how does a system naturally change over time

Answer:

B has greater potential

Explanation:

We know;

Potential Energy (PE) = mgh

where, m=mass of body

g=acceleration due to gravity

h=height of body

From the formula,

PE is directly proportional to the mass of the body

so the body with greater mass has greater potential.

Answer:

she need to use 20 cm thick

Explanation:

given data

wants the attic insulated = R-30

purchased = 10 cm thick

solution

as per given we can say that

10 cm is for the R 15

but she want for R 30

so  

R 30 thickness = [tex]\frac{30}{15} \times 10[/tex]  

R 30 thickness = 20 cm

so she need to use 20 cm thick

Answer:

The speed of the electron is 1.371 x 10⁶ m/s.

Explanation:

Given;

wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m

the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J

The energy of the incident light is given by;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

f = c / λ

[tex]E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{130*10^{-9}} \\\\E = 15.291*10^{-19} \ J[/tex]

Photo electric effect equation is given by;

E = W₀ + K.E

Where;

K.E is the kinetic energy of the emitted electron

K.E = E - W₀

K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J

K.E = 8.563 x 10⁻¹⁹ J

Kinetic energy of the emitted electron is given by;

K.E = ¹/₂mv²

where;

m is mass of the electron = 9.11 x 10⁻³¹ kg

v is the speed of the electron

[tex]v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2*8.563*10^{-19}}{9.11*10^{-31}}}\\\\v = 1.371 *10^{6} \ m/s[/tex]

Therefore, the speed of the electron is 1.371 x 10⁶ m/s.

Answer:

D. It does not emit greenhouse gases.

Explanation:

D. It does not emit greenhouse gases.

Answer:

The answer is "45.79 m/s"

Explanation:

Given values:

diameter= 25 cm

w= 3500 rpm

Formula:

[tex]\boxed{v=w \times r} \ \ \ \ \ \ _{where} \ \ \ w = \frac{rad}{s} \ \ \ and \ \ \ r = meters[/tex]

Calculating r:

[tex]r= \frac{diameter}{2}[/tex]

  [tex]=\frac{25}{2}\\\\=12.5 \ cm[/tex]

converting value into meters: [tex]12.5 \times 10^{-2} \ \ meter[/tex]

calculating w:

[tex]w= diameter \times \frac{2\pi}{60}\\[/tex]

   [tex]= 3500 \times \frac{2\times 3.14}{60}\\\\= 3500 \times \frac{2\times 314}{6000}\\\\= 35 \times \frac{314}{30}\\\\= 35 \times \frac{314}{30}\\\\=\frac{10990}{30}\\\\=\frac{1099}{3}\\\\=366.33[/tex]

w= 366.33 [tex]\ \ \frac{rad}{s}[/tex]

Calculating v:

[tex]v= w\times r\\[/tex]

  [tex]= 366.33 \times 12.5 \times 10^{-2}\\\\= 366.33 \times 12.5 \times 10^{-2}\\\\= 4579.125 \times 10^{-2}\\\\\boxed{=45.79 \ \ \frac{m}{s}}[/tex]

Explanation:

Distance = speed × time

d = (340 m/s) (9.6 s)

d = 3264 m

Answer:

The value of de Broglie wavelength is 14.0 pm

Explanation:

Given;

mass of proton, m = 1.673 x 10⁻²⁷ kg

velocity of the proton, v = 2.83 x 10⁴ m/s

De Broglie wavelength is given as;

[tex]\lambda = \frac{h}{mv}[/tex]

where;

h is planck's constant = 6.626 x 10⁻³⁴ kgm²/s

m is mass of the proton

v is the velocity of the proton

[tex]\lambda = \frac{6.626*10^{-34}}{(1.673*10^{-27})(2.83*10^4})} \\\\\lambda = 1.40 *10^{-11} \ m\\\\\lambda = 14.0 \ pm[/tex]

Therefore, the value of de Broglie wavelength is 14.0 pm

Answer:

Explanation:

150/5  = 30

30mph per 1 second

Answer:

2 cm.

Explanation:

Data obtained from the question include the following:

Original Length (L₁ ) = 10 m

Initial temperature (T₁) = 0°C

Final temperature (T₂) = 400°C

Linear expansivity (α) = 5×10¯⁶ /°C

Increase in length (ΔL) =..?

Next, we shall determine the temperature rise (ΔT).

This can be obtained as follow:

Initial temperature (T₁) = 0°C

Final temperature (T₂) = 400°C

Temperature rise (ΔT) =..?

Temperature rise (ΔT) = T₂ – T₁

Temperature rise (ΔT) = 400 – 0

Temperature rise (ΔT) = 400°C

Thus, we can obtain the increase in length of the airplane by using the following formula as illustrated below:

Linear expansivity (α) = increase in length (ΔL) /Original Length (L₁ ) × Temperature rise (ΔT)

α = ΔL/(L₁ × ΔT)

Original Length (L₁ ) = 10 m

Linear expansivity (α) = 5×10¯⁶ /°C

Temperature rise (ΔT) = 400°C

Increase in length (ΔL) =..?

α = ΔL/(L₁ × ΔT)

5×10¯⁶ = ΔL/(10 × 400)

5×10¯⁶ = ΔL/4000

Cross multiply

ΔL = 5×10¯⁶ × 4000

ΔL = 0.02 m

Converting 0.02 m to cm, we have:

1 m = 100 cm

Therefore, 0.02 m = 0.02 × 100 = 2 cm.

Therefore, the length of the plane will increase by 2 cm.

Answer:

Explanation:

Time is not part of the Work equation. That's the only conclusion that you can come to.

Work = Force * distance.

Not enough information is given to go any further. You don't have enough information to calculate the distance.

We don't know if the box can be moved or not. It says heavy. 50 N is really not very much.

I would guess that you are intended to answer that the box didn't move, but it's really hard to tell.

Answer:

The maximum wavelength of incident light that can produce photoelectrons from silver is 423.5 nm.

Explanation:

Given;

work function of silver, Φ = 2.93 eV = 2.93 x 1.602 x 10⁻¹⁹ J = 4.6939 x 10⁻¹⁹ J

Apply Einstein Photo electric effect;

E = K.E + Ф

Where;

E is the energy of the incident light

K.E is the kinetic of electron

Ф is the work function of silver surface

For the incident light to have maximum wavelength, the kinetic energy of the electron will be zero.

E = Ф

hf = Ф

[tex]h\frac{c}{\lambda} = \phi[/tex]

where;

c is speed of light = 3 x 10⁸ m/s

h is Planck's constant, = 6.626 x 10⁻³⁴ J/s

λ is the wavelength of the incident light

[tex]\lambda = \frac{hc}{\phi}\\\\\lambda =\frac{6.626*10^{-34} *3*10^8}{4.6939*10^{-19}} \\\\\lambda = 4.235 *10^{-7} \ m\\\\\lambda = 423.5 *10^{-9} \ m\\\\\lambda = 423.5 \ nm[/tex]

Therefore, the maximum wavelength of incident light that can produce photoelectrons from silver is 423.5 nm.

Answer:

Hi I hope this is correct!

Explanation:

To find average velocity you can use the formula av = (v1 + v2) / 2

*I converted everything into m/s because that it usually the measurement for velocity*

v1 = initial velocity = 8.49376 m/s , v2 = final velocity = 33.528 m/s

av = 8.49376 + 33.528 / 2

    = 21.01088 m/s

*If you were required to leave the final answer in mph here it is

av = 19 + 75 / 2

    = 47 mph

Hope this helps! Best of luck <3

Explanation:

hope it helps you

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Answer:

english

Explanation:

Answer:

A = 6.8 km²

Explanation:

A) The sail should be reflective. This is so that, it can produce the maximum radiation pressure.

B) let's begin with the formula used to calculate the average solar sail in orbit around the sun. Thus;

F_rad = 2IA/c

I is given by the formula;

I = P/(4πr²)

Thus;

F_rad = (2A/c) × (P/(4πr²)) = PA/2cπr²

Where;

A is the area of the sail

r is the distance of the sail from the sun

c is the speed of light = 3 × 10^(8) m/s

P is total power output of the sun = 3.90 × 10^(26) W

Now,F_rad = F_g

Where F_g is gravitational force.

Thus;

PA/2cπr² = G•m•M_sun/r²

r² will cancel out to givw;

PA/2cπ = G•m•M_sun

Making A the subject, we have;

A = (2•c•π•G•m•M_sun)/P

Now, m = 1.06 × 10⁴ kg and M_sun has a standard value of 1.99 × 10^(30) kg

G is gravitational constant and has a value of 6.67 × 10^(-11) Nm²/kg²

Thus;

A = (2 × 3 × 10^(8) × π × 6.67 × 10^(-11) × 1.06 × 10^(4) × 1.99 × 10^(30))/(3.90 × 10^(26))

A = 6.8 × 10^(6) m² = 6.8 km²

Answer:

A

 [tex]\lambda = 361.45 \ m[/tex]

B

[tex]k = 0.01739 \ rad/m[/tex]

C

 [tex]w = 5.22 *10^{6} \ rad/s[/tex]

D

[tex]E = 0.01446 \ N/C[/tex]

Explanation:

From the question we are told that

   The frequency is [tex]f = 83 0 \ kHz = 830 *10^{3} \ Hz[/tex]

    The  magnetic field amplitude is  [tex]B = 4.82*10^{-11} \ T[/tex]

Generally wavelength is mathematically represented as

        [tex]\lambda = \frac{c}{f}[/tex]

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

     =>  [tex]\lambda = \frac{3.0*10^{8}}{ 830 *10^{3}}[/tex]

     =>  [tex]\lambda = 361.45 \ m[/tex]

Generally the wave number is mathematically represented as

        [tex]k = \frac{2 \pi }{\lambda }[/tex]

=>     [tex]k = \frac{2 * 3.142 }{ 361.45 }[/tex]

=>    [tex]k = 0.01739 \ rad/m[/tex]

Generally the angular frequency is mathematically represented as

     [tex]w = 2 * \pi * f[/tex]

=>  [tex]w = 2 * 3.142 * 830*10^{3}[/tex]

=>   [tex]w = 5.22 *10^{6} \ rad/s[/tex]

The the electric-field amplitude is mathematically represented as

     [tex]E = B * c[/tex]

=>    [tex]E = 4.82 *10^{-11} * 3.0*10^{8}[/tex]

=>     [tex]E = 0.01446 \ N/C[/tex]

This question involves the concepts of wavelength, frequency, wave number, and electric field.

a) The wavelength is "361.44 m".

b) The wave number is "0.0028 m⁻¹".

c) The angular frequency is "5.22 x 10⁶ rad/s".

d) The electric field amplitude is "0.0145 N/C".

a)

The wavelength can be given by the following formula:

[tex]c=f\lambda[/tex]

where,

c = speed of light = 3 x 10⁸ m/s

f = frequency = 830 KHz = 8.3 x 10⁵ Hz

λ = wavelength = ?

Therefore,

[tex]3\ x\ 10^8\ m/s=(8.3\ x\ 10^5\ Hz)\lambda\\\\\lambda=\frac{3\ x\ 10^8\ m/s}{8.3\ x\ 10^5\ Hz}\\\\[/tex]

λ = 361.44 m

b)

The wave number can be given by the following formula:

[tex]wave\ number = \frac{1}{\lambda} = \frac{1}{361.44\ m}[/tex]

wave number = 0.0028 m⁻¹

c)

The angular frequency is given as follows:

[tex]\omega = 2\pi f = (2)(\pi)(8.3\ x\ 10^5\ Hz)[/tex]

ω = 5.22 x 10⁶ rad/s

d)

The electric field amplitude can be given by the following formula:

[tex]\frac{E}{B} = c\\\\c(B)=E\\\\E = (3\ x\ 10^8\ m/s)(4.82\ x\ 10^{-11}\ T)\\[/tex]

E = 0.0145 N/C

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Answer:

The average power dissipated is 72 W.

Explanation:

Given;

peak voltage of the AC circuit, V₀ = 120 V

resistance of the resistor, R = 100 -ohm

The average power dissipated by the resistor is given by;

[tex]P_{avg} = \frac{1}{2} I_oV_o= I_{rms}V_{rms} = \frac{V_{rms}^2}{R}[/tex]

where;

[tex]V_{rms}[/tex] is the root-mean-square-voltage

[tex]V_{rms} = \frac{V_o}{\sqrt{2}} \\\\V_{rms} = \frac{120}{\sqrt{2}}\\\\V_{rms} = 84.853 \ V[/tex]

The average power dissipated by the resistor is calculated as;

[tex]P_{avg} = \frac{V_{rms}^2}{R}\\\\P_{avg} = \frac{84.853^2}{100}\\\\P_{avg} = 72 \ W[/tex]

Therefore, the average power dissipated is 72 W.

a) y(max)  = 337.76 m

b) t₁ = 5.30 s  the time for y maximum

c)t₂ =  13.60 s  time for y = 0 time when the fly finish

d) vₓ = 30 m/s        vy = - 81.32 m/s

e)x = 408 m

Equations for projectile motion:

v₀ₓ = v₀ * cosα          v₀ₓ = 60*(1/2)     v₀ₓ = 30 m/s   ( constant )

v₀y = v₀ * sinα           v₀y = 60*(√3/2)     v₀y = 30*√3  m/s

a) Maximum height:

The following equation describes the motion in y coordinates

y  =  y₀ + v₀y*t - (1/2)*g*t²      (1)

To find h(max), we need to calculate t₁ ( time for h maximum)

we take derivative on both sides of the equation

dy/dt  = v₀y  - g*t

dy/dt  = 0           v₀y  - g*t₁  = 0    t₁ = v₀y/g

v₀y = 60*sin60°  = 60*√3/2  = 30*√3

g = 9.8 m/s²

t₁ = 5.30 s  the time for y maximum

And y maximum is obtained from the substitution of t₁  in equation (1)

y (max) = 200 + 30*√3 * (5.30)  - (1/2)*9.8*(5.3)²

y (max) = 200 + 275.40 - 137.64

y(max)  = 337.76 m

Total time of flying (t₂)  is when coordinate y = 0

y = 0 = y₀  + v₀y*t₂ - (1/2)* g*t₂²

0 = 200 + 30*√3*t₂  - 4.9*t₂²            4.9 t₂² - 51.96*t₂ - 200 = 0

The above equation is a second-degree equation, solving for  t₂

t =  [51.96 ±√ (51.96)² + 4*4.9*200]/9.8

t =  [51.96 ±√2700 + 3920]/9.8

t =  [51.96 ± 81.36]/9.8

t = 51.96 - 81.36)/9.8         we dismiss this solution ( negative time)

t₂ =  13.60 s  time for y = 0 time when the fly finish

The components of the velocity just before striking the ground are:

vₓ = v₀ *cos60°       vₓ = 30 m/s  as we said before v₀ₓ is constant

vy = v₀y - g *t        vy = 30*√3  - 9.8 * (13.60)

vy = 51.96 - 133.28         vy = - 81.32 m/s

The sign minus means that vy  change direction

Finally the horizontal distance is:

x = vₓ * t

x = 30 * 13.60  m

x = 408 m

Answer:

d' = 75.1 cm

Explanation:

It is given that,

The actual depth of a shallow pool is, d = 1 m

We need to find the apparent depth of the water in the pool. Let it is equal to d'.

We know that the refractive index is also defined as the ratio of real depth to the apparent depth. Let the refractive index of water is 1.33. So,

[tex]n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{1\ m}{1.33}\\\\d'=0.751\ m[/tex]

or

d' = 75.1 cm

So, the apparent depth is 75.1 cm.

Answer:

8.28 V

Explanation:

Using,

N2/N1 = V2/V1.................. Equation 1

Where N2/N1 = Turn ratio of the transformer, V1 = primary/input voltage, V2 = output/secondary voltage

make V2 the subject of the equation

V2 = (N2/N1)V1............ Equation 2

Given: N2/N1 = 2:29 = 2/29, V1 = 120 V

Substitute these values into equation 2

V2 = (2/29)120

V2 = 8.28 V

Hence the rms output voltage of the transformer = 8.28 V

Answer: F

Out of the page.

Explanation:

For an electron with a charge of -e, the magnitude of the force on it is F = BeV

Where

F = force on the electron

e = charge ( electrons )

V = velocity

B = magnetic field

F is the force acting on all the electrons in a wire which gives rise to the F = BIL

Where

I = current

L = length of the wire

The force F is always at the right angle to the particle's velocity and its direction can be found using the left hand rule.

When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is directed out of the page.

Answer:

[tex]v_B/v_A=\sqrt{3}[/tex]

Explanation:

Consider the two kinematic equations for velocity and position of an object falling due to the action of gravity:

[tex]v=-g\,t\\ \\position=-\frac{1}{2} g\,t^2[/tex]

Therefore, if we consider [tex]t_A[/tex] the time for the object to reach point A, and [tex]t_B[/tex] the time for it to reach point B, then:

[tex]v_A=-g\,t_A\\v_B=-g\,t_B\\\frac{v_B}{v_A}= \frac{-g\,t_B}{-g\,t_A} =\frac{t_B}{t_A}[/tex]

Let's work in a similar way with the two different positions at those different times, and for which we have some information;

[tex]x_A=-x=-\frac{1}{2}\, g\,t_A^2\\x_B=-3\,x=-\frac{1}{2}\, g\,t_B^2\\ \\\frac{x_B}{x_A} =\frac{t_B^2}{t_A^2} \\\frac{t_B^2}{t_A^2}=\frac{-3\,x}{-x} \\\frac{t_B^2}{t_A^2}=3\\(\frac{t_B}{t_A})^2=3[/tex]

Notice that this quotient is exactly the square of the quotient of velocities we are looking for, therefore:

[tex](\frac{t_B}{t_A})^2=3\\(\frac{v_B}{v_A})^2=3\\ \frac{v_B}{v_A}=\sqrt{3}[/tex]

Answer:

From fastest speed to slowest speed, the electromagnetic waves are ranked as(up to down):

d. X-ray

c. Green light

a. Yellow light

f. Infrared wave

b. FM radio wave

e. AM radio wave

Explanation:

Electromagnetic waves are waves produced as a result of vibrations between an electric field and a magnetic field. The waves have three properties and these properties are frequency, speed and wavelength, which are related by the relationship below

V = Fλ

where:\

V = speed (velocity)

F = frequency

λ = wavelength.

From the relationship above, it is seen that the speed of a wave is directly proportional to its frequency. The higher the frequency, the higher the speed. Therefore, from the list given, the waves with  the highest to lowest frequencies/ from left to right are:

X-ray (3×10¹⁹ Hz to 3×10¹⁶Hz), Green light (5.66×10¹⁴Hz), Yellow light (5.17×10¹⁴Hz), Infrared wave (3×10¹¹Hz), FM radio wave (10.8×10⁸Hz to 8.8×10⁷Hz), AM radio wave (1.72 × 10⁶Hz to 5.5×10⁵Hz).

This corresponds to the speed from highest to lowest from left to right.

Answer:

The intensity is  [tex]I_2 = 1654 \ W/m^2[/tex]

Explanation:

From the question we are told that

    The intensity of the unpolarized light is  [tex]I_o = 4000 \ W/m^2[/tex]

    The  angle between the ideal polarizing sheet is  [tex]\theta = 0.429 \ rad = 0.429 * 57.296 = 24.58^o[/tex]

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

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

substituting values

               [tex]I_1 = \frac{4000}{2}[/tex]

                [tex]I_1 = 2000 \ W/m^2[/tex]

Then the intensity of  incident light emerging from the second polarizer is mathematically represented by Malus law as

                 [tex]I_2 = I_1 cos^2 (\theta )[/tex]

substituting values

                [tex]I_2 = 2000 * [cos (24.58)]^2[/tex]

                [tex]I_2 = 1654 \ W/m^2[/tex]

We have that the total distance covered by the runner is

[tex]d_t=20.723miles[/tex]

The total distance covered by the runner is a sum of all miles covered by the runner

Therefore

With

[tex]d_t[/tex]=Total distance

[tex]d_t=d_1+d_2+d_3\\\\d_t=9.5+8.89+2.333[/tex]

[tex]d_t=20.723miles[/tex]

in conclusion

The total distance covered by the runner is

[tex]d_t=20.723miles[/tex]

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Answer:

The moment of inertia is  [tex]I= 312.09 \ kg \cdot m^2[/tex]

Explanation:

From the question we are told that

    The  mass of the platform is   m =  137 kg

     The radius is  r  =  1.53 m

    The mass of the person is  [tex]m_p = 68.7 \ kg[/tex]

    The distance of the person from the center is  [tex]d_c =1.19 \ m[/tex]

    The mass of the dog is  [tex]m_d = 25.9 \ kg[/tex]

     The distance of the dog from the person [tex]d_d = 1.45 \ m[/tex]

Generally the moment of inertia of the system is mathematically represented as

      [tex]I = I_1 + I_2 + I_3[/tex]

Where [tex]I_1[/tex] is the moment of inertia of the platform which mathematically represented as

          [tex]I_1 = \frac{m * r^2}{2}[/tex]

substituting values

           [tex]I_1 = \frac{ 137 * (1.53)^2}{2}[/tex]

           [tex]I_1 = 160.35 \ kg\cdot m^2[/tex]

Also  [tex]I_2[/tex]  is the moment of inertia of the person about the axis which is mathematically represented as

          [tex]I_2 = m_p * d_c^2[/tex]

substituting values

          [tex]I_2 = 68.7 * 1.19^2[/tex]

          [tex]I_2 = 97.29 \ kg \cdot m^2[/tex]

Also  [tex]I_3[/tex]  is the moment of inertia of the dog about the axis which is mathematically represented as

          [tex]I_3 = m_d * d_d^2[/tex]

substituting values

          [tex]I_3 = 25.9 * 1.45^2[/tex]

          [tex]I_3 = 54.45 \ kg \cdot m^2[/tex]

Thus  

        [tex]I= 160.35 + 97.29 + 54.45[/tex]

        [tex]I= 312.09 \ kg \cdot m^2[/tex]

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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Answer:

v₀(1 + B²L²t/mR)

Explanation:

We know that the force on the loop is F = BIL where B = magnetic field strength, I = current and L = length of side of loop. Now the current in the loop I = ε/R where ε = induced e.m.f in the loop = BLv₀ where v₀ = velocity of loop and r = resistance of loop

F = BIL = B(BLv₀)L/R = B²L²v₀/R  

Since F = ma where a = acceleration of loop and m = mass of loop

a = F/m = B²L²v₀/mR

Using v = u + at where u = initial velocity of loop = v₀, t = time after t = 0 and v = velocity of loop after time t = 0

Substituting the value of a and u into v, we have

v = v₀ + B²L²v₀t/mR

= v₀(1 + B²L²t/mR)

So the velocity of the loop after time t is v = v₀(1 + B²L²t/mR)

The expression for the loop's velocity as a function of time as it enters the magnetic field is v = v₀(1 + B²L²t/mR).

As we know that

Force on the loop

F = BIL

here

B = magnetic field strength,

I = current

and L = length of side of loop.

Now

the current in the loop I = ε/R

where

ε = induced e.m.f in the loop = BLv₀

where v₀ = velocity of loop

and r = resistance of loop

So,

F = BIL = B(BLv₀)L/R = B²L²v₀/R  

Also, F = ma where a = acceleration of loop and m = mass of loop

Now

a = F/m = B²L²v₀/mR

We have to use

v = u + at

where

u = initial velocity of loop = v₀,

t = time after t = 0

and v = velocity of loop after time t = 0

So, it be like

v = v₀ + B²L²v₀t/mR

= v₀(1 + B²L²t/mR)

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