A foot is 12 inches and a mile is 5280I ft exactly. A centimeter is exactly 0.01m or mm. Sammy is 5 feet and 5.3tall. what is Sammy's height in inches?

Answer:

Explanation:

In case of soap film , light gets reflected from denser medium , hence interference takes place between two waves , one reflected from upper and second from lower surface . For destructive interference the condition is

2μt = nλ where μ is refractive index of water , t is thickness , λ is wavelength of light and n is an integer .

2 x 1.34 x t = 1  x 580

t = 216.42 nm .

Thickness must be 216.42 nm .

Answer:

c. because the rod polarizes the metal.

Explanation:

Bringing the negatively charged rod close to the neutral metal ball causes the neutral metal ball to be polarized with induced positive charge on it. The polarizing of the formally neutral metal ball is due to the negative charge on the metal rod (bodies induce a charge opposite of their own charge on a nearby neutral body). The ball and rod then attract themselves because bodies with opposite charges attract each other, unlike bodies with same charges that repel each other.

Answer:

a) k = 95.54 N / m,   c =   19.55 , b)      m₃ = 0.9078 kg

Explanation:

In a simple harmonic movement with friction, we can assume that this is provided by the speed

          fr = -c v

when solving the system the angular value remains

          w² = w₀² + (c / 2m)²

They give two conditions

1) m₁ = 1 kg

     f₁ = 1.1 Hz

the angular velocity is related to frequency

         w = 2π f₁

Let's find the angular velocity without friction is

         w₂ = k / m₁

we substitute

        (2π f₁)² = k / m₁ + (c / 2m₁)²

2) m₂ = 2 kg

    f₂ = 0.8 Hz

        (2π f₂)² = k / m₂ + (c / 2m₂)²

we have a system of two equations with two unknowns, so we can solve it

we solve (c / 2m)² is we equalize the expression

           (2π f₁)² - k / m₁ = (2π f₂²) 2 - k / m₁

           k (1 / m₂ - 1 / m₁) = 4π² (f₂² - f₁²)

           k = 4π² (f₂² -f₁²) / (1 / m₂ - 1 / m₁)

a) Let's calculate

           k = 4 π² (0.8² -1.1²) / (½ -1/1)

           k = 39.4784 (1.21) / (-0.5)

           k = 95.54 N / m

now we can find the constant of friction

              (2π f₁) 2 = k / m₁ + (c / 2m₁)²

           c2 = ((2π f₁)² - k / m₁) 4m₁²

           c2 = (4ππ² f₁² - k / m₁) 4 m₁²

let's calculate

           c² = (4π² 1,1² - 95,54 / 1) 4 1²

           c² = (47.768885 - 95.54) 8

           c² = -382.1689

           c =   19.55    

b) f₃ = 0.2 Hz

   m₃ =?

              (2πf₃)² = k / m₃ + (c / 2m₃) 2

we substitute the values

              (4π² 0.2²) = 95.54 / m₃ + 382.1689 2/4 m₃²

              1.579 = 95.54 / m₃ + 95.542225 / m₃²

let's call

              x = 1 / m₃

              x² = 1 / m₃²

- 1.579 + 95.54 x + 95.542225 x² = 0

              60.5080 x² + 60.5080 x -1 = 0

                x² + x - 1.65 10⁻² = 0

                  x = [1 ±√ (1- 4 (-1.65 10⁻²)] / 2

                  x = [1 ± 1.03] / 2

                  x₁ = 1.015 kg

                  x₂ = -0.015 kg

Since the mass must be positive we eliminate the second results

                  x₁ = 1 / m₃

                 m₃ = 1 / x₁

                  m₃ = 1 / 1.1015

Answer:

7066kg/m³

Explanation:

The forces in these cases (air and water) are: Fa =mg =ρbVg Fw =(ρb −ρw)Vg where ρw = 1000 kg/m3 is density of water and ρb is density of the block and V is its density. We can find it from this two equations:

Fa /Fw = ρb / (ρb −ρw) ρb = ρw (Fa /Fa −Fw) =1000·(1* 21.2 /21.2 − 18.2)

= 7066kg/m³

Explanation:

Answer:

The volume of the block is 306 cm³

The density of the block is 7.07 g/cm³

Explanation:

Given;

weight of block in air, [tex]W_a[/tex] = 21.2 N

Weight of block in water, [tex]W_w[/tex] = 18.2 N

Mass of the block in air;

[tex]W_a = mg[/tex]

21.2 = m x 9.8

m = 21.2 / 9.8

m = 2.163 kg

mass of the block in water;

[tex]W_w = mg[/tex]

18.2 = m x 9.8

m = 18.2 / 9.8

m = 1.857 kg

Apply Archimedes principle

Mass of object in air  - mass of object in water = density of water   x  volume                  of object

2.163 kg - 1.857 kg = 1000 kg/m³ x Volume of block

0.306 kg = 1000 kg/m³ x Volume of block

Volume of the block = [tex]\frac{0.306 \ kg}{1000 \ kg/m^3}[/tex]

Volume of the block = 3.06 x 10⁻⁴ m³

Volume of the block = 306 cm³

Determine the density of the block

[tex]Density = \frac{mass}{volume} \\\\Density =\frac{2163 \ g}{306 \ cm^3} \\\\Density = 7.07 \ g/cm^3[/tex]

Answer:

   v₂ = 9 v

Explanation:

For this exercise in fluid mechanics, let's use the continuity equation

           v₁ A₁ = v₂ A₂

where v is the velocity of the fluid, A the area of ​​the pipe and the subscripts correspond to two places of interest.

The area of ​​a circle is

           A = π R²

let's use the subscript 1 for the starting point and the subscript 2 for the part with the constraint

In this case v₁ = v and the area is

            A₁ = π R²

in the second point

           A₂= π (R / 3)²

we substitute in the continuity equation

           v π R² = v₂ π R² / 9

            v = v₂ / 9

            v₂ = 9 v

Answer:

A thermos bottle works well because:

A) Its glass walls are thin

Answer:

A thermos bottle works well because:

C

Vacuum reduces heat radiation

Answer:

29.5°

Explanation:

To find the distance d

d = 1E10^-2/8500lines

= 1.17x 10-6m

But wavelength in first order maximum is 577nm

and M = 1

So

dsin theta= m. Wavelength

Theta= sin^-1 (m wavelength/d)

= Sin^-1 ( 1* 577 x10^-8m)/1.17*10^-6

= 493*10^-3= sin^-1 0.493

Theta = 29.5°

Answer:

s=((vf+vi)/2)t vf is final velocity and vi is initial velocity

Answer:

The speed of the wave is 19.4 m/s

Explanation:

The wave's crest to crest distance (the wavelength of this rope's wave) λ= 9.70 m

The bridge is shaken twice, meaning that two wavelengths passed a given point on the rope per sec. The frequency of a wave is the amount of that wave that passes a given point in a second.

this means that the frequency f = 2 Hz

The speed of a wave = fλ = 9.70 x 2 = 19.4 m/s

Answer:

0.367A = Current of both resistors

For resistor 1: 1.89W; For resistor 2: 11.3W

Explanation:

When the resistors are connected in series, the equivalent resistance is the sum of both resistors, that is:

R = 14.0Ω + 84.0Ω = 98.0Ω

Using Ohm's law, we can find the current of the circuit (Is the same for both resistors):

V = RI

V / R = I

36.0V / 98.0Ω = I

Power is defined as:

P = I²*R

For resistor 1:

P = 0.367A²*14.0Ω = 1.89W

For resistor 1:

P = 0.367A²*84.0Ω = 11.3W

Answer:

A. We have that radius r = 4.00m intensity I = 8.00 W/m^

total power = power/ Area ( 4πr2)= 8.00 w/m^2( 4π ( 4.00 m)2=1607.68 W

b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2

c) E = total power x time= 1607 . 68 W x 1s= 1607.68 J

Answer:

Argument in favor of less total energy consumption if the store is kept at a low temperature

Explanation:

Have in mind that if the store has numerous refrigerators and freezers, the energy consumption of those machines have to be included into the analysis.

Recall that the efficiency (or Coefficient Of Performance - COP) of a frezzer or refrigerator is inversely proportional to the temperature difference between the inside of th machine and the environment where it is operation, therefore the smaller the difference, the highest their efficiency. Therefore, the cooler the environment (the temperature at which the store is kept) the better performance of the running refrigerators and freezers.

Answer:

A. 110Hz,220Hz, 330 Hz

Explanation:

for organ open at open both ends;

the length of the organ for the fundamental frequency, L = A---->N + N----->A

A---->N  = λ /4 and N----->A = λ /4

L = λ /4 + λ /4 = λ /2

[tex]L = \frac{\lambda}{2} \\\\\lambda = 2L[/tex]

λ  = 2 x 1.5m = 3.0 m

Wave equation is given by;

V = Fλ

Where;

V is the speed of sound

F is the frequency of the wave

F = V/ λ

F₀ = V / 2L

Where;

F₀  is the fundamental frequency

F₀ = 330 / 2(1.5)

F₀ = 330 / 3

F₀ = 110 Hz

the length of the organ for the first overtone, L = A---->N + N----->A + A----->N +  N----->A

L = 4λ /4

L = λ

λ = 1.5 m

F₁ = 330 / 1.5

F₁ = 220 Hz

Thus, F₁ = 2F₀

For open organ at one end

the length of the organ for the fundamental frequency, L = N------A

L = λ /4

λ = 4L

F₀ = V/4L

F₀ = 330 / (4 x 0.75)

F₀ = 110 Hz

the length of the organ for the first overtone, L = N-----N + N-----A

L = λ/2 + λ / 4

L = 3λ /4

F₁ = 3F₀

F₁ = 3 x 110

F₁ = 330 Hz

Thus the fundamental frequency for both organs is 110 Hz,

The first overtone for the organ open at both ends is 220 Hz

The first overtone for the organ open at one end is 330 Hz

The correct option is "A. 110Hz,220Hz, 330 Hz"

The correct option is option (A)

the frequencies produced by the pipes are (A) 110Hz,220Hz, 330 Hz

(I) For an organ pipe open at open both ends the frequency of different modes is given by:

F =  nv/2L

where

F is the frequency

L is the length of the organ pipe

v is the speed of the wave

and, n is the mode of frequency

the fundamental frequency corresponds to n = 1, given by:

F₀ = v/2L

F₀ = 330 / 2(1.5)

F₀ = 330 / 3

F₀ = 110 Hz

The first overtone corresponds to n = 2, the second overtone corresponds to n = 3, and so on...

F₁ =2v/2L

F₁ = 330 / 1.5

F₁ = 220 Hz

Thus, F₁ = 2F₀

The difference between successive overtones is F₀

(II) For an organ pipe open at one end the frequency of different modes is given by:

F =  nv/4L

where

F is the frequency

L is the length of the organ pipe

v is the speed of the wave

and, n is the mode of frequency

the fundamental frequency corresponds to n = 1, given by:

F₀ = V/4L

F₀ = 330 / (4 x 0.75)

F₀ = 110 Hz

For an organ pipe open at one end, only those overtones are present which correspond to odd n, that is n = 3,5,...so:

F₁ = 3F₀

F₁ = 3 x 110

F₁ = 330 Hz

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Answer:

In an electromagnetic wave in free space, the ratio of the magnitudes of electric and magnetic field vectors E and B is equal:  speed of light(c)

Explanation:

Generally the ratio of the E(electric field ) and  the B(magnetic field ) is  equal to the speed of the electromagnetic wave i.e the speed of  light (c) the value is

    [tex]c = 3.0 *10^{8} \ m/s[/tex]

Answer:

. The loop is pushed to the right, away from the magnetic field

Explanation

This decrease in magnetic strength causes an opposing force that pushes the loop away from the field

Answer:

[tex]v=1.24\times 10^8\ m/s[/tex]

Explanation:

Given that,

The refractive index of benzene is 2.419

We need to find the speed of light in benzene. The ratio of speed of light in vacuum to the speed of light in the medium equals the refractive index. So,

[tex]n=\dfrac{c}{v}\\\\v=\dfrac{c}{n}\\\\v=\dfrac{3\times 10^8}{2.419}\\\\v=1.24\times 10^8\ m/s[/tex]

So, the speed of light in bezene is [tex]1.24\times 10^8\ m/s[/tex].

Answer:

787528.7 J

Explanation:

Work done: This can be defined as the product of force and distance along the direction of force. The S.I unit of work is Joules (J).

From the question,

W = Tcos∅(d)............. Equation 1

Where W = work done, T = tension in the rope, ∅ = the angle of the rope to the horizontal, d = distance.

But,

d = v(t)..................... equation 2

Where v = velocity, t = time

Substitute equation 2 into equation 1

W = Tcos∅(vt)............. Equation 3

Given: T = 210 N, ∅ = 23°, v = 7 km/h = 1.94 m/s, t = 35 min = 2100 s

Substitute into equation 3

W = 210(cos23°)(1.94×2100)

W = 787528.7 J

Answer:

[tex]y = 0.0394 \ m[/tex]

Explanation:

From the question we are told that

        The  distance of the screen is  [tex]D = 2.20 \ m[/tex]

       The distance of separation of the slit is  [tex]d = 0.0328 \ mm = 0.0328*10^{-3} \ m[/tex]

        The  wavelength of light is  [tex]\lambda = 588 \ nm = 588 *10^{-9} \ m[/tex]

Generally the condition for constructive interference is

            [tex]dsin\theta = n * \lambda[/tex]

=>        [tex]\theta = sin^{-1} [ \frac{ n * \lambda }{d } ][/tex]

here n = 1 because we are considering the central diffraction peak

=>        [tex]\theta = sin^{-1} [ \frac{ 1 * 588*10^{-9} }{0.0328*10^{-3} } ][/tex]

=>       [tex]\theta = 1.0274 ^o[/tex]

Generally the width of central diffraction peak on a screen is mathematically evaluated as

           [tex]y = D tan (\theta )[/tex]

substituting values

        [tex]y = 2.20 * tan (1.0274)[/tex]

        [tex]y = 0.0394 \ m[/tex]

Answer:

counterclockwise

Explanation:

given data

area = 25 cm²

solution

We know that a changing magnetic field induces the current and induced emf is express as

[tex]\epsilon = -N \frac{d \phi }{dt}[/tex]     ..................................1

and we will get here direction of the induced current in the loop that is express by the Lens law that state that the direction of induces current is such that the magnetic flux due to the induced current opposes the change in magnetic flux due to the change in magnetic field

so when magnetic field decrease and point coming out of the paper.

so induced current in the loop will be counterclockwise

Answer:

The electric field is [tex]E = 5.25 V/m[/tex]

Explanation:

From the question we are told that

    The peak magnitude of the magnetic field is  [tex]B = 17.5 nT = 17.5 *10^{-9}\ T[/tex]

Generally the peak magnitude of the electric field is mathematically represented as

         [tex]E = c * B[/tex]

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

So

       [tex]E = 3.0 *10^{8} * 17.5 *10^{-9}[/tex]

       [tex]E = 5.25 V/m[/tex]

The peak magnitude of the electric field will be "5.25 V/m".

According to the question,

Magnetic field's peak magnitude, B = 17.5 nT or,

                                                           = 17.5 × 10⁻⁹ T

Speed of light, c = 3.0 × 10⁸ m/s

We know the relation,

→ E = c × B

By substituting the values, we get

      = 3.0 × 10⁸ × 17.5 × 10⁻⁹

      = 5.25 V/m

Thus the above approach is appropriate.

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Answer:

8.65x10^3m

Explanation:

See attached file

Answer:

a)  t = 2 10⁻² s ,  t = 2.4 10-1 s

Explanation:

In this exercise they indicate the delay in two signals

a) A signal travels on an electrical cable, between New York and London.

The wave formed in this wire for the signal. This wave travels at the speed of light, c = 3 108 m / s, so the delay is very small

                t = d / c

               t = 6000 10³/3 10⁸

               t = 2 10⁻² s

b) The signal points to a satellite in geostationary orbit

                   distance traveled = √ (2 36000 10³)² + 6000²

                  distance = 72 10⁶ m

the waiting time is

                t = d / c

                t = 72 10⁶/3 10⁸

               t = 2.4 10-1 s

    we can see that the signal sent by the satellites has more delay because its distance is much greater

Answer:

Some parts of your question is missing attached below is the missing parts and the answer provided is pertaining to your question alone

answer : -6661.59 volts

Explanation:

The total electric potential can be calculated using this relation

V = k [tex](\frac{q1}{r1} + \frac{q2}{r2})[/tex]

q 1 = 1.62 uc

r1 = 4.00 m

q2 = -5.73 uc

r2 = 5.00 m  

k = 8.99 * 10^9 N.m^2/c^2

insert the given values into the above equation

V = ( 8.99 * 10^9 ) * [tex](\frac{1.62*10^{-6} }{4} + \frac{-5.73*10^{-6} }{5})[/tex]  =  -6661.59 volts

Answer:

The number of turns of the inductor is 2000 turns.

Explanation:

Given;

emf of the inductor, E = 0.8 V

the rate of change of current with time, dI/dt = 10 A/s

steady current in the solenoid, I = 0.2 A

flux per turn, Ф = 8.0 μWb per

Determine the inductance of the solenoid, L

E = L(dI/dt)

L = E / (dI/dt)

L = 0.8 / (10)

L = 0.08 H

The inductance of the solenoid is given by;

[tex]L = \frac{\mu_o N^2 A}{l}[/tex]

Also, the magnetic field of the solenoid is given by;

[tex]B = \frac{\mu_o NI}{l}[/tex]

I is 0.2 A

[tex]B = \frac{\mu_oN(0.2)}{l} = \frac{0.2\mu_o N}{l}[/tex]

[tex]\frac{B}{0.2 } = \frac{\mu_o N}{l}[/tex]

[tex]L = \frac{\mu_o N^2 A}{l} \\\\L = \frac{\mu_o N }{l} (NA)\\\\L = \frac{B}{0.2} (NA)\\\\L = \frac{BA}{0.2} (N)[/tex]

But Ф = BA

[tex]L = \frac{\phi N}{0.2} \\\\\phi N = 0.2 L\\\\N = \frac{0.2 L}{\phi} \\\\N = \frac{0.2 *0.08}{8*10^{-6}}\\\\N = 2000 \ turns[/tex]

Therefore, the number of turns of the inductor is 2000 turns.

This question involves the concepts of magnetic flux, magnetic field, and inductance.

The inductor has "2000" turns.

The magnetic field due to an inductor coil is given as follows:

[tex]B=\frac{\mu_o NI}{L}\\\\[/tex]

where,

B = magnetic field

μ₀ = permeability of free space \

N = No. of turns

I = current = 0.2 A

L = length of inductor

Therefore,

[tex]\frac{\mu_oN}{L}=\frac{B}{0.2\ A}---------- eqn(1)[/tex]

Now, the inductance of a solenoid is given by the following formula:

[tex]E = L\frac{dI}{dt}\\\\L = \frac{E}{\frac{dI}{dt}}[/tex]

The inductance of solenoid can also be given using the following formula:

[tex]L = \frac{\mu_o N^2A}{L}[/tex]

comparing both the formulae, we get:

[tex]\frac{E}{\frac{dI}{dt}}= \frac{\mu_oN^2A}{L}\\\\E=\frac{dI}{dt}\frac{\mu_oN}{l}(NA)\\\\using\ eqn (1):\\\\E=\frac{dI}{dt}\frac{B}{0.2}(NA)\\\\[/tex]

where,

BA = magnetic flux = [tex]\phi[/tex] = 8 μWb/turn = 8 x 10⁻⁶ Wb/turn

N = No. of turns = ?

E = E.M.F = 0.8 volts

[tex]\frac{dI}{dt}[/tex] = rate of change in current = 10 A/s

Therefore,

[tex]0.8=(10)\frac{8\ x\ 10^{-6}}{0.2}N\\\\N=\frac{(0.8)(0.2)}{8\ x\ 10^{-5}}[/tex]

N = 2000 turns

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The attached picture shows the magnetic flux.

Answer:

W = 122.3 J

Explanation:

First, we need to find out the force applied to the smaller piston. We know that the pressure applied to smaller piston must be equally transmitted to the larger piston. Therefore,

P₁ = P₂

F₁/A₁ = F₂/A₂

F₂ = F₁(A₂/A₁)

where,

F₁ = Force of Larger Piston = Weight of car = mg = (1500 kg)(9.8 m/s²)

F₁ = 14700 N

F₂ = Force applied to smaller piston = ?

A₁ = Area of larger piston = πd₁²/4

A₂ = Area of smaller piston = πd₂²/4

Therefore,

F₂ = (14700 N)[(πd₂²/4)/(πd₁²/4)]

F₂ = (14700 N)(d₂²/d₁²)

where,

d₁ = diameter of large piston = 50 cm

d₂ = diameter of small piston = 4 cm

Therefore,

F₂ = (14700 N)[(4 cm)²/(50 cm)²]

F₂ = 94.08 N

Now, for the work done on the car:

Work Done = W = F₂ d

where,

d = displacement of car = 1.3 m

Therefore,

W = (94.08 N)(1.3 m)

W = 122.3 J

Answer:

the correct statement is

* atoms cannot be created or destroyed

Explanation:

The Datlon atomic model was proposed in 1808 and represents atoms as the smallest indivisible particle of matter, they were the building blocks of matter and are represented by solid spheres.

Based on the previous descriptive, the correct statement is

* atoms cannot be created or destroyed

Answer:

the Answer is b hope it help

Explanation:

Answer:

Explanation:

The Force experienced by the wire in the uniform magnetic field is expressed as F = BILsin∝ where;

B is the magnetic field (in Tesla)

I is the current (in amperes)

L is the length of the wire (in meters)

∝ is the angle that the conductor makes with the magnetic field.

Given parameters

L = 0.56 m

I = 2.6A

F = 0.24N

∝  = 90°

Required

magnitude of the magnetic field (B)

Substituting the given values into the formula given above we will have;

F = BILsin∝

0.24 = B * 2.6 * 0.56 sin90°

0.24 =  B * 2.6 * 0.56 (1)

0.24 = 1.456B

1.456B = 0.24

Dividing both sides by 1.456 will give;

1.456B/1.456 = 0.24/1.456

B ≈ 0.165Tesla

Hence the magnitude of the magnetic field is approximately 0.165Tesla

Answer:

The tension on string when the speed was raised is 134.53 N

Explanation:

Given;

Tension on the string, T = 120 N

initial speed of the transverse wave, v₁ = 170 m/s

final speed of the transverse wave, v₂ = 180 m/s

The speed of the wave is given as;

[tex]v = \sqrt{\frac{T}{\mu} }[/tex]

where;

μ is mass per unit length

[tex]v^2 = \frac{T}{\mu} \\\\\mu = \frac{T}{v^2} \\\\\frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}[/tex]

The final tension T₂ will be calculated as;

[tex]T_2 = \frac{T_1 v_2^2}{v_1^2} \\\\T_2 = \frac{120*180^2}{170^2} \\\\T_2 = 134.53 \ N[/tex]

Therefore, the tension on string when the speed was raised is 134.53 N

Answer:

a) 318.2 W/m^2

b) 2.5 x 10^-4 J

c) 1.55 x 10^-8 v/m

Explanation:

Power of laser P = 1 mW = 1 x 10^-3 W

exposure time t = 250 ms = 250 x 10^-3 s

If beam diameter = 2 mm = 2 x 10^-3 m

then

cross-sectional area of beam A = [tex]\pi d^{2} /4[/tex] = (3.142 x [tex](2*10^{-3} )^{2}[/tex])/4

A = 3.142 x 10^-6 m^2

a) Intensity I = P/A

where P is the power of the laser

A is the cros-sectional area of the beam

I = ( 1 x 10^-3)/(3.142 x 10^-6) = 318.2 W/m^2

b) Total energy delivered E = Pt

where P is the power of the beam

t is the exposure time

E = 1 x 10^-3 x 250 x 10^-3 = 2.5 x 10^-4 J

c) The peak electric field is given as

E = [tex]\sqrt{2I/ce_{0} }[/tex]

where I is the intensity of the beam

E is the electric field

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

[tex]e_{0}[/tex] = 8.85 x 10^9 m kg s^-2 A^-2

E = [tex]\sqrt{2*318.2/3*10^8*8.85*10^9}[/tex]  = 1.55 x 10^-8 v/m

(a)  The intensity of laser beam is  [tex]318.2 \;\rm W/m^{2}[/tex].

(b)  The total energy delivered before the blink reflex shuts the eye is [tex]2.5 \times 10^{-4} \;\rm J[/tex].

(c)  The required value of peak electric field in the laser beam is [tex]1.55 \times 10^{-8} \;\rm V/m[/tex].

Given data:

The power of laser is, [tex]P=1 \;\rm mW = 1 \times 10^{-3} \;\rm W[/tex].

The exposure time is, [tex]t = 250\;\rm ms = 250 \times 10^{-3} \;\rm s[/tex].

The beam diameter is, [tex]d = 2 \;\rm mm = 2 \times 10^{-3} \;\rm m[/tex].

a)

The standard expression for the intensity of beam is given as,

I = P/A

Here, P is the power of the laser  and A is the cross-sectional area of the beam. And its value is,

[tex]A =\pi /4 \times d^{2}\\\\A =\pi /4 \times (2 \times 10^{-3})^{2}\\\\A =3.142 \times 10^{-6} \;\rm m^{2}[/tex]

Then intensity is,

[tex]I = (1 \times 10^{-3})/(3.142 \times 10^{-6})\\\\I =318.2 \;\rm W/m^{2}[/tex]

Thus, the intensity of laser beam is [tex]318.2 \;\rm W/m^{2}[/tex].

(b)

The expression for the total energy delivered is given as,

E = Pt

Solving as,

[tex]E = 1 \times 10^{-3} \times (250 \times 10^{-3})\\\\E = 2.5 \times 10^{-4} \;\rm J[/tex]

Thus, the total energy delivered before the blink reflex shuts the eye is [tex]2.5 \times 10^{-4} \;\rm J[/tex].

(c)

The expression for the peak electric field is given as,

[tex]E = \sqrt{\dfrac{2I}{c \times \epsilon_{0}}}[/tex]

Solving as,

[tex]E = \sqrt{\dfrac{2 \times 318.2}{(3 \times 10^{8}) \times (8.85 \times 10^{9})}}\\\\E =1.55 \times 10^{-8} \;\rm V/m[/tex]

Thus, the required value of peak electric field in the laser beam is [tex]1.55 \times 10^{-8} \;\rm V/m[/tex].

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