Monochromatic light is incident on a pair of slits that are separated by 0.220 mm. The screen is 2.60 m away from the slits. (Assume the small-angle approximation is valid here.)
(a) If the distance between the central bright fringe and either of the adjacent bright fringes is 1.97 cm, find the wavelength of the incident light.
(b) At what angle does the next set of bright fringes appear?

Answer:

a) m = 0.626 kg , b) T = 2.09 s , c)   a = 1.0544 m / s²

Explanation:

In a spring mass system the equation of motion is

        x = A cos (wt + Ф)

with      w = √(k / m)

a) velocity is defined by

        v = dx / dt

        v = - A w sin (wt + Ф)          (1)

give us that the speed is

        v = 26.1 m / s

for the point

        x = a / 2

the range of motion is a = 11.0 cm

       x = 11.0 / 2

       x = 5.5 cm

Let's find the time it takes to get to this distance

       wt + Ф = cos⁻¹ (x / A)

       wt + Ф = cos 0.5

        wt + Ф = 0.877

In the exercise they do not indicate that the body started its movement with any speed, therefore we assume that for the maximum elongation the body was released, therefore the phase is zero f

       Ф = 0

       wt = 0.877

       t = 0.877 / w

we substitute in equation 1

       26.1 = -11.0 w sin (w 0.877 / w)

        w = 26.1 / (11 sin 0.877))

        w = 3.096 rad / s

from the angular velocity equation

       w² = k / m

       m = k / w²

       m = 6 / 3,096²

       m = 0.626 kg

b) angular velocity and frequency are related

       w = 2π f

frequency and period are related

        f = 1 / T

we substitute

        w = 2π / T

        T = 2π / w

        T = 2π / 3,096

        T = 2.09 s

c) maximum acceleration

 the acceleration of defined by

        a = dv / dt

        a = - Aw² cos (wt)

the acceleration is maximum when the cosine is ±1

         a = A w²

          a = 11  3,096²

        a = 105.44 cm / s²

we reduce to m / s

        a = 1.0544 m / s²

Answer:

B. 1.6 x 10⁻³ m

Explanation:

The magnetic field at the center of the loop is given by;

[tex]B = \frac{\mu_o I }{2R}[/tex]

Where;

μ₀ is the permeability of free space

I is the current in the loop

R is the radius of the circular loop

B is the magnetic field

Given;

diameter of the loop = 1cm

radius of the loop, r = 0.5 cm = 0.005 m

magnetic field, B = 2.5mT = 2.5 x 10⁻³ T

The current in the loop is calculated as;

[tex]I = \frac{2BR}{\mu_o} \\\\I = \frac{2*2.5*10^{-3}*0.005}{4\pi*10^{-7}} \\\\I = 19.89 \ A[/tex]

The magnetic at a distance from the long straight wire is calculated as;

[tex]B = \frac{\mu_o I}{2\pi d}[/tex]

where;

d is the distance from the wire;

[tex]d = \frac{\mu_o I}{2\pi B} \\\\d = \frac{4\pi *10^{-7} * 19.89}{2\pi *2.5*10^{-3}} \\\\d = 1.6 *10^{-3} \ m[/tex]

Therefore, the distance from the wire where the magnetic field is 2.5 mT is 1.6 x 10⁻³ m.

B. 1.6 x 10⁻³ m

This question involves the concepts of the magnetic field due to a loop and a  current-carrying wire and current.

A long straight wire carrying the same current as the loop of wire has a magnetic field of 2.5 mT at a distance of b "B. 1.5 x 10⁻³ m".

The magnetic field at the center of a loop of wire is given by the following formula:

[tex]B=\frac{\mu_o I}{2r}[/tex]

where,

B = Magnetic Field = 2.5 mT = 2.5 x 10⁻³ T

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

I = current = ?

r = radius = diameter/2 = 1 cm/2 = 0.5 cm = 0.005 m

Therefore,

[tex]I = \frac{(2.5\ x\ 10^{-3}\ T)(2)(0.005\ m)}{4\pi\ x\ 10^{-7}\ N/A^2}[/tex]

I = 19.9 A

Now, the magnetic field at a distance from the straight wire is given by the following formula:

[tex]B=\frac{\mu_o I}{2\pi R}[/tex]

where,

R = distance from wire = ?

Therefore,

[tex]R = \frac{(4\pi \ x \ 10^{-7}\ N/A^2)(19.9\ A)}{2\pi(2.5\ x\ 10^{-3}\ T)}[/tex]

R = 1.6 x 10⁻³ m

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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:

covalent

Explanation:

covalent bonds share electrons

Answer:

The  period is [tex]T = 0.00255 \ s[/tex]

Explanation:

From the question we are told that

  The  frequency is  [tex]f = 392 \ Hz[/tex]

Generally the period is mathematically represented as  

           [tex]T = \frac{1}{f}[/tex]

=>       [tex]T = \frac{1}{ 392}[/tex]

=>       [tex]T = 0.00255 \ s[/tex]

Answer:

m = 22,877 kg / s

Explanation:

Let's solve this exercise in parts, first look for the amount of heat generated by the plant and then the amount of water to dissipate this heat

The plant generates a power of 300 MW at a rate of 40%, let's use a direct ratio rule to find the heat. If the power is 400 MW it corresponds to 40%, what heat (Q) corresponds to the other 60%

           Q = 300 60% / 40%

           Q = 450 MW

having the amount of heat generated we can use the calorimeter equation,

           Q = m [tex]c_{e}[/tex] [tex](T_{f} - T_{o})[/tex]

            m = Q / c_{e} (T_{f} - T_{o})

let's use the maximum temperature change allowed

           (T_{f} - T_{o}) = 5

the specific heat of sea water is 3934 J / kg ºC, note that it is less than that of pure water, due to the salts dissolved in sea water

power and energy are related

              W = Q / t

               Q = W t

let's calculate

             m = 450 10⁶ / (3934 5)

             m = 22,877 kg / s

Answer:

A. 2.2*10^-2m

Explanation:

Using

Area = length x L/ uo xN²

So A = 0.7m * 25 x 10^-3H /( 4π x10^-7*

3000²)

A = 17.5*10^-3/ 1.13*10^-5

= 15.5*10^-2m²

Area= π r ²

15.5E-2/3.142 = r²

2.2*10^2m

Explanation:

Explanation:

In order of Increasing Wavelength of the Electromagnetic Spectrum :

B) X rays

D) Visible light

A) Microwave

C) Radio Waves

Electromagnetic waves in order of decreasing wavelength  is X-rays,visible light,microwaves and radio waves.

The electromagnetic radiation consists of waves made up of electromagnetic field which are capable of propogating through space and carry the radiant electromagnetic energy.

The radiation are composed of electromagnetic waves which are synchronized oscillations of electric and magnetic fields . They are created due to change which is periodic in electric as well as magnetic fields.

In vacuum ,all the electromagnetic waves travel at the same speed that is with the speed of air.The position of an electromagnetic wave in an electromagnetic spectrum is characterized by it's frequency or wavelength.They are emitted by electrically charged particles which undergo acceleration and subsequently interact with other charged particles.

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Explanation:

Given that,

Separation between two long parallel conductors, r = 11 cm = 0.11 m

Current in first wire, [tex]I_1=3\ A[/tex]

Current in second wire, [tex]I_2=8\ A[/tex]

(a) The magnetic field created by I₁ at the location of I₂ is given by :

[tex]B_{12}=\dfrac{\mu_o I_1}{2\pi r}\\\\B_{12}=\dfrac{4\pi \times 10^{-7}\times 3I_1}{2\pi \times 0.11}\\\\B_{12}=5.45\times 10^{-6}\ T[/tex]

(b) Magnetic force per unit length exerted by [tex]I_1[/tex] on [tex]I_2[/tex] is given by :

[tex]\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi r}\\\\\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times 3\times 8}{2\pi \times 0.11}\\\\\dfrac{F}{l}=4.36\times 10^{-5}\ N/m[/tex]

(c) The magnetic field created by I₂ at the location of I₁ is given by :

[tex]B_{21}=\dfrac{\mu_o I_2} {2\pi r}\\\\B_{12}=\dfrac{4\pi \times 10^{-7}\times 8}{2\pi \times 0.11}\\\\B_{12}=1.45\times 10^{-5}\ T[/tex]

Hence, this is the required solution.

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: Wavelength can be defined as the distance between two successive crests or troughs of a wave. It is measured in the direction of the wave.

Explanation:

Wavelength refers to the length or distance between two identical points of neighboring cycles of a wave signal traveling in space or in any physical medium. ... The wavelength of a signal is inversely proportional to its frequency, that is, the higher the frequency, the shorter the wavelength.

Answer

A)184.9J

B)=3.63m/s

C) Zero

Explanation:

A)potential energy of the child at the initial position, measured relative the her potential energy at the bottom of the motion, is

U=Mgh

Where m=28kg

g= 9.8m/s

h= difference in height between the initial position and the bottom position

We are told that the rope is L = 2.30 m long and inclined at 45.0° from the vertical

h=L-Lcos(x)= L(1-cosx)=2.30(1-cos45)

=0.674m

Her Potential Energy will now

= 28× 9.8×0.674

=184.9J

B)we can see that at the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:

E= 0.5mv^2

where

m = 28.0 kg is the mass of the child

v is the speed of the child at the bottom position

Solving the equation for v, we find

V=√2k/m

V=√(2×184.9/28

=3.63m/s

C)we can find work done by the tension in the rope is given using expresion below

W= Tdcosx

where W= work done

T is the tension

d = displacement of the child

x= angle between the directions of T and d

In this situation, we have that the tension in the rope, T, is always perpendicular to the displacement of the child, d. x= 90∘ and cos90∘=0 hence, the work done is zero.

Answer:

Magnitude of momentum = q × B0 × [d^2 + 2L^2] / 2d.

Explanation:

So, from the question, we are given that the charge = q, the momentum = p.

=> From the question We are also given that, "initially, there is movement along the x-axis which then enters a region where a uniform magnetic field* B = (B0)(k) which then extends over a width x = L, the distance = d in the +y direction as it traverses the field."

Momentum,P = mass × Velocity, v -----(1).

We know that for a free particle the magnetic field is equal to the centrepetal force. Thus, we have the magnetic field = mass,.m × (velocity,v)^2 / radius, r.

Radius,r = P × v / B0 -----------------------------(2).

Centrepetal force = q × B0 × v. ----------(3).

(If X = L and distance = d)Therefore, the radius after solving binomially, radius = (d^2 + 2 L^2) / 2d.

Equating Equation (2) and (3) gives;

P = B0 × q × r.

Hence, the Magnitude of momentum = q × B0 × [d^2 + 2L^2] / 2d.

Explanation:

We know that rate of flow through a cross section :

[tex]v1 \times a1 = v2 \times a2[/tex]

5 m/s * 1.76cm^2 = v2 * 7.06cm^2

[tex]v2 = 1.24 \: m {s}^{ - 1} [/tex]

At a point where the pipe diameter is 3.00 cm, the velocity is 1.25 m/s.

Fluid Flow, a branch of fluid dynamics, is concerned with fluids. It involves the movement of a fluid under the influence of uneven forces. As long as unbalanced pressures are applied, this motion will persist.

Given parameters:

Initial velocity of the water: u = 5.00 m/s

Initial diameter of the pipe: d = 1.50 cm.

Final diameter of the pipe: D = 3.00 cm.

Final velocity of the water: v = ?

In fluid motion:

velocity×(diameter)² = constant

Hence, initial velocity × ( initial diameter)² = final velocity × ( final diameter)²

ud² = vD²

v = u (d/D)²

v= 5 × (1.50/3.0)²

v= 5/2²

v= 5/4

v= 1.25 m/s.

Hence, at a point where the pipe diameter is 3.00 cm, the velocity is 1.25 m/s.

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Answer:

The answer is A

Explanation:

Answer:

implausible or untestable scientific claims

Answer:

The induced current is clockwise

Answer:

Explanation:

The force experienced by the moving electron in the magnetic field is expressed as F = qvBsinθ where;

q is the charge on the electron

v is the velocity of the electron

B is the magnetic field strength

θ is the angle that the velocity of the electron make with the magnetic field.

Given parameters

F =  1.40*10⁻¹⁶ N

q = 1.6*10⁻¹⁹C

v = 3.94*10³m/s

B = 1.23T

Required

Angle that the velocity of the electron make with the magnetic field

Substituting the given parameters into the formula:

1.40*10⁻¹⁶ =  1.6*10⁻¹⁹ * 3.94*10³ * 1.23 * sinθ

1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁹⁺³sinθ

1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁶sinθ

sinθ = 1.40*10⁻¹⁶/7.75392 * 10⁻¹⁶

sinθ = 1.40/7.75392

sinθ = 0.1806

θ = sin⁻¹0.1806

θ₁ = 10.4⁰

Since sinθ is positive in the 1st and 2nd quadrant, θ₂ = 180-θ₁

θ₂ = 180-10.4

θ₂ = 169.6⁰

Hence, the angle that the velocity of the electron make with the magnetic field are 10.4⁰ and 169.6⁰

Answer:

B.

Explanation:

The water collects in the ocean; it is then evaporated by the sun. After evaporation the water turns into water vapor, it then condenses to form clouds.

The ocean water prior to the part of the water cycle should be option B.

The ocean water should be collected back in the ocean prior to the part of the water cycle.

Because this should be done when it is evaporated by the sun.  When the evaporation is done so the water should be transformed into water vapor.

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Answer:

Explanation:

Suppose initially the plane was horizontal and light was reflected back at some angle θ from the normal .

Now the reflecting surface is twisted so that is becomes inclined at angle alpha .

The reflected light will be deviated from its original direction by angle

2 x alpha .

Similarly when the reflecting surface is further twisted so that it becomes inclined at angle beta then again the reflected beam will deviated by angle

2 x beta

Hence angle between these two reflected beam

= 2 beta - 2 alpha

= 2 ( β - α )

So, angular separation between the rays reflected from the two surfaces

= 2 ( β - α ) .

Answer:

A) 2

Explanation:

Given;

magnetic field of the coil, B = 1.6 T

frequency of the coil, f = 75 Hz

maximum emf developed in the coil, E = 56.9 V

area of the coil, A = 0.15 m x 0.25 m = 0.0375 m²

The maximum emf in the coil is given by;

E = NBAω

Where;

N is the number of turns

ω is the angular velocity = 2πf = 2 x 3.142 x 75 = 471.3 rad/s

N = E / BAω

N = 56.9 / (1.6 x 0.0375 x 471.3)

N = 2 turns

Therefore, the value of N is 2

A) 2

Answer:

298rad/s and 116.96 ohms

Explanation:

Given an L-R-C series circuit where

L = 0.450 H,

C=2.50×10^−5F, and resistance R= 0

In this situation we have a simple LC circuit with angular frequency

Wo = 1√LC

= 1/√(0.450)(2.50×10^-5)

= 1/√0.00001125

= 1/0.003354

= 298rad/s

B) Now we need to find the value of R such that it gives a 10% decrease in angular frequency.

Wi/W° = (100-10)/100

Wi/W° = 90/100

Wi/W° = 0.90 ............... 1

Angular frequency of oscillation

The complete aspect of the solution is attached, please check.

a. The angular frequency of the circuit when R = 0 Ohms is 294.12 rad/s.

b. The value R must have to give a decrease in angular frequency of 10.0 % compared to the initial value is equal to 116.96 Ohms.

Given the following data:

a. To determine the angular frequency of the circuit when R = 0 Ohms:

Mathematically, the angular frequency of a LC circuit is given by the formula:

[tex]\omega = \frac{1}{\sqrt{LC} } \\\\\omega =\frac{1}{\sqrt{0.450 \times 2.50\times 10^{-5}}} \\\\\omega =\frac{1}{\sqrt{1.125 \times 10^{-5}}} \\\\\omega = \frac{1}{0.0034} \\\\\omega = 294.12\;rad/s[/tex]

b. To find the value R must have to give a decrease in angular frequency of 10.0 % compared to the value calculated above:

The mathematical expression is given as follows:

[tex]\frac{\omega_f}{\omega_i} = \frac{100-10}{100} \\\\\frac{\omega_f}{\omega_i} =\frac{90}{100} \\\\\frac{\omega_f}{\omega_i} =0.9[/tex]

[tex](\frac{\omega_f}{\omega_i})^2 = 1 - \frac{R^2C}{4L} \\\\0.90^2=1 - \frac{R^2C}{4L}\\\\R=\sqrt{\frac{4L(1-0.81)}{C}} \\\\R=\sqrt{\frac{4\times 0.450 \times (0.19)}{2.50\times 10^{-5}}}\\\\R = \sqrt{\frac{0.342}{2.50\times 10^{-5}} }\\\\R =\sqrt{13680}[/tex]

R = 116.96 Ohms.

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Explanation:

Using Equations of Motion :

[tex]s = ut + \frac{1}{2} g {t}^{2} [/tex]

Height = 0 * 4 + 4.9 * 16

Height = 78.4 m

Answer:

E = 2.48 eV

Explanation:

The energy of a photon is given by the following formula:

E = hυ

where,

E = Energy of Photon = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

υ = frequency of photon = c/λ

Therefore,

E = hc/λ

where,

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of light = 500 nm = 5 x 10⁻⁷ m

Therefore,

E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(5 x 10⁻⁷ m)

E = (3.97 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)

E = 2.48 eV

A photon of visible light that has a wavelength of 500 nm, has an energy of 2.48 eV.

We can calculate the energy (E) of a photon with a wavelength (λ) of 500 nm using the Planck's-Einstein relation.

[tex]E = \frac{h \times c}{\lambda } = \frac{(6.63 \times 10^{-34}J.s ) \times (3.00 \times 10^{8}m/s )}{500 \times 10^{-9}m } = 3.98 \times 10^{-19} J[/tex]

where,

We can convert 3.98 × 10⁻¹⁹ J to eV using the conversion factor 1 J = 6.24 × 10¹⁸ eV.

[tex]3.98 \times 10^{-19} J \times \frac{6.24 \times 10^{18} eV }{1J} = 2.48 eV[/tex]

A photon of visible light that has a wavelength of 500 nm, has an energy of 2.48 eV.

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Answer:

b. approaches infinity

Explanation:

Because Capacitive reactance is given as Xc = 1/ωC

​

So we can see that the value of capacitive reactance and therefore its overall impedance (in Ohms) decreases to zero as the frequency increases acting like a short circuit.

Same as the frequency approaches zero or DC, the capacitors reactance increases to infinity, acting like an open circuit which is why capacitors block DC

Answer:

Not slow down or speed up.

Explanation:

Hitting the puck accelerates the speed of the puck from zero to the speed with which it leaves at the instance they lose contact. Since there is no friction between the puck and the ice, there will be no force decelerating or accelerating the hockey puck, allowing the puck to move away and remain in motion without speeding up or slowing down indefinitely theoretically.

Answer:

The magnitude of the magnetic flux through the loop is 0.0982 T.m²

Explanation:

Given;

magnitude of magnetic field, B = 0.5 T

radius of the loop, r = 0.25 m

Area of the loop is given by;

A = πr²

A = 3.142 x (0.25)²

A = 0.1964 m²

The magnitude of the magnetic flux through the loop is given by;

Ф = BA

Where;

B is the magnitude of the magnetic field

A is area of the field

Ф = 0.5 x 0.1964

Ф = 0.0982 T.m²

Therefore, the magnitude of the magnetic flux through the loop is 0.0982 T.m²

Answer:

Amount of charge

Explanation:

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a. I've attached a plot of the surface. Each face is parameterized by

• [tex]\mathbf s_1(x,y)=x\,\mathbf i+y\,\mathbf j[/tex] with [tex]0\le x\le2[/tex] and [tex]0\le y\le6-x[/tex];

• [tex]\mathbf s_2(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf k[/tex] with [tex]0\le u\le2[/tex] and [tex]0\le v\le\frac\pi2[/tex];

• [tex]\mathbf s_3(y,z)=y\,\mathbf j+z\,\mathbf k[/tex] with [tex]0\le y\le 6[/tex] and [tex]0\le z\le2[/tex];

• [tex]\mathbf s_4(u,v)=u\cos v\,\mathbf i+(6-u\cos v)\,\mathbf j+u\sin v\,\mathbf k[/tex] with [tex]0\le u\le2[/tex] and [tex]0\le v\le\frac\pi2[/tex]; and

• [tex]\mathbf s_5(u,y)=2\cos u\,\mathbf i+y\,\mathbf j+2\sin u\,\mathbf k[/tex] with [tex]0\le u\le\frac\pi2[/tex] and [tex]0\le y\le6-2\cos u[/tex].

b. Assuming you want outward flux, first compute the outward-facing normal vectors for each face.

[tex]\mathbf n_1=\dfrac{\partial\mathbf s_1}{\partial y}\times\dfrac{\partial\mathbf s_1}{\partial x}=-\mathbf k[/tex]

[tex]\mathbf n_2=\dfrac{\partial\mathbf s_2}{\partial u}\times\dfrac{\partial\mathbf s_2}{\partial v}=-u\,\mathbf j[/tex]

[tex]\mathbf n_3=\dfrac{\partial\mathbf s_3}{\partial z}\times\dfrac{\partial\mathbf s_3}{\partial y}=-\mathbf i[/tex]

[tex]\mathbf n_4=\dfrac{\partial\mathbf s_4}{\partial v}\times\dfrac{\partial\mathbf s_4}{\partial u}=u\,\mathbf i+u\,\mathbf j[/tex]

[tex]\mathbf n_5=\dfrac{\partial\mathbf s_5}{\partial y}\times\dfrac{\partial\mathbf s_5}{\partial u}=2\cos u\,\mathbf i+2\sin u\,\mathbf k[/tex]

Then integrate the dot product of f with each normal vector over the corresponding face.

[tex]\displaystyle\iint_{S_1}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{6-x}f(x,y,0)\cdot\mathbf n_1\,\mathrm dy\,\mathrm dx[/tex]

[tex]=\displaystyle\int_0^2\int_0^{6-x}0\,\mathrm dy\,\mathrm dx=0[/tex]

[tex]\displaystyle\iint_{S_2}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,0,u\sin v)\cdot\mathbf n_2\,\mathrm dv\,\mathrm du[/tex]

[tex]\displaystyle=\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=-8[/tex]

[tex]\displaystyle\iint_{S_3}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^6\mathbf f(0,y,z)\cdot\mathbf n_3\,\mathrm dy\,\mathrm dz[/tex]

[tex]=\displaystyle\int_0^2\int_0^60\,\mathrm dy\,\mathrm dz=0[/tex]

[tex]\displaystyle\iint_{S_4}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,6-u\cos v,u\sin v)\cdot\mathbf n_4\,\mathrm dv\,\mathrm du[/tex]

[tex]=\displaystyle\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=\frac{40}3+6\pi[/tex]

[tex]\displaystyle\iint_{S_5}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^{\frac\pi2}\int_0^{6-2\cos u}\mathbf f(2\cos u,y,2\sin u)\cdot\mathbf n_5\,\mathrm dy\,\mathrm du[/tex]

[tex]=\displaystyle\int_0^{\frac\pi2}\int_0^{6-2\cos u}12\,\mathrm dy\,\mathrm du=36\pi-24[/tex]

c. You can get the total flux by summing all the fluxes found in part b; you end up with 42π - 56/3.

Alternatively, since S is closed, we can find the total flux by applying the divergence theorem.

[tex]\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\iiint_R\mathrm{div}\mathbf f(x,y,z)\,\mathrm dV[/tex]

where R is the interior of S. We have

[tex]\mathrm{div}\mathbf f(x,y,z)=\dfrac{\partial(3x)}{\partial x}+\dfrac{\partial(x+y+2z)}{\partial y}+\dfrac{\partial(3z)}{\partial z}=7[/tex]

The integral is easily computed in cylindrical coordinates:

[tex]\begin{cases}x(r,t)=r\cos t\\y(r,t)=6-r\cos t\\z(r,t)=r\sin t\end{cases},0\le r\le 2,0\le t\le\dfrac\pi2[/tex]

[tex]\displaystyle\int_0^2\int_0^{\frac\pi2}\int_0^{6-r\cos t}7r\,\mathrm dy\,\mathrm dt\,\mathrm dr=42\pi-\frac{56}3[/tex]

as expected.