A 0.100-kg metal rod carrying a current of 15.0 A glides on two horizontal rails 0.550 m apart and 2.0 m long,
(a) If the coefficient of kinetic friction between the rod and rails is 0.120, what vertical magnetic field is required to keep the rod moving at a constant speed?
(b) If the friction between the rod and rail is reduced zero, the rod will accelerate. If the rod starts from rest at the one end of the rails, what is the speed of the rod at the other end of the rails for this frictionless situation? Use the same field value you calculated in part (a).

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:

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:

As opposed to n-type semiconductors, p-type semiconductors have a larger hole concentration than electron concentration.

Explanation:

In p-type semiconductors, holes are the majority carriers and electrons are the minority carriers. A common p-type dopant for silicon is boron or gallium. hope this you :)

Answer:

Here are notes I took on semiconductor conductivity :

________________________________________________________
-A p-type semiconductor is made of a material in which electrical conduction is due to the movement of a positive charge.

-Examples of p-type dopants - boron, aluminum, gallium, indium, and thallium

Explanation:

In this case, indium only correct option being a dopant of a P-type semiconductor. Other options are N-type dopants.

Hopefully its correct !! <3

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:

energy carried by the current is given by the pointyng vector

Explanation:

The current is defined by

       i = dQ / dt

this is the number of charges per unit area over time.

The movement of the charge carriers (electrons) is governed by the applied potential difference, when the filament has a movement the drag speed of these moving electrons should change slightly.

But the energy carried by the current is given by the pointyng vector of the electromagnetic wave

            S = 1 / μ₀ EX B

It moves at the speed of light and its speed depends on the properties of the doctor and is not disturbed by small changes in speed, therefore the current in the circuit does not change due to this movement

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:

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:

a. 365; b. 3; c. 78; d. 1.343 rad; e. 12; f. 10.66

Explanation:

Assume that the function is

[tex]D(x) = 3 \sin \left (\dfrac{2\pi}{365}(x - 78) \right ) + 12[/tex]

The general formula for a sinusoidal function is

      y = A sin(B(x - C))+ D

   |A| = amplitude

     B = frequency

2π/B = period, P

     C = horizontal shift (phase shift)

     D = vertical shift

By comparing the two formulas, we find

|A| = 3

 B = 2π/365

 C = 78

 D = 12

a. Period

P = 2π/B = 2π/(2π/365) = 2π × 365/2π = 365

The period is 365.

b. Amplitude

|A| = 3

The amplitude is 3.  

c. Horizontal shift

C= 78

The horizontal shift is 78.

d. Phase shift  (φ)

Ths phase shift is the horizontal shift expressed in radians.

φ = C × 2π/365 = 78 × 2π/365 ≈ 1.343

The phase shift is 1.343 rad.

e. Vertical shift

D = 12

The vertical shift is 12.

f. Hours of sunlight on Feb 21

Feb 21 is the 52nd day of the year, so x = 51 (the number of days after Jan 1),

[tex]\begin{array}{rcl}D(x) &=& 3 \sin \left (\dfrac{2\pi}{365}(x - 78) \right ) + 12\\\\&=& 3 \sin (0.01721(51 - 78) ) + 12\\&=& 3\sin(-0.4648) + 12\\&=& 3(-0.4482) + 12\\\&=& -1.345 + 12\\& = & \textbf{10.66 h}\\\end{array}[/tex]

There will be 10.66 h of sunlight on Feb 21 of any given year.

The figure below shows the graph of the function from 0 ≤ x ≤ 365.

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:

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:

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:

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]

Answer:

statement 1 with answer C

statement 2 with answer F

statement 3 with answer B

Statement 1 with E

Statement 2 with A

Statement 3 with D

Explanation:

In this exercise you are asked to relate each with the answers

In general, in the optics diagram,

* Ray 1 is a horizontal ray that after stopping by the optical system goes to the focal point

* Ray 2 is a ray that passes through the intercept point between the optical axis and the system and does not deviate

* Ray 3 is a ray that passes through the focal length and after passing the optical system, it comes out horizontally.

With these statements, let's review the answers

statement 1 with answer C

statement 2 with answer F

statement 3 with answer B

Statement 1 with E

Statement 2 with A

Statement 3 with D

Answer:

A. The sound wave will reflect off Buildings and automobiles.

Explanation:

This is because the sound waves would more likely propagate through diffraction through buildings and transmission through the air. It is also more likely to be absorbed by buildings than for multiple reflections to occur off buildings and automobiles. In the process of reflection, these materials would absorb the sound energy thereby reducing its ability to reflect.

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: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.

To learn more about the resistance, refer to the link;

https://brainly.com/question/20708652

#SPJ2

Answer:

covalent

Explanation:

covalent bonds share electrons

Answer:

so the probability will be = 0.062

Standard deviation =  0.8925

Explanation:

The probability of rain = 15% = 15/100= 0.15

and the probability of no rain=q = 1-p= 1-0.15= 0.85

The number of trials = 7

so the probability will be

7C3 * ( 0.15)^3 (0.85)^4= 35* 0.003375 * 0.52200 =0.06166= 0.062

Taking this as binomial as the p and q are constant and also the trials are independent .

For a binomial distribution

Standard deviation = npq= 0.15 *0.85 *7= 0.8925

Answer:

Explanation:

The center of a lens is known as its optical center. All light rays incident on a particular lens converges at a points a point known as the principal focus or the focal point after reflecting. Note that all light incident on a reflecting surface must all converge at this focal point after reflection.

The distance measured from the center of this lens to its principal focus (otherwise known as focal point) is known as the focal length of the lens.

Based on the explanation above, it cam be concluded that the distance from the center of a lens to the location where parallel rays converge or appear to converge is called the FOCAL length.

Answer:

X and Y are two uncharged metal spheres on insulating stands, and are in contact with each other. A positively charged rod R is brought close to X as shown in Figure (a).

The figure shows two spheres on stands and the positively charged rod. The sphere on the left is marked X. The sphere on the right is marked Y. The spheres are in contact with each other. The rod is marked R and it is located to the left of sphere X.

Sphere  Y  is now moved away from  X , as in Figure (b).

The figure shows two spheres on stands and the positively charged rod. The sphere on the left is marked X. The sphere on the right is marked Y and it is moved away from sphere X. The rod is marked R and it is located to the left of sphere X.

What are the final charge states of X and Y?

Both X and Y are neutral.

X is neutral and Y is positive.

X is positive and Y is neutral.

X is negative and Y is positive.

Both X and Y are negative.

Explanation:

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:

Velocity of wave (V) = 5.2 × 10² m/s

Explanation:

Given:

Per unit length mass (U) = 4.80 × 10⁻³ kg/m

Tension (T)= 1,300 N

Find:

Velocity of wave (V)

Computation:

Velocity of wave (V) = √T / U

Velocity of wave (V) = √1300 / 4.80 × 10⁻³

Velocity of wave (V) = √ 270.84 × 10³

Velocity of wave (V) = 5.2 × 10² m/s

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:

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:

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:

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

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:

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]

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:

The  value [tex]y = 0.0349 \ m[/tex]

Explanation:

From the question we are told that

   The  distance of the screen is  [tex]D = 2.30 \ m[/tex]

   The  width of the slit is  [tex]d = 0.0368 \ nm = 0.0368 *10^{-3} \ m[/tex]

   The  wavelength is  [tex]\lambda = 558 \ nm = 558 *10^{-9} \ m[/tex]

The  width of the central diffraction peak is  mathematically represented as

        [tex]k = 2 * y[/tex]

Where  y is the distance from the center to the high peak which  is mathematically represented as

       [tex]y = \frac{\lambda * D }{d }[/tex]

substituting values

      [tex]y = \frac{ 558 *10^{-8} * 2.30 }{0.0368 *10^{-3} }[/tex]

      [tex]y = 0.0349 \ m[/tex]