tex]I_3 = \frac{I_o}{8}[/tex]

Now the percentage of the intensity of light that emerged with respect to the intensity of  the unpolarized light is

      [tex]k = \frac{\frac{I_o}{8} }{I_o } * 100[/tex]

     [tex]k = 12.5 \%[/tex]

The percentage of light that gets through the three successive Polaroid filters is; 12.5%

We are given;

Angle of transmission axis; θ = 45°

Formula for intensity of light from first polarizer is;

I₁ = ¹/₂I₀

Formula for intensity of light from second polarizer is;

I₂ = I₁cos²θ

Formula for intensity of light from third polarizer is;

I₃ = I₂cos²(90 - θ)

Combining the 3 equations;

Put ¹/₂I₀ for I₁ in second formula to get;

I₂ = ¹/₂I₀cos²θ

Put ¹/₂I₀cos²θ for I₂ in third formula to get;

I₃ = ¹/₂I₀cos²θ*cos²(90 - θ)

Plugging in 45° for θ gives;

I₃ = ¹/₂I₀cos²45*cos²(90 - 45)

⇒ I₃ = ¹/₂I₀cos²45*cos²45

⇒ I₃ = ¹/₂I₀cos⁴45

Now, cos 45 in surd form is 1/√2. Thus;

I₃ = ¹/₂I₀(1/√2)⁴

I₃ = ¹/₂I₀(¹/₄)

I₃ = ¹/₈I₀

I₃/I₀ = ¹/₈

I₃/I₀ = 0.125

In percentage form, we have;

I₃/I₀ = 12.5%

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Answer:

Option (A) : 1m

Explanation:

According to Hooke's law:

F (spring elastic force) =

k( spring const.) * x(displacement)

Case-1

2 N = k * 0.4m

k = 5

Case- 2

5 N = 5 * x

x ( displacement) = 1 m

The displacement of the spring if a 5-N force was applied is equal to 1m. Therefore, option (1) is correct.

The strain and stress are proportional to each other, and this is called Hooke’s Law. Hooke’s law states that the strain is proportional to the stress applied within the elastic limit of the material.

When the materials are stretched, the atoms or molecules deform and when the stress is removed, they will return to their original state.

The mathematical equation for Hooke's law is as follows:

F = –kx

where F is the force, x is displacement, and k is the spring constant in N/m.

Given, F = 2N and x = 0.4m

F = -kx

2 N = - k (0.4m)

k = 5 N/m where the negative sign is omitted.

Now, the spring constant of the spring, k = 5 N/m and F = 5N

F = -kx

5 N = - (5 N/m)(x)

x = - 1m

Therefore, the displacement of the spring is 1 m.

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Answer:

30cm

Explanation:

assume that the eyes are substantially above the water so that sin(theta) is approximately theta.

( small angle approximation).

The point at which a ray leaving the fish hits the surface of the water is x to the side of the centreline and the depth of the water is d

x/d = sin( angle of incidence)

if the apparent depth of the water is h then

x/h = sin( angle of refraction)

and applying snells law

1 sin ( theta air) = 1.33 sin( theta water)

1 * x/h = 1.33 * x/d

d/h = 1.33

or h/d = 1/1.33

h/39 = 1.33

h = 39 /1.33 so that is the apparent depth of the stream assuming:-

1. Your eyes are almost directly overhead

and

2. your eyes are a significant distance above the surface of the water.

x/d = 1.33 x/h

h/d =39/1.3

= 30cm

Answer:

E = 1336.71875 J

Explanation:

We are given;. Capacitance of Capacitor 1; C1 = 10.7 µF

Capacitor 2; C2 = 23.0 µF

Capacitor 3; C3 = 29.3 µF

Supply voltage;V = 125 V

Formula for capacitance in series is;

Capacitors in series circuit: C(eq) = 1/C(1) +1/C(2) +1/C(3) .......

Thus, equivalent capacitance is;

C(eq) = (1/10.7) + (1/23) + (1/29.3) = 0.1711 µF = 0.1711 × 10^(6) F

Now, the formula for maximum energy stored is;

E = ½ × C(eq) × V²

E = ½ × 0.1711 × 10^(-6) × 125²

E = 1336.71875 J

Answer:

photon density = 1.0 × [tex]10^{16}[/tex] photon/m³

Explanation:

given data

frequency f = 100 mhz = 100 × [tex]10^{6}[/tex] Hz

we consider here intensity I = 0.2 W/m²

solution

we take here plank constant is h i.e = 6.626 × [tex]10^{-34}[/tex] s

and take energy density is E

so here

E × C = I  

E = [tex]\frac{I}{C}[/tex]   ................1

here C = 3 × [tex]10^{8}[/tex] m/s

so photon density is

photon density = [tex]\frac{I}{C} \times \frac{1}{f \times h}[/tex]     ...............2

photon density = [tex]\frac{0.2}{3 \times 10^8} \times \frac{1}{100 \times 10^6 \times 6.626 \times 10^{-34} }[/tex]

photon density = 1.0 × [tex]10^{16}[/tex] photon/m³

Answer:

  λ= 5.4379 10⁻⁷ m = 543.79 nm

Explanation:

The phenomenon of diffraction is described by the expression for destructive diffraction is

           a sin θ = (m + 1/2) λ

           λ = a sin θ / (m + 1/2)

let's reduce the magnitudes to the SI system

           a = 15 um = 15 10⁻⁶ m

           m = 2

           θ = 5.2º

Let's calculate

          λ = 15 10⁻⁶ sin 5.2 / (2 +1/2)

         λ = 5.4379 10⁻⁷ m

Let's reduce to nm

          λ= 5.4379 10⁻⁷ m = 543.79 nm

Answer:

a)   I = I₀/2, b)  I = I₀/2 cos² θ

Explanation:

To answer these questions, let's analyze a little the way of working of a polarized

* When a non-polarized light hits a polarizer, the electric field that is not in the direction of the polarizer is absorbed, so the transmitted light is

          i = I₀ / 2

and is polarized in the direction of the polarizer

* when a polarized light reaches the analyzer it must comply with Malus's law

          I = I₁ cos² θ

where the angle is between the polarized light and the analyzer.

With this, let's answer the questions

a) When a polarizer is placed in the non-polarized light path, half of it is absorbed and only the light that has polarization in the direction of the polarizer is transmitted with an intensity of

                  I = I₀/2

b) when a polarizer and an analyzer are fitted, the intensity of the light transmitted by the analyzer is

                I = I₀/2 cos² θ

where the final value depends on the angle between the polarizer and the analyzer.

Let's look at two extreme cases

θ = 0          I = Io / 2

θ = 90º      I = 0

The bullet's vertical velocity at time [tex]t[/tex] is

[tex]v=1400\dfrac{\rm m}{\rm s}-gt[/tex]

where [tex]g=9.80\frac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity.

At its highest point, the bullet's vertical velocity is 0, which happens

[tex]0=1400\dfrac{\rm m}{\rm s}-gt\implies t=\dfrac{1400\frac{\rm m}{\rm s}}g\approx\boxed{142.857\,\mathrm s}[/tex]

(or about 140 s, if you're keeping track of significant figures) after being fired.

Answer:

there are over 100 billion stars in our galaxy.

Answer:

Explanation:

Laser angle with water surface is given by: Tan α = 1/2.0= 0.5/

α = 26.56°

Laser angle with Normal = 90 - 26.56 = 63.44 °

Assuming a red laser, refractive index in water is 1.331.

Angle of refraction in water is given by:

Ref Ind = Sin i / Sin r

1.331 = Sin 63.44 / Sin r

Sin r = 0.8945 / 1.331 = 0.6721

Angle r = 42.22°

For the path in water:

Tan 42.22 = x / 3.2

x = 2.9m where x is the lateral displacement of the laser ince it hits the water

So the goggles are 2.0 + 2.9 = 4.9 m from edge of pool

Answer:

[tex]q = -461532.5 \ C[/tex]

Explanation:

From the question we are told that

     The  electric filed is  [tex]E = 102 \ N/C[/tex]  

Generally according to Gauss law

=>   [tex]E A = \frac{q}{\epsilon_o }[/tex]

Given that  the electric field is pointing downward  , the equation become

    [tex]- E A = \frac{q}{\epsilon_o }[/tex]

Here   [tex]q[/tex] is the excess charge on the surface of the earth

          [tex]A[/tex] is the surface  area of the of the earth which is mathematically represented as

     [tex]A = 4\pi r^2[/tex]

Where r is the radius of the earth which has a value [tex]r = 6.3781*10^6 m[/tex]

 substituting values

    [tex]A = 4 * 3.142 * (6.3781*10^6 \ m)^2[/tex]

    [tex]A =5.1128 *10^{14} \ m^2[/tex]

So

   [tex]q = -E * A * \epsilon _o[/tex]

Here [tex]\epsilon_o[/tex] s the permitivity of free space with value

          [tex]\epsilon_o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

substituting values

     [tex]q = -102 * 5.1128 *10^{14} * 8.85 *10^{-12}[/tex]

     [tex]q = -461532.5 \ C[/tex]

Answer:

[tex]f=1.13s^{-1}=1.13Hz[/tex]

Explanation:

Hello,

In this case, a frequency stands for a rate in which some action is done per unit of time. In this case, for the heartbeat, since 68 actions (heartbeats) occur in 1.0, the frequency turns out:

[tex]f=\frac{68}{1.0min}=68min^{-1}[/tex]

Or as most commonly used in Hz ([tex]s^{-1}[/tex]):

[tex]f=68\frac{1}{min} *\frac{1min}{60s}=1.13s^{-1}=1.13Hz[/tex]

Best regards.

Answer:

a) 360.7 J/s

b) 16.23 °C

c) 34.48 g

Explanation:

The mass of the person = 80 kg

The person is a perfect emitter, ε = 1

surface area of the person = 2.5 m^2

a) If he emits radiation at 37 °C, [tex]T_{out}[/tex] = 37 + 273 = 310 K

and receives radiation at 13 °C, [tex]T_{in}[/tex] = 13 + 273 = 286 K

Rate of energy loss E = Aεσ([tex]T^{4} _{out}[/tex] - [tex]T^{4} _{in}[/tex] )

where σ = 5.67 x 10^-8 J/(s m^2 K^4)

substituting values, we have

E = 2.5 x 1 x 5.67 x 10^-8 x ([tex]310^{4}[/tex] - [tex]286^{4}[/tex]) = 360.7 J/s

b) If they have specific heat about equal to that of water = 1 Cal/kg-°C

but 1 Cal = 1 kcal = 10^3 cal

specific heat of person is therefore = 10^3 cal/kg-°C

heat loss = 360.7 J/s = 360.7 x 3600 = 1298520 J/hr

heat lost in 1 hour = 1 x 1298520 = 1298520 J

This heat lost = mcΔT

where ΔT is the temperature fall

m is the mass

c is the specific heat equivalent to that of water

the specific heat is then = 10^3 cal/kg-°C

equating, we have

1298520 = 80 x 10^3 x ΔT

1298520 = 80000ΔT

ΔT = 1298520/80000 = 16.23 °C

c) 1298520 J = 1298520/4184 = 310.35 Cal

density of fat = 9 Cal/g

gram of fat = 310.35/9 = 34.48 g

Answer:

Kindly check explanation

Explanation:

Length of race = 5km

Maximum speed = 45 yards

Converting from yards to kilometer :

1km = 1093.613 yards

x = 45 yards

(1093.613 * x) = 45

x = 45 / 1093.613

x = 0.0411480 km

Where x = maximum length for which he can maintain his maximum speed expressed in kilometers.

Therefore, with the available information, it can be concluded that Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.

Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.

Length of race = 5km

Maximum speed = 45 yards

Converting from yards to kilometer :

1km = 1093.613 yards

x = 45 yards

[tex](1093.613 \times x) = 45[/tex]

[tex]x = 45 \div 1093.613[/tex]

x = 0.0411480 km

here x represent maximum length for which he can maintain his maximum speed expressed in kilometers.

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Answer:

ΔE = GMm/24R

Explanation:

centripetal acceleration a = V^2 / R = 2T/mr

T= kinetic energy

m= mass of satellite, r= radius of earth

= gravitational acceleration = GM / r^2

Now, solving for the kinetic energy:

T = GMm / 2r = -1/2 U,

where U is the potential energy

So the total energy is:

E = T+U = -GMm / 2r

Now we want to find the energy difference as r goes from one orbital radius to another:

ΔE = GMm/2 (1/R_1 - 1/R_2)

So in this case, R_1 is 3R (planet's radius + orbital altitude) and R_2 is 4R

ΔE = GMm/2R (1/3 - 1/4)

ΔE = GMm/24R

Answer:

The  angular acceleration is  [tex]\alpha = 3.235 \ rad/s ^2[/tex]

Explanation:

From the question we are told that

    The moment of inertia is  [tex]I = 0.034\ kg \cdot m^2[/tex]

     The  net torque is  [tex]\tau = 0.11\ N \cdot m[/tex]

Generally the net torque is mathematically represented as

           [tex]\tau = I * \alpha[/tex]

Where [tex]\alpha[/tex] is the angular acceleration so  

        [tex]\alpha = \frac{\tau }{I}[/tex]

substituting values

         [tex]\alpha = \frac{0.1 1}{ 0.034}[/tex]

        [tex]\alpha = 3.235 \ rad/s ^2[/tex]

Answer:

0.61 m

Explanation:

The smallest observable length by the radar must be at least equal to or greater than the wavelength of the radar.

using the relationship

c = fλ

where

c is the speed of light in vacuum = 3 x 10^8 m/s

f is the frequency of the wave = 495 MHz = 4.95 x 10^8 Hz

λ is the wavelength = ?

λ = c/f = (3 x 10^8)/(4.95 x 10^8) = 0.61 m

answer to your question is 0.6m

Answer:

The magnification is  [tex]m = -2.15[/tex]

Explanation:

From the question we are told that

   The  radius is  [tex]r = 14.0 \ m[/tex]

    The  focal length eyepiece is  [tex]f_e = 3.25 \ m[/tex]

Generally the objective focal length is mathematically represented as

        [tex]f_o = \frac{r}{2}[/tex]

=>     [tex]f_o = \frac{14}{2}[/tex]

=>     [tex]f_o = 7 \ m[/tex]

The  magnification is mathematically represented as

      [tex]m = - \frac{f_o }{f_e }[/tex]

=>    [tex]m = - \frac{7 }{ 3.25 }[/tex]

=>   [tex]m = -2.15[/tex]

Explanation:

Answer:

[tex]\theta_1 = 0.400^o[/tex]

[tex]\theta_2 =0.378^o[/tex]

Explanation:

From the question we are told that

    The  number of slits per cm is  k =  [tex]161\ slits\ per\ cm = 161 \ slits\ per\ 0.01 m[/tex]

    The order of the maxima is  n =  1

    The wavelength are  [tex]\lambda_1 = 434 nm = 434 *10^{-9} \ m \ \ \ , \lambda_2 = 410nm = 410 *10^{-9} \ m[/tex]

The  spacing between the slit is mathematically represented as

           [tex]d = \frac{ 0.01}{k}[/tex]

=>       [tex]d = \frac{ 0.01}{161}[/tex]

=>         [tex]d = 6.211 *10^{-5} \ m[/tex]

Generally the condition for constructive interference is  

        [tex]n\lambda = d \ sin \theta[/tex]

At  [tex]\lambda_1[/tex]

      [tex]\theta _1 = sin^{-1} [ \frac{1 * 434 *10^{-9}}{6.211 *10^{-5}} ][/tex]

      [tex]\theta_1 = 0.400^o[/tex]

At  [tex]\lambda_2[/tex]

       [tex]\theta _2 = sin^{-1} [ \frac{1 * 410 *10^{-9}}{6.211 *10^{-5}} ][/tex]

       [tex]\theta_2 =0.378^o[/tex]

Answer:

9.43*10^3 year

Explanation:

For this question, we ought to remember, or know that the half life of carbon 14 is 5730, and that would be vital in completing the calculation

To start with, we use the formula

t(half) = In 2/k,

if we make k the subject of formula, we have

k = in 2/t(half), now we substitute for the values

k = in 2 / 5730

k = 1.21*10^-4 yr^-1

In(A/A•) = -kt, on rearranging, we find out that

t = -1/k * In(A/A•)

The next step is to substitite the values for each into the equation, giving us

t = -1/1.21*10^-4 * In(5.4/15.3)

t = -1/1.21*10^-4 * -1.1041

t = 0.943*10^4 year

Answer:

0 J

Explanation:

Since work done W = ∫F.dr and F(x, y, z)= z²i + 4xyj + 5y²k and dr = dxi + dyj + dzk

F.dr = (z²i + 4xyj + 5y²k).(dxi + dyj + dzk) = z²dx + 4xydy + 5y²dz

W = ∫F.dr = ∫z²dx + 4xydy + 5y²dz = z²x + 2xy² + 5y²z

We now evaluate the work done for the different regions

W₁ = work done from (0,0,0) to (1,0,0)

W₁ = {z²x + 2xy² + 5y²z}₀₀₀¹⁰⁰ = 0²(1) + 2(1)(0)² + 5(0)²(0) - [(0)²(0) + 2(0)(0)² + 5(0)²(0)] = 0 - 0 = 0 J

W₂ = work done from (1,0,0) to (1,5,1)

W₂ = {z²x + 2xy² + 5y²z}₁₀₀¹⁵¹ =   (1)²(1) + 2(1)(5)² + 5(5)²(1) - [0²(1) + 2(1)(0)² + 5(0)²(0)] =  1 + 50 + 125 - 0 = 176 J

W₃ = work done from (1,5,1) to (0,5,1)

W₃ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁵¹ =   1²(0) + 2(0)(5)² + 5(5)²(1) - [(1)²(1) + 2(1)(5)² + 5(5)²(1)]  = 125 - (1 + 50 + 125) = 125 - 176 = -51 J

W₄ = work done from (0,5,1) to (0,0,0)

W₄ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁰⁰ =   (0)²(0) + 2(0)(0)² + 5(0)²(0) - [1²(0) + 2(0)(5)² + 5(5)²(1)] = 0 - 125 = -125 J

The total work done W is thus

W = W₁ + W₂ + W₃ + W₄

W = 0 J + 176 J - 51 J - 125 J

W = 176 J - 176 J

W = 0 J

The total work done equals 0 J

Answer:

Explanation:

[tex]Force = 25N\\Area = 0.05m^2\\\\Pressure = \frac{Force}{Area}\\ \\Pressure = \frac{25}{0.05}\\\\ Pressure = 500 Pascals[/tex]

Answer:

a

    [tex]n = 1.119 *10^{18} \ photons[/tex]

b

  [tex]P = 1.6 \ W[/tex]

Explanation:

From the question we are told that

    The wavelength is  [tex]\lambda = 2780 nm = 2780 *10^{-9} \ m[/tex]

     The  energy  is  [tex]E = 80 mJ = 80 *10^{-3} \ J[/tex]

This energy is mathematically represented as

     [tex]E = \frac{n * h * c }{\lambda }[/tex]

Where  c is the speed of light with a value  [tex]c = 3.0 *10^{8} \ m/s[/tex]

             h is the Planck's  constant with the value  [tex]h = 6.626 *10^{-34} \ J \cdot s[/tex]

             n is the number of pulses

So

      [tex]n = \frac{E * \lambda }{h * c }[/tex]

substituting values

       [tex]n = \frac{80 *10^{-3} * 2780 *10^{-9}}{6.626 *10^{-34} * 3.0 *10^{8} }[/tex]

       [tex]n = 1.119 *10^{18} \ photons[/tex]

Given that the pulses where emitted 20 times in one second then the period of the pulse is

       [tex]T = \frac{1}{20}[/tex]

      [tex]T = 0.05 \ s[/tex]

Hence the average power of photons in one 80-mJ pulse during 1 s is mathematically represented as

       [tex]P = \frac{E}{T}[/tex]

substituting values

       [tex]P = \frac{ 80 *10^{-3}}{0.05}[/tex]

        [tex]P = 1.6 \ W[/tex]

Answer: i think it is d. none of them.

Explanation: The speed of light in a vacuum is 186,282 miles per second and so when you look and the answer choices and the question it doesnt make any since.