g suppose he used an alpha particle with an energy of 8.3 MeV, what would be the speed of this alpha particle

Answer:

     I₂ = 143.79

Explanation:

To solve this problem, work them in two parts. A first one where we look for the intensity of the incident light in the set and a second one where we silence the light transmuted by the other set,

Let's start with the set of three curling irons

Beautiful light falls on the first polarized is not polarized, therefore only half the radiation passes

              I₁ = I₀ / 2

this light reaches the second polarized and must comply with the Mule law

             I₂ = I₁ cos² tea

The angle between the first polarized and the second is Tea = 29.0º

             I₂ = I / 2 cos² 29

The light that comes out of the third polarized is

              I₃ = I₂ cos² tea

the angle between the third - second polarizer is

             tea = 58-29

             tea = 29th

               I3 = (I₀ / 2 cos² 29) cos² 29

indicate the output intensity

                I3 = 110

we clear

              I₀ = 2I3 / cos4 29

              I₀ = 2 110 / cos4 29

              I₀ = 375.96 W / cm²

Now we have the incident intensity in the new set of three polarizers

back to the for the first polarizer

                 I₁ = I₀ / 2

when passing the second polarizer

                     I₂ = I1 cos² 29

                    I2 = IO /2 cos²29

let's calculate

                I₂ = 375.96 / 2 cos² 29

               I₂ = 143.79

CHECK COMPLETE QUESTION BELOW

you stood on a planet having a mass four times that of earth mass and a radius two times of earth radius , you would weigh?

A) four times more than you do on Earth.

B) two times less than you do on Earth.

C) the same as you do on Earth

D) two times more than you do on Earth

Answer:

OPTION C is correct

The same as you do on Earth

Explanation :

According to law of gravitation :

F=GMm/R^2......(a)

F= mg.....(b)

M= mass of earth

m = mass of the person

R = radius of the earth

From law of motion

Put equation b into equation a

mg=GMm/R^2

g=GMm/R^2

g=GM/R^2

We know from question a planet having a mass four times that of earth mass and a radius two times of earth radius if we substitute we have

m= 4M

r=(2R)^2=4R^2

g= G4M/4R^2

Then, 4in the denominator will cancel out the numerator we have

g= GM/R^2

Therefore, g remain the same

Answer:

Static friction is greater than kinetic friction, but they both act opposite the applied force.

Explanation:

Newton's 3rd law states that every action has and equal but opposite reaction.

If an object has static friction, that means it stays in one spot, and it takes a great amount of force to get it moving.

Once the object is moving it has kinetic friction, but it's easier to keep it moving unless you are trying to stop it.

The equal but opposite reaction to something moving it is stopping it, and the equal but opposite reaction to stopping something is moving it.

The same amount of force used to move/stop something is used to stop/move it.

Static friction is greater than kinetic friction, but they both act opposite the applied force.

Friction is the force that opposes motion. Frictional force always acts in opposition to the direction of motion.

There are two kinds of friction;

Since more forces tend to act on a body at rest and prevent it from getting into motion than the forces that tend to stop an already moving body, it follows that static friction is greater than kinetic friction. Both act in opposite direction to the applied force.

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Answer:

The intensity of light from the first polarizer  is [tex]I_1 = 26 W/m^2[/tex]

Explanation:

  The intensity of the unpolarized light is  [tex]I_o = 52.0 \ W/m^2[/tex]

Generally the intensity of light that emerges from the first polarized light is

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

 substituting values

             [tex]I_1 = \frac{52. 0}{2 }[/tex]

             [tex]I_1 = 26 W/m^2[/tex]

Answer:

The vision becomes blurred due to the refractive defects of the eye. There are mainly three common refractive defects of vision. These are (i) myopia or near-sightedness, (ii) Hypermetropia or far – sightedness, and (iii) Presbyopia. These defects can be corrected by the use of suitable spherical lenses.

Answer:

d. 107 cm above the water surface.

Explanation:

The refractive index of water and air = 1.333

The real height of the girl's sole above water = 80.0 cm

From the water, the apparent height of the girl's sole will be higher than it really is in reality by a factor that is the refractive index.

The boy in the water will therefore see her feet as being

80.0 cm x 1.333 = 106.64 cm above the water

That is approximately 107 cm above the water

Answer: E = 39.54 N/C

Explanation: Electric field can be determined using surface charge density:

[tex]E = \frac{\sigma}{2\epsilon_{0}}[/tex]

where:

σ is surface charge density

[tex]\epsilon_{0}[/tex] is permitivitty of free space ([tex]\epsilon_{0} = 8.85.10^{-12}[/tex][tex]C^{2}/N.m^{2}[/tex])

Calculating resulting electric field:

[tex]E=E_{1} - E_{2}[/tex]

[tex]E = \frac{\sigma_{1}-\sigma_{2}}{2\epsilon_{0}}[/tex]

[tex]E = \frac{[0.3-(-0.4)].10^{-9}}{2.8.85.10^{-12}}[/tex]

[tex]E=0.03954.10^{3}[/tex]

E = 39.54

The resulting Electric Field at any point is 39.54N/C.

The magnitude of the resulting electric field at any point should be  28.2 N/C.

Since the Charge of uniform density (0.30 nC/m2) should be allocated over the xy plane, and charge of uniform density (−0.40 nC/m2)should be allocated over the yz plane.

So,

E1

= σ1/2ε0

= 0.30e-9/(2*8.85e-12)

= 16.949 N/C

So, direction of E1 is +z

Now

E2 = σ2/2ε0

= 0.40e-9/(2*8.85e-12)

= 22.6 N/C

So,  direction of E2 is -x

Now

E = √(E1*E1+E2*E2)

= √(16.949*16.949+22.6*22.6)

= 28.2 N/C

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The image in the attachment describes the situation of the fishing rod.

Answer: F = 10.8 N

Explanation: The image shows a fishing rod attached to an axis. To stay in equilibrium, Torque must be equal for the force of magnitude 18N and for the unknow force.

Torque (τ) is a measure of a force's tendency to cause rotation and, in physics, defined as:

τ = F.r.sin(θ)

F is the force acting on the object;

r is distance between where the torque is measured to where the force is applied;

θ is the angle between F and r;

For the fishing rod:

[tex]\tau_{1} = \tau_{2}[/tex]

[tex]F_{1}.r_{1}.sin(\theta) = F_{2}.r_{2}.sin(\theta)[/tex]

Assuming part (1) is related to unknown force:

[tex]F = \frac{F_{2}.r_{2}.sin(\theta}{r_{1}.sin(\theta) }[/tex]

Replacing the corresponding values:

[tex]F = \frac{18*1.8*sin(30)}{3*sin(30)}[/tex]

[tex]F = \frac{18*1.8}{3}[/tex]

F = 10.8

The fishing line exert on the the rod a force of 10.8N.

Complete question is;

The place you get your hair cut has two nearly parallel mirrors 6.50 m apart. As you sit in the chair, your head is 3.00 m from the nearer mirror. Looking toward this mirror, you first see your face and then, farther away, the back of your head. (The mirrors need to be slightly nonparallel for you to be able to see the back of your head, but you can treat them as parallel in this problem.) How far away does the back of your head appear to be?

Answer:

13 m

Explanation:

We are given;

Distance between two nearly parallel mirrors; d = 6.5 m

Distance between the face and the nearer mirror; x = 3 m

Thus, the distance between the back-head and the mirror = 6.5 - 3 = 3.5m

Now, From the given values above and using the law of reflection, we can find the distance of the first reflection of the back of the head of the person in the rear mirror.

Thus;

Distance of the first reflection of the back of the head in the rear mirror from the object head is;

y' = 2y

y' = 2 × 3.5

y' = 7

The total distance of this image from the front mirror would be calculated as;

z = y' + x

z = 7 + 3

z = 10

Finally, the second reflection of this image will be 10 meters inside in the front mirror.

Thus, the total distance of the image of the back of the head in the front mirror from the person will be:

T.D = x + z

T.D = 3 + 10

T.D = 13m

Given that,

Radius of curvature = 1.4 m

Wavelength = 520 nm

Refraction indexes = 1.6

We know tha,

The condition for constructive interference as,

[tex]t=(m+\dfrac{1}{2})\dfrac{\lambda}{2}[/tex]

Where, [tex]\lambda=wavelength[/tex]

We need to calculate the radius of first bright fringes

Using formula of radius

[tex]r_{1}=\sqrt{2tR}[/tex]

Put the value of t

[tex]r_{1}=\sqrt{2\times(m+\dfrac{1}{2})\dfrac{\lambda}{2}\times R}[/tex]

Put the value into the formula

[tex]r_{1}=\sqrt{2\times(0+\dfrac{1}{2})\dfrac{520\times10^{-9}}{2}\times1.4}[/tex]

[tex]r_{1}=0.603\ mm[/tex]

We need to calculate the radius of second bright fringes

Using formula of radius

[tex]r_{2}=\sqrt{2\times(m+\dfrac{1}{2})\dfrac{\lambda}{2}\times R}[/tex]

Put the value into the formula

[tex]r_{1}=\sqrt{2\times(1+\dfrac{1}{2})\dfrac{520\times10^{-9}}{2}\times1.4}[/tex]

[tex]r_{1}=1.04\ mm[/tex]

Hence, The radius of first bright fringe is 0.603 mm

The radius of second bright fringe is 1.04 mm.

Answer:

1,000 Hz hope this helps.

Explanation:

The sound said to be audible if it comes in the range of audible sound range. The audible sound is the specific frequency range of sound, which can be heard by human ears. The audible sound frequency range is 20Hz−20,000H

Complete question is;

One long wire carries a current of 30 A along the entire x axis. A second long wire carries a current of 40 A perpendicular to the xy plane and passes through the point (0, 4, 0) m. What is the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis?

Answer:

B_net = 50 × 10^(-7) T

Explanation:

We are told that the 30 A wire lies on the x-plane while the 40 A wire is perpendicular to the xy plane and passes through the point (0,4,0).

This means that the second wire is 4 m in length on the positive y-axis.

Now, we are told to find the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis.

This means that the position we want to find is half the length of the second wire.

Thus, at this point the net magnetic field is given by;

B_net = √[(B1)² + (B2)²]

Where B1 is the magnetic field due to the first wire and B2 is the magnetic field due to the second wire.

Now, formula for magnetic field due to very long wire is;

B = (μ_o•I)/(2πR)

Thus;

B1 = (μ_o•I_1)/(2πR_1)

Also, B2 = (μ_o•I_2)/(2πR_2)

Now, putting the equation of B1 and B2 into the B_net equation, we have;

B_net = √[((μ_o•I_1)/(2πR_1))² + ((μ_o•I_2)/(2πR_2))²]

Now, factorizing out some common terms, we have;

B_net = (μ_o/2π)√[((I_1)/R_1))² + ((I_2)/R_2))²]

Now,

μ_o is a constant and has a value of 4π × 10^(−7) H/m

I_1 = 30 A

I_2 = 40 A

Now, as earlier stated, the point we are looking for is 2 metres each from wire 2 end and wire 1.

Thus;

R_1 = 2 m

R_2 = 2 m

So, let's calculate B_net.

B_net = ((4π × 10^(−7))/2π)√[(30/2)² + (40/2)²]

B_net = 50 × 10^(-7) T

Answer:a) 4+8+24=36

B) 1/4+1/8+1/24=10

C) yu will connect them in parallel connection.

D) you will connect two in parallel then the remaining one in series to the ons connected in parallel.

Explanation:

(a)The maximum resistance that can be produced using all three resistors will be 36 ohms.

(b)The minimum resistance that can be produced using all three resistors will be 10 ohms.

(c)The three resistors to obtain a resistance of 10.0 Ω will be in the parallel connection.

(d) You connect these three resistors to obtain a resistance of 8.00 Ω will be in parallel. Two will be linked in parallel, and the last one will be connected in series to the two that are connected in parallel.

Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. Resistance is the enemy of the flow of current.

The maximum resistance that can be produced using all three resistors is obtained by adding all the given resistance;

[tex]\rm R_{max}=(4 +8+24 )\ ohms \\\\ R_{max}=36 \ ohms[/tex]

The minimum resistance that can be produced using all three resistors is obtained when connected in the parallel.

[tex]\rm R_{min}=\frac{1}{4} +\frac{1}{8} +\frac{1}{24} \\\\ R_{min}=10 \ ohm[/tex]

(c)The three resistors to obtain a resistance of 10.0 Ω will be in the parallel connection.

(d) You connect these three resistors to obtain a resistance of 8.00 Ω will be in parallel. Two will be linked in parallel, and the last one will be connected in series to the two that are connected in parallel.

Hence,the maximum resistance that can be produced using all three resistors will be 36 ohms.

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Answer:

About 23 meters

Explanation:

To do this, you'll want to apply one of the kinematic equations to find the time it takes for the cabin to reach max velocity from rest. (Use the max velocity as V_f and V_i=0)

Then, you can find the distance travelled during the acceleration by equating the acceleration to the change in distance of the time squared.

My work is in the attachment, comment if you have any questions.

Answer:

x₂ = 0.1715 d

1) false

2) True

3) True

4) false

5) True

Explanation:

The field electrifies a vector quantity, so we can add the creative field by these two charges

             E₂-E₁ = 0

             k q₂ / r₂² - k q₁ / r1₁²= 0

             q₂ / r₂² = q₁ / r₁²

suppose the sum of the fields is zero at a place x to the right of zero

          r₂ = d + x

          r₁ = d -x

we substitute

           q₂ / (d + x)² = q₁ / (d-x)²

we solve the equation

           q₂ / q₁ (d-x)² = (d + x) ²

let's replace the value of the charges

       q₂ / q₁ = + 2q / + q = 2

          2 (d²- 2xd + x²) = d² + 2xd + x²

          x² -6xd + d² = 0

we solve the quadratic equation

          x = [6d ± √ (36d² - 4 d²)] / 2

          x = [6d ± 5,657 d] / 2

          x₁ = 5.8285 d

          x₂ = 0.1715 d

      with the total field value zero it is between the two loads the correct solution is x₂ = 0.1715 d

this value remains on the positive part of the x axis, that is, near charge 1

now let's examine the different proposed outcomes

1) false

2) True

3) True

4) false

5) True

Answer:

c. the work done by the environment on the system equals the changein internal energy.

Explanation:

Adiabatic process:

When the boundary of a system is perfectly insulated, it means that the energy can not flow from the system and into the system ,these system is known as adiabatic system.

When the energy transfer in the system is zero ,then these type of process is known as adiabatic process.

From the first law of thermodynamics

Q= ΔU + W

Q=Heat transfer

ΔU=Change in internal energy

W=Work transfer

In adiabatic process , Q= 0

Therefore

0=ΔU +W

W=- ΔU

Negative sign indicates that ,the work done by the environment.

Therefore the correct option will be (c).

Answer:

We apply force to move the brick.

Explanation:

Let me first of define a force .

A force is something applied to an object or thing to change it's internal or external state.

Now if a brick is resting on smooth wood inclined at 30° to the horizontal for us to overcome the friction which is also a force we have to apply a force greater than the gravity force acting on the body and then depending on the direction of the applied force the angle to apply it also.

Answer:

The  radius of the earth is [tex]r = 6365.4 \ km[/tex]

Explanation:

From the question we are told that

     The distance at  Alexandria is  [tex]d_a = 800 \ km = 800 *10^{3} \ m[/tex]

      The angle of the sun is  [tex]\theta = 7.2 ^o[/tex]

So we want to first obtain the circumference of the earth

   So let assume that the earth is  circular ([tex]360 ^o[/tex])

  Now from question we know that the sun made an angle of [tex]7.2 ^o[/tex] so with this we will obtain how many  [tex](7.2 ^o)[/tex]  are in [tex]360^o[/tex]

 i.e    [tex]N = \frac{360}{7.2}[/tex]

=>      [tex]N = 50[/tex]

     With this  value we can evaluate the circumference as

             [tex]c = 50 * 800[/tex]

              [tex]c = 40000 \ km[/tex]

Generally circumference is mathematically represented as

        [tex]c = 2\pi r[/tex]

         [tex]40000 = 2 * 3.142 * r[/tex]

=>        [tex]r = 6365.4 \ km[/tex]

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.

Given that,

Energy [tex]H=2.7\times10^{31}\ W[/tex]

Surface temperature = 11000 K

Emissivity e =1

(a). We need to calculate the radius of the star

Using formula of energy

[tex]H=Ae\sigma T^4[/tex]

[tex]A=\dfrac{H}{e\sigma T^4}[/tex]

[tex]4\pi R^2=\dfrac{H}{e\sigma T^4}[/tex]

[tex]R^2=\dfrac{H}{e\sigma T^4\times4\pi}[/tex]

Put the value into the formula

[tex]R=\sqrt{\dfrac{2.7\times10^{31}}{1\times5.67\times10^{-8}\times(11000)^4\times 4\pi}}[/tex]

[tex]R=5.0\times10^{10}\ m[/tex]

(b). Given that,

Radiates energy [tex] H=2.1\times10^{23}\ W[/tex]

Temperature T = 10000 K

We need to calculate the radius of the star

Using formula of radius

[tex]R^2=\dfrac{H}{e\sigma T^4\times4\pi}[/tex]

Put the value into the formula

[tex]R=\sqrt{\dfrac{2.1\times10^{23}}{1\times5.67\times10^{-8}\times(10000)^4\times4\pi}}[/tex]

[tex]R=5.42\times10^{6}\ m[/tex]

Hence, (a). The radius of the star is [tex]5.0\times10^{10}\ m[/tex]

(b). The radius of the star is [tex]5.42\times10^{6}\ m[/tex]

Answer:

The maximum temperature increase is [tex]\Delta T = 0.0497 \ ^oC[/tex]

Explanation:

From the question we are told that

    The mass of the bullet is [tex]m = 17.0 \ g =0.017 \ kg[/tex]

     The  speed is  [tex]v_1 = 785 \ m/s[/tex]

     The mass of the water is  [tex]m_w = 13.5 \ kg[/tex]

     The velocity it emerged with is  [tex]v_2 = 534 \ m/s[/tex]

Generally due to the fact that energy can nether be created nor destroyed but transferred from one form to another then  

the change in kinetic energy of the bullet =  the heat gained by the water

 So

 The change in kinetic energy of the water is  

          [tex]\Delta KE = \frac{1}{2} m (v_1^2 - v_2 ^2 )[/tex]

substituting values  

        [tex]\Delta KE =0.5 * 0.017 * (( 785)^2 - (534) ^2 )[/tex]

        [tex]\Delta KE = 2814.1 \ J[/tex]

Now the heat gained by the water is

     [tex]Q = m_w* c_w * \Delta T[/tex]

Here [tex]c_w[/tex] is the specific heat of water which has a value  [tex]c_w = 4190 J/kg \cdot K[/tex]

So  since   [tex]\Delta KE = Q[/tex]  

we have that

          [tex]2814.1 = 13.5 * 4190 * \Delta T[/tex]

          [tex]\Delta T = 0.0497 \ ^oC[/tex]

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.

Answer:

Binding Energy = 2.24 eV

Explanation:

First, we need to find the energy of the photon of light:

E = hc/λ

where,

E = Energy of Photon = ?

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

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of light = 400 nm = 4 x 10⁻⁷ m

Therefore,

E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(4 x 10⁻⁷ m)

E = (4.97 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)

E = 3.1 eV

Now, from Einstein's Photoelectric Equation:

E = Binding Energy + Kinetic Energy

Binding Energy = E - Kinetic Energy

Binding Energy = 3.1 eV - 0.86 eV

Binding Energy = 2.24 eV

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 and Explanation:

Data provided as per the question is as follows

Speed at point A = 20 m/s

Acceleration at point C = [tex]5 m/s^2[/tex]

[tex]r_A = 25 m[/tex]

The calculation of the magnitude of the acceleration at A is shown below:-

Centripetal acceleration is

[tex]a_c = \frac{v^2}{r}[/tex]

now we will put the values into the above formula

= [tex]\frac{20^2}{25}[/tex]

After solving the above equation we will get

[tex]= 16 m/s^2[/tex]

Tangential acceleration is

[tex]= \sqrt{ac^2 + at^2} \\\\ = \sqrt{16^2 + 5^2}\\\\ = 16.703 m/s^2[/tex]

Answer:

The number of molecules is 4.574 x 10¹¹ Molecules

Explanation:

Given;

pressure in the vacuum system, P = 1 x 10⁻⁹ Pa

volume of the vessel, V = 1.9 m³

temperature of the system, T = 28°C = 301 K

Apply ideal gas law;

[tex]PV= nRT = NK_BT[/tex]

Where;

n is the number of gas moles

R is ideal gas constant = 8.314 J / mol.K

[tex]K_B[/tex] is Boltzmann's constant, = 1.38 x 10⁻²³ J/K

N is number of gas molecules

N = (PV) / ([tex]K_B[/tex]T)

N = (1 x 10⁻⁹ X 1.9) / ( 1.38 x 10⁻²³  X 301)

N = 4.574 x 10¹¹ Molecules

Therefore, the number of molecules is 4.574 x 10¹¹ Molecules

Answer:

63 cm

Explanation:

Mathematically;

The focal length of a double convex lens is given as;

1/f = (n-1)[1/R1 + 1/R2]

where n is the refractive index of the medium given as 1.52

R1 and R2 represents radius of curvature which are given as 60cm and 72cm respectively.

Plugging these values into the equation, we have:

1/f = (1.52-1)[1/60 + 1/72)

1/f = 0.0158

f = 1/0.0158

f = 63.29cm which is approximately 63cm

Answer:

covalent

Explanation:

covalent bonds share electrons

Answer:

The  distance of the proton is  [tex]r_p =10 \ cm[/tex]

Explanation:

Generally the force experience by the electron is mathematically represented as

         [tex]F_e = \frac{q * \lambda_e }{ 2 \pi * \epsilon_o * r_e}[/tex]

 Where  [tex]\lambda _e[/tex] is the charge density of the charge wire before it is doubled

Also the force experience by the proton is mathematically represented as

         [tex]F_p = \frac{q * \lambda_p }{ 2 \pi * \epsilon_o * r_p}[/tex]

Given that the charge density is doubled i.e [tex]\lambda_p = 2 \lambda_e[/tex] and that the the force are  equal then

      [tex]\frac{q * \lambda_e }{ 2 \pi * \epsilon_o * r_e} = \frac{q * 2 \lambda_e }{ 2 \pi * \epsilon_o * r_p}[/tex]

      [tex]\frac{ \lambda_e }{ r_e} = \frac{ 2 \lambda_e }{ r_p}[/tex]

      [tex]r_p * \lambda_e =2 \lambda_e * r_e[/tex]

       [tex]r_p =2 r_e[/tex]

Now given from the question that  [tex]r_e[/tex] the distance of the electron from the charged wire is  5 cm

 Then  

        [tex]r_p =2 (5)[/tex]

         [tex]r_p =10 \ cm[/tex]