A rotating viscometer consists of two concentric cylinders –an inner cylinder of radius Rirotating at angular velocity (rotation rate) ωi, and a stationary outer cylinder of inside radius Ro. In the tiny gap between the two cylinders is the fluid of viscosity μ. The length of the cylinders (into the page) is L. L is large such that end effects are negligible (we can treat this as a two-dimensional problem). Torque (T) is required to rotate the inner cylinder at a constant speed. (a) Showing all of your work and algebra, generate an approximate expression for T as a function of the other variables.

(b) Explain why your solution is only an approximation. In particular, do you expect the velocity prole in the gap to remain linear as the gap becomes larger and larger (i.e., if the outer radiusR0 were to increase, all else staying the same)?

Answer:

[tex]5.48\cdot 10^{-14} N[/tex]

Explanation:

When a charged particle is moving in a region with a magnetic field, the particle experiences a force perpendicular to its direction of motion. The magnitude of this force is given by

[tex]F=qvB sin \theta[/tex]

where

q is the charge of the particle

v is its velocity

B is the strength of the magnetic field

[tex]\theta[/tex] is the angle between the direction of v and B

In this problem we have:

[tex]q=1.6\cdot 10^{-19}C[/tex] is the charge of the proton

[tex]v=0.261 c[/tex] is the speed of the proton, where

[tex]c=2.998\cdot 10^8 m/s[/tex] is the speed of light

[tex]B=0.00667 T[/tex] is the strength of the magnetic field

[tex]\theta=139^{\circ}[/tex] is the angle between the direction of the proton and the magnetic field

Substituting, we find the magnitude of the force:

[tex]F=(1.6\cdot 10^{-19})(0.261\cdot 2.998\cdot 10^8)(0.00667)(sin 139^{\circ})=5.48\cdot 10^{-14} N[/tex]

The magnitude of the magnetic force acting on the proton is [tex]5.46*10^{-14} N[/tex]

When a charge particle is moving in magnetic field.

Then force is given as,  [tex]F=qvBsin(\theta)[/tex]

where

Given that, [tex]q=1.6*10^{-19}C,v=0.261*3*10^{8}=7.8*10^{7} ,B=0.00667T,\theta=139[/tex]

Substitute all values in above relation.

         [tex]F=1.6*10^{-19}*7.8*10^{7} *0.00667T*sin(139)\\\\F=5.46*10^{-14} N[/tex]

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Answer:a low pressure system is moving into the area

Explanation:

This is the only reasonable answer

Answer: a low pressure system is moving into the area

Explanation: It says it on study island

Answer:

Explanation:

Given that,

Metal of mass

M = 0.065kg

Initial temperature of metal

θm = 210°C

The metal is drop into a beaker which contain liquid of mass

m = 0.377 kg

Initial temperature of water

θw = 26°C

The final mixture temperature is

θf = 28.14°C

Specific heat capacity of water

Cw = 4186 J/kg°C

Since the metal is hotter than the water, then the metal will lose heat, while the water will gain heat, we assume that no heat is loss by the beaker.

So,

Heat Loss = Heat gain

Now, heat loss by metal

H(loss) = M•Cm•∆θ

Where M is mass of meta

Cm is specific capacity of metal, which we are looking fro

So,

H(loss) = 0.065 × Cm × (θi - θf)

H(loss) = 0.065 × Cm × (210-28.14)

H(loss) = 11.821 •Cm

Now, Heat gain by water

H(gain) = m•Cw•∆θ

H(gain) = m•Cw•(θf - θi)

Where

m is mass of water and Cw is specific heat capacity of water

H(gain) = 0.377 ×4286 × (28.14-26)

H(gain) = 3457.86

So, H(loss) = Heat(gain)

11.821 •Cm = 3457.86

Cm = 3457.86/11.821

Cm = 292.52 J/Kg°C

The specific heat capacity of the metal ball is 292.52 J/Kg°C

Answer:

320.86J/kgC

Explanation:

To find the specific heat of the substance you take into account that the heat lost by the metal is gained by the water, that is:

[tex]Q_1=-Q_2[/tex]

Furthermore the heat is given by:

[tex]Q_1=m_1c(T_1-T)\\\\Q_2=m_2c(T_2-T)[/tex]

m1: mass of the metal

m2: mass of the water

c: specific heat

T: equilibrium temperature

T1: temperature of the metal

T2: temperature of water

By replacing all these values you can calculate c of the metal:

[tex]m_1c_1(T_1-T)=-m_2c_2(T_2-T)\\\\c_1(0.065kg)(210-28.4)\°C=-(0.377kg)(4186J/kg\°C)(26-28.4)\°C\\\\c_1=320.86\frac{J}{kg\°C}[/tex]

Hence, the specific heat of the metal is 320.86J/kgC

Answer:

Explanation:

solution solved below

Answer: The ratio of their radii (Ratio of the radius of iron wire to that of copper wire) is 2.43

Explanation: Please see the attachments below

Final answer:

The ratio of the radius of an iron wire to that of a copper wire that have the same potential difference and current is approximately 1.9.

Explanation:

To find the ratio of the radii of an iron wire and a copper wire with the same potential difference and current, we can use Ohm's law and the formulas for resistance and resistivity.

The resistance is given by R = rho * (L/A), where rho is the resistivity, L is the length, and A is the cross-sectional area of the wire. Since the potential difference and current are the same for both wires, we can equate the resistances and cross-sectional areas, resulting in rho_iron * (L/A_iron) = rho_copper * (L/A_copper). Rearranging the equation, we get (A_iron/A_copper) = (rho_iron/rho_copper).

Since the radii are related to the areas of the wires by the equation A = pi * r^2, we can substitute A_iron = pi * (r_iron)^2 and A_copper = pi * (r_copper)^2 into the equation. This gives us (pi * (r_iron)^2)/(pi * (r_copper)^2) = (rho_iron/rho_copper). By canceling out the pi terms, we find (r_iron/r_copper)^2 = (rho_iron/rho_copper). Simplifying further, we get r_iron/r_copper = sqrt(rho_iron/rho_copper).

Substituting the given resistivities into the equation, we have r_iron/r_copper = sqrt((1.0 * 10^-7) / (1.7 * 10^-8)). Evaluating the expression, we find r_iron/r_copper ≈ 1.9. Therefore, the ratio of the radius of the iron wire to that of the copper wire should be approximately 1.9.

To determine the distance at which a candle that emits a power of 0.20 W would be visible, we can use the concept of luminosity and the inverse square law. The distance at which the candle would be visible is approximately 2.1 million kilometers.

To determine the distance at which a candle that emits a power of 0.20 W would be visible, we can use the concept of luminosity and the inverse square law. The luminosity of the Sun is 3.8 * 10^26 W. The candle's luminosity can be calculated using the ratio of its power to the power of the Sun. Luminosity is inversely proportional to the distance squared, so we can set up an equation with the ratio of the candle's luminosity to the Sun's luminosity equal to the ratio of the distance at which the candle is visible to the distance at which the Sun is barely visible:

(0.20 W) / (3.8 * 10^26 W) = (6.6 * 10^16 m)^2 / (x)^2

Cross-multiplying and solving for x, we find that the distance at which the candle would be visible is approximately 2.1 * 10^9 m, or 2.1 million kilometers.

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Approximately 4785 meters is the distance at which a candle would just be visible.

To solve this problem, we need to use the inverse square law for the intensity of light. The apparent brightness of a light source decreases with the square of the distance from the observer. The formula relating the brightness (intensity  I) at a distance d is given by:

[tex]\[ I = \frac{P}{4 \pi d^2} \][/tex]

First, we find the intensity I at the distance where the sun-like star is barely visible:

[tex]\[ I = \frac{P_{\text{star}}}{4 \pi d_{\text{star}}^2} \][/tex]

Plugging in the values:

[tex]\[ I = \frac{3.8 \times 10^{26}}{4 \pi (6.6 \times 10^{16})^2} \][/tex]

Now we need to find the distance [tex]\( d_{\text{candle}} \)[/tex] at which the candle with power [tex]\( P_{\text{candle}} \)[/tex] would have the same intensity I:

[tex]\[ I = \frac{P_{\text{candle}}}{4 \pi d_{\text{candle}}^2} \][/tex]

Setting the intensities equal:

[tex]\[ \frac{P_{\text{star}}}{4 \pi d_{\text{star}}^2} = \frac{P_{\text{candle}}}{4 \pi d_{\text{candle}}^2} \][/tex]

Solving for [tex]\( d_{\text{candle}} \)[/tex]:

[tex]\[ \frac{P_{\text{star}}}{d_{\text{star}}^2} = \frac{P_{\text{candle}}}{d_{\text{candle}}^2} \][/tex]

[tex]\[ d_{\text{candle}}^2 = d_{\text{star}}^2 \frac{P_{\text{candle}}}{P_{\text{star}}} \][/tex]

[tex]\[ d_{\text{candle}} = d_{\text{star}} \sqrt{\frac{P_{\text{candle}}}{P_{\text{star}}}} \][/tex]

Plugging in the values:

[tex]\[ d_{\text{candle}} = 6.6 \times 10^{16} \sqrt{\frac{0.20}{3.8 \times 10^{26}}} \][/tex]

[tex]\[ d_{\text{candle}} = 6.6 \times 10^{16} \sqrt{\frac{0.20}{3.8 \times 10^{26}}} \][/tex]

[tex]\[ d_{\text{candle}} = 6.6 \times 10^{16} \sqrt{\frac{2 \times 10^{-1}}{3.8 \times 10^{26}}} \][/tex]

[tex]\[ d_{\text{candle}} = 6.6 \times 10^{16} \sqrt{5.26 \times 10^{-28}} \][/tex]

[tex]\[ d_{\text{candle}} = 6.6 \times 10^{16} \times 7.25 \times 10^{-14} \][/tex]

[tex]\[ d_{\text{candle}} = 4.785 \times 10^3 \][/tex]

[tex]\[ d_{\text{candle}} \approx 4785 \, \text{meters} \][/tex]

Thus, a candle that emits a power of 0.20 W would just be visible at a distance of approximately 4785 meters.

Answer:

1.56 - 1.67

Explanation:

Refractive index of any material is given as the ratio of the speed of light in a vacuum to the speed of light in that medium.

Mathematically, it is given as:

n = c/v

Where c is the speed of light in a vacuum and v is the speed of light in the medium.

Given that the speed of light in the optical medium varies from 1.8 * 10^8 m/s to 1.92 * 10^8 m/s, we can find the range of the refractive index.

When the speed is 1.8 * 10^8 m/s, the refractive index is:

n = (3 * 10^8) / (1.8 * 10^8)

n = 1.67

When the speed is 1.92 * 10^8 m/s, the refractive index is:

n = (3 * 10^8) / (1.92 * 10^8)

n = 1.56

Therefore, the range of values of the refractive index of the optical medium is 1.56 - 1.67.

Final answer:

The range of the index of refraction for visible light in this optical medium is approximately 1.56 to 1.67.

Explanation:

The index of refraction, n, of a material can be calculated using the equation n = c/v, where c is the speed of light in a vacuum and v is the speed of light in the material.

For this specific optical medium, the speed of light varies from 1.80 × 10^8 m/s for violet light to 1.92 × 10^8 m/s for red light. To calculate the range of the index of refraction, we need to determine the ratio of the speed of light in a vacuum to the speed of light in the material for both violet and red light.

The range of the index of refraction, n, can be calculated as:

For violet light: n = c/v = (3.00 × 10^8 m/s) / (1.80 × 10^8 m/s) ≈ 1.67

For red light: n = c/v = (3.00 × 10^8 m/s) / (1.92 × 10^8 m/s) ≈ 1.56

Therefore, the range of the index of refraction for visible light in this optical medium is approximately 1.56 to 1.67.

Answer:

The path-length difference is [tex]dsin\theta=1.08*10^{-3}mm[/tex]

The angle is  [tex]\theta = 0.5157^o[/tex]

Explanation:

  From the question we are told that

             The distance of separation is  d = 0.12 mm = [tex]0.12*10^{-3} m[/tex]

             The distance from the screen is  D = 0.63 m

              The wavelength is [tex]\lambda = 540nm = 540 *10^{-9}m[/tex]

From the question we can deduce that the the two  maxima's are at the m=0 and m=2

   Now the path difference for this second maxima is mathematically represented as

                   [tex]d sin \theta = m \lambda[/tex]

  Where d[tex]dsin\theta[/tex] is the path length difference

Substituting values

        [tex]dsin \theta = 2 * 540*10^{-9}[/tex]

                 [tex]dsin\theta = 1.08*10^{-6}m[/tex]

converting to mm

               [tex]dsin\theta = 1.08*10^{-6} * 1000 mm[/tex]

                        [tex]dsin\theta=1.08*10^{-3}mm[/tex]

To obtain the angle we make [tex]\theta[/tex] the subject

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

 Substituting values

             [tex]\theta = sin ^{-1} [\frac{1.08*10^{-6}}{0.12*10^-3} ][/tex]

               [tex]\theta = 0.5157^o[/tex]

The Pathlength difference between the waves at second maximum on the screen is; 1.08 × 10^(-6) m

We are given;

Distance between two slits; d = 0.12 mm = 0.12 × 10^(-3) m

Distance of slit from screen; D = 0.6 m

We want to find the path length at second maxima m = 2

λ = 540 nm = 540 × 10^(-9) m

Formula for Pathlength is;

dsin θ = mλ

Where mλ is the Pathlength difference.

Since at m = 0, the pathlength is zero,

Thus;

Pathlength difference = (2 × 540 × 10^(-9)) - 0

Pathlength difference = 1.08 × 10^(-6) m

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Answer:

The acceleration is   [tex]a = 3.45*10^{3} m/s^2[/tex]

Explanation:

 From the question we are told that

         The radius is  [tex]d = 6.5 cm = \frac{6.5}{100} = 0.065 m[/tex]

           The magnitude of the magnetic field is  [tex]B = 5.5 T[/tex]

           The rate at which it decreases is  [tex]\frac{dB}{dt} = 24.5G/s = 24.5*10^{-4} T/s[/tex]

             The distance from the center of field is  [tex]r = 1.5 cm = \frac{1.5}{100} = 0.015m[/tex]

  According to Faraday's law

          [tex]\epsilon = - \frac{d \o}{dt}[/tex]

and   [tex]\epsilon = \int\limits {E} \, dl[/tex]

 Where  the magnetic flux [tex]\o = B* A[/tex]

             E is the electric field  

             dl is a unit length

 So

         [tex]\int\limits {E} \, dl = - \frac{d}{dt} (B*A)[/tex]

         [tex]{E} l = - \frac{d}{dt} (B*A)[/tex]

Now [tex]l[/tex] is the circumference of the circular loop formed by the magnetic field and it mathematically represented as  [tex]l = 2\pi r[/tex]

A is the area  of the circular loop formed by the magnetic field and it mathematically represented as  [tex]A= \pi r^2[/tex]

So

    [tex]{E} (2 \pi r)= - \pi r^2 \frac{dB}{dt}[/tex]

    [tex]E = \frac{r}{2} [ - \frac{db}{dt} ][/tex]  

Substituting values  

    [tex]E = \frac{0.015}{2} (24*10^{-4})[/tex]

         [tex]E = 3.6*10^{-5} V/m[/tex]

The negative signify the negative which is counterclockwise

  The force acting on the proton is mathematically represented as

                       [tex]F_p = ma[/tex]

        Also       [tex]F_p = q E[/tex]

So

           [tex]ma = qE[/tex]

 Where m is the mass of the the proton which has a value of  [tex]m = 1.67 *10^{-27} kg[/tex]

 [tex]q = 1.602 *10^{-19} C[/tex]

     So

            [tex]a =\frac{1.60 *10^{-19} *(3.6 *10^{-5}) }{1.67 *10^{-27}}[/tex]

               [tex]a = 3.45*10^{3} m/s^2[/tex]

Answer:

0.253 m/s

Explanation:

As the conductor falls down, the magnetic flux throug the coil formed by the conductor and the the rest of the circuit changes, therefore an electromotive force is induced in the rod; its magnitude is given by

[tex]E=BvL[/tex]

where

B = 0.052 T is the strength of the magnetic field

v is the speed at which the rod is falling

L = 21 cm = 0.21 m is the length of the rod

Due to this electromotive force, a current is also induced in the rod and the circuit, and this current is given by

[tex]I=\frac{E}{R}[/tex]

where

[tex]R=0.0011 \Omega[/tex] is the resistance of the rod

So the current is

[tex]I=\frac{BvL}{R}[/tex] (1)

At the same time, we know that a current-carrying wire in a magnetic field experiences a force, which is given by

[tex]F_B = IBL[/tex] (2)

where in this case:

I is the induced current given by eq(1)

B is the strength of the magnetic field

L is the lenght of the rod

Inserting eq(1) into (2), we find that the magnetic force on the rod is:

[tex]F_B=\frac{BvL}{R}\cdot BL = \frac{B^2 L^2 v}{R}[/tex]

However, there is another force acting on the rod: the force of gravity, given by

[tex]F_g=mg[/tex]

where

[tex]m=2.8 g = 0.0028 kg[/tex] is the mass of the rod

[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity

The falling rod will reach its terminal speed when its net acceleration becomes zero; this occurs when the net force on it is zero, so when the magnetic force is balanced by the force of gravity, so when

[tex]F_B = F_g[/tex]

So

[tex]\frac{B^2 L^2v}{R}=mg[/tex]

And solving for v, we find the terminal speed:

[tex]v=\frac{mgR}{B^2L^2}=\frac{(0.0028)(9.8)(0.0011)}{(0.052)^2(0.21)^2}=0.253 m/s[/tex]

Answer: A. The total number of protons and the total number of neutrons both remain the same.

Explanation:

Nuclear fission is the process in which a large nucleus splits into two smaller nuclei with the release of energy. In other words, fission is the process in which a nucleus is divided into two or more fragments, and neutrons and energy are released.

In fission processes, the total number of protons and neutrons both remain the same because nucleons are neither created nor destroyed, they just rearrange into new nuclei.

In fission processes, it is true that the overall number of protons and the total number of neutrons remains the same after the event. This is because nucleons, which include both protons and neutrons, are neither created nor destroyed during fission; they simply rearrange themselves into new nuclei. The mass number, which is the sum of protons and neutrons, is preserved, and though there can be a conversion between protons and neutrons (e.g., via beta decay), the total count of nucleons stays constant. However, the total mass of the product nuclei is less than the mass of the reactants due to the release of nuclear energy as a result of the conversion of mass to energy according to Einstein's equation E=mc2.

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Final answer:

The question relates to thermodynamics in physics, focusing on gas expansion or compression in a piston and its related work, heat transfer, and temperature change.

Explanation:

The question involves the concept of thermodynamics, which is a branch of physics dealing with heat, work, and energy transfer. When gas in a piston expands or compresses, it can perform work on its surroundings, and there may also be a transfer of heat between the system and its surroundings. The specifics of the temperature change, work done, and energy transfer depend on factors such as initial and final volume, pressure, and the heat capacity of the system or the environment it's in contact with.

Answer:

515.025J

A woman lifts a 35-kg child a distance of 1.5 m and carries her forward for 6.5 m.

How much work does the woman do in lifting the child?​

Explanation:

Given;

Mass m = 35kg

Distance of lifting l = 1.5m

Acceleration due to gravity g = 9.81 m/s^2

Workdone = mgl = 35×1.5 × 9.81 = 515.025 J

The change in potential energy is equal to the work that must be done, to lift the child a certain distance. Therefore, the work done by the woman in lifting the child is:

[tex]W=\Delta U\\W=mgh_f-mgh_0\\W=mg(h_f-h_0)\\W=35kg*1.5m(1.5m-0m)\\W=515.03J[/tex]

Answer:

Explanation:

To find this speed find the attached document below;

The speed of a comet can be calculated at aphelion and perihelion by using the vis-viva equation, which involves the gravitational constant, the mass of the Sun, the distance of the comet from the Sun at specific points, and the semi-major axis of the comet's orbit.

To find the speed of a comet at aphelion (ra) and perihelion (rp), we will use the conservation of energy and angular momentum principles for orbital motion. The specific mechanical energy (the sum of kinetic and potential energy) in an elliptical orbit around a much larger body, like the sun, is constant at any point along the orbit. This allows us to create an equation relating the speed of the comet at its closest and furthest points from the Sun.

Using the vis-viva equation:

v = √GM(rac{2}{r} -  rac{1}{a})

We can calculate the speed at perihelion (vp) and aphelion (va) using the given distances, where a is the semi-major axis (the average of aphelion and perihelion distances) and r is the comet's distance to the Sun at a specific point:

For the given values of rA = 5.00 x 109 km and rP = 8.00 x 107 km, we calculate a, and then use these values to find the comet's speeds at its closest and furthest points from the Sun.

Answer:

d. Planets formed from the disk of gas and dust that surrounded the Sun, and such disks are common around young stars.

Explanation:

A star passes through various stages before it becomes a full fledged star with its own planetary system. The same is with Sun as well. It was born out of a Nebula. Nebula is a cloud of dust and gases. The dust and gases start accumulating to form what we call as a protostar.

A lot of material at this stage is thrown out from the young star, this material forms a disk around the star. This disk is known as proto-planetary disk. The gases and dust of this disk then coalesce together to make planets and other objects of the star system.

We have observed such disks in the Orion nebula.

The solar nebula theory suggests that planets are formed from the disk of gas and dust that surrounds young stars. Such disks are commonly observed around stars, leading to the inference that planet formation is a universal occurrence.

The solar nebula theory, or nebular hypothesis, explains the formation of our solar system from a nebular cloud of gas and dust. According to this theory, planets are formed from the disk of gas and dust that surrounds a young star, also known as protoplanetary disk. This leads to the implication that planets are common because such disks are commonly found around young stars.

When a star condenses from a nebula, it forms a hot, spinning disk of gas and dust around it. Over time, particles within this disk begin to collide and stick together, gradually forming planetesimals and eventually, planets. Since we observe many stars surrounded by these disks, we can infer that planet formation is a common process in the universe.

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Answer:

a)  λ = 435 nm , c) c) λ = 4052 nm, d) λ= 95 nm

Explanation:

A) To carry out this excitation, the energy of the laser must be greater than or equal to the energy of the transition of the hydrogen atom, whose states of energy are described by the Bohr model.

        En = -13,606 / n²    [eV]

therefore the energy of the transition is

          ΔE = E₅ -E₂

          ΔE = 13.606 (1 / n₂² - 1 / n₅²)

         ΔE = 13.606 (1/2² - 1/5²)

         ΔE = 2,85726 eV

now let's use Planck's equation

          E = h f

the speed of light is related to wavelength and frequencies

        c = λ f

        f = c /λ

       E = h c /λ

       λ = h c / E

let's reduce the energy to the SI system

      E = 2,85726 eV (1.6 10⁻¹⁹ J / 1 eV) = 4.5716 10⁻¹⁹ J

let's calculate

       λ = 6,626 10⁻³⁴ 3 10⁸ / 4,5716 10⁻¹⁹

       λ = 4.348 10⁺⁷ m (10⁹ nm / 1 m)

       λ = 435 nm

B) photon emission processes from this state with n = 5 to the base state n = 1, can give transition

    initial state n = 5

    final state   n = 4

      ΔE = 13.606 (1/4² - 1/5²)

      ΔE = 0.306 eV

      λ = h c / E

      λ = 4052  nm

      n = 5

 final        ΔE (eV)     λ (nm)

 level   

   4             0.306      4052

   3             0.9675     1281

   2             2,857       435

   1            13.06           95

  n = 4

  3              0.661      1876

  2              2,551       486

  1              11,905       104

n = 3

  2              1.89         656

  1             12.09         102.5

n = 2

 1               10.20         121.6

c) λ = 4052 nm

d)  λ= 95 nm

The monochromatic laser has a wavelength of 434 nm. When excited hydrogen atoms fall back to the ground state, 10 different wavelengths can be observed ranging from 97.3 nm to 121.6 nm.

A monochromatic laser is used to excite hydrogen atoms from the n=2 state to the n=5 state. The wavelength of the laser can be calculated using the Rydberg formula for hydrogen, 1 / λ = RH (1/n1^2 - 1/n2^2), where RH is the Rydberg constant for hydrogen (1.097×10^7 m^-1), n1 is the lower energy level, and n2 is the higher energy level. After calculating, we get λ = 434 nm.

Eventually, as excited hydrogen atoms fall back to the ground state, there are 10 different transitions possible, corresponding to 10 different wavelengths of light.

The longest wavelength λ_max is observed when the electron falls from n=2 to n=1 state, λ_max = 121.6 nm.

The shortest wavelength λ_min is observed when an electron falls from n=5 to n=1, λ_min = 97.3 nm.

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Answer:

(a) 9.375 N/m

(b) 0.5024 m/s

(c) 0.01 kg/s

Explanation:

mass of head, m = 15 g = 0.015 kg

frequency, f = 4 Hz

Time period, T = 1 / f = 0.25 s

Let k is the spring constant.

(a)

The formula for the time period is

[tex]T=2\pi\sqrt{\frac{m}{K}}[/tex]

[tex]0.25=2\times 3.14 \sqrt{\frac{0.015}{K}}[/tex]

[tex]0.04=\sqrt{\frac{0.015}{K}}[/tex]

K = 9.375 N/m

(b)

Amplitude, A = 2 cm

Let ω is the angular velocity.

Maximum velocity, v = A ω = A x 2πf

v = 0.02 x 2 x 3.14 x 4 = 0.5024 m/s

(c)

Let b is the damping constant.

A(t = 4s) = 0.5 cm

Ao = 2 cm

Using the formula of damping

[tex]\frac{A}{A_{0}}=e^{-\frac{bt}{2m}}[/tex]

[tex]\frac{0.5}{2}}=e^{-\frac{b\times 4}{2\times 0.015}}[/tex]

[tex]0.25=e^{-133.3 b}[/tex]

Taking natural log on both the sides

ln (0.25) = - 133.3 b

- 133.3 b = - 1.386

b = 0.01 kg/s

This question involves the concepts of simple harmonic motion, spring constant, and amplitude.

a) The spring constant of the spring is "9.47 N/m".

b) The maximum speed of the head is "0.5 m/s".

c) The damping constant is "0.01 kg/s".

a)

We can find the spring constant of the spring by using the formula of frequency in the simple harmonic motion:

[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]

where,

f = frequency = 4 Hz

k = spring constant = ?

m = mass = 15 g = 0.015 kg

Therefore,

[tex]4\ Hz=\frac{1}{2\pi}\sqrt{\frac{k}{0.015\ kg}}\\\\(16\ Hz^2)(4\pi^2)(0.015\ kg)=k\\[/tex]

k = 9.47 N/m

b)

Maximum speed is simply given by the following formula:

[tex]v=A\omega[/tex]

where,

v = maximum speed = ?

A = Amplitude = 2 cm = 0.02 m

ω = angular freuency = 2πf

Therefore,

[tex]v=A(2\pi f)=(0.02\ m)(2\pi)(4\ Hz)[/tex]

v = 0.5 m/s

c)

using the following equation to find out the damping constant:

[tex]ln(\frac{A}{A_o})=-\frac{bt}{2m}[/tex]

where,

A = amplitude at t = 4 s = 0.5 cm

A₀ = initial amplitude = 2 cm

b = damping constant = ?

t = time = 4 s

m = mass = 15 g = 0.015 kg

Therefore,

[tex]ln(\frac{0.5\ cm}{2\ cm})=-\frac{b(4\ s)}{2*0.015\ kg}[/tex]

[tex]\frac{(-1.386)(2)(0.015\ kg)}{4\ s}=-b[/tex]

b = 0.01 kg/s

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Answer:

The magnitude of the force transmitted to his leg bones is 1413.6 N

Explanation:

Recall that force is defined as the change in linear momentum per unit time:

[tex]F=\frac{P_f-P_i}{\Delta t}[/tex]

We can use this formula to find the force transmitted to his legs. We know that the final momentum ([tex]P_f[/tex]) is 0 since the person is not moving on the floor, but we need to find what the person's momentum was an instant before he touches the ground. Since we know he person's mass, all we need for the initial momentum is his velocity.

For such we use conservation of energy in free fall, knowing that he jumped from 1.7 meters:

[tex]Potential \,\,Energy \,\,at \,\,the \,\,top \,\,of \,\,the\,\, jump = U_i=m * g * h = 60*9.8*1.7 \,J\\Kinetic \,\,Energy \,\,when \,\,touching \,\,ground =KE_f= \frac{1}{2} m*v^2=\frac{60\,kg}{2} v^2\\\\KE_f=U_i\\\frac{60\,kg}{2} v^2=60*9.8*1.7 \,J\\v^2=2*9.8*1.7 \,\frac{m^2}{s^2} \\v=0.589\,\frac{m}{s}[/tex]

Now with this velocity, we know the [tex]P_i[/tex] (initial momentum) just before impact.

[tex]P_i=60 \,kg * 0.589 \frac{m}{s} =35.34 \,kg\,\frac{m}{s}[/tex]

And since the impact lasted 0.025 seconds, we can find the force using the first formula we recalled:

[tex]F=\frac{P_f-P_i}{\Delta t}=\frac{0-35.34_i}{\0.025}: N= -1413.6\,N[/tex]

so the magnitude of the force is 1413.6 N

Answer: A cold front occurs when cold, denser air replaces the rising, less dense air mass. The reason this front brings in the rain is that as the rising warm air cools (as it rises to the cooler upper atmosphere) the moisture in it condenses into clouds that precipitate down as rain or snow.

Answer:

cold front

Explanation:

Answer:

Explanation:

Given that,

Surface area A= 17m²

The speed at the top v" = 66m/s

Speed beneath is v' =40 m/s

The density of air p =1.29kg/m³

Weight of plane?

Assuming that,

the height difference between the top and bottom of the wind is negligible and we can ignore any change in gravitational potential energy of the fluid.

Using Bernoulli equation

P'+ ½pv'²+ pgh' = P'' + ½pv''² + pgh''

Where

P' is pressure at the bottom in N/m²

P" is pressure at the top in N/m²

v' is velocity at the bottom in m/s

v" is velocity at the top in m/s

Then, Bernoulli equation becomes

P'+ ½pv'² = P'' + ½pv''²

Rearranging

P' — P'' = ½pv"² —½pv'²

P'—P" = ½p ( v"² —v'²)

P'—P" = ½ × 1.29 × (66²-40²)

P'—P" = 1777.62 N/m²

Lift force can be found from

Pressure = force/Area

Force = ∆P ×A

Force = (P' —P")×A

Since we already have (P'—P")

Then, F=W = (P' —P")×A

W = 1777.62 × 17

W = 30,219.54 N

The weight of the plane is 30.22 KN

Answer:

Weight of plane ; W = 30219.54 N

Explanation:

For us to determine the lift force of the system, let's multiply the pressure difference with the effective wing surface area given that the area is obtained by Bernoulli equation. Thus,

P_b + (1/2)ρ(v_b)² + ρg(y_b) = P_t + (1/2)ρ(v_t)² + ρg(y_t)

Now, since the flight is level, the height is constant.

Thus, (y_b) = (y_t)

So, we now have;

P_b + (1/2)ρ(v_b)² = P_t + (1/2)ρ(v_t)²

Rearranging, we have ;

P_b - P_t = (1/2)ρ(v_t)² - (1/2)ρ(v_b)²

P_b - P_t = (1/2)ρ[(v_t)² - (v_b)²]

Now, weight is given by the formula;

W = (P_b - P_t) •A

Thus,

W = (1/2)ρ[(v_t)² - (v_b)²] •A

From the question,

Density; ρ = 1.29 kg/m³

Velocity over top of wings; v_t = 66 m/s

Velocity beneath the wings; v_b = 40 m/s

Surface Area; A = 17 m²

Thus;

W = (1/2)1.29[(66)² - (40)²] •17

W = (1/2)•1.29•17[2756]

W = 30219.54 N

The total volume of bound charge is zero.

Explanation:

We have to the volume and surface bounded charge densities.

   ρb = - Δ . p = - Δ .k ([tex]x^{X}[/tex] +[tex]y^{Y}[/tex] +[tex]x^{Y}[/tex])

                      = - 3k

 On the top of the cube the surface charge density is

                     σb = p . z

                           = [tex]\frac{ka}{2}[/tex]

By symmetry this holds for all the other sides. The total bounded charge should be zero

        Qtot = (-3k)a³ + 6 . [tex]\frac{ka}{2}[/tex] . a² = 0

               σb = -3K σb = [tex]\frac{ka}{2}[/tex]

            Qtot = 0

Hence,  the total volume of bound charge is zero.

Answer:

Explanation:

Solution is attached below

Answer:

8.049 MW

Explanation:

The expression for gravitational potential energy is given as

Ep = mgh............. Equation 1

Ep = gravitational potential energy, m = mass of water, h = height, g = acceleration due to gravity.

Given: m = 58.4×10³ kg, h = 20.1 m, g = 9.81 m/s²

Substitute into equation 1

Ep =  58.4×10³(20.1)(9.81)

Ep = 1.6098×10⁷ J.

If one half the gravitational potential energy of the water were converted to electrical energy

Electrical energy = Ep/2

Electrical energy = (1.6098×10⁷)/2

Electrical energy = 8.049×10⁶ J

In one seconds,

The power generated = 8.049×10⁶ W

Power generated = 8.049 MW

The power of the eye for a person seeing an object which is distanced a distance 7.25 cm is 63.75 diptore

Explanation:

To find the power of the eye

we have the formula,

P=1/f= 1/d0 +1/di

Where,

do denotes the distance between eyes length and the object

di  denotes the distance between eyes length and the image

Given data

do=7.25 cm di=2.00 cm

substitute in the formula

P=1/f= 1/d0 +1/di

P= 1/0.0725 +1/0.02=13.79

P=63.79 D

The power of the eye for a person seeing an object which is distanced a distance 7.25 cm is 63.75 diptore

Answer:

So, the target proton's speed is  777.82 m/s

And, the projectile proton's speed is  777.82 m/s

Explanation:

as per the system, it conserves the linear momentum,

so along x axis :

Mp V1 (i) = Mp V1 (f) cos θ1 + Mp V2 (f) cos θ2

along y axis :

0 = -Mp V1 (f) sin θ1 + Mp V2 (f) sin θ2

let us assume before collision it was moving on positive x axis, hence target angle will be θ2 = 45° from x axis

V2(f) = V1 (i) sin θ1 / ( cosθ2 sin θ1 + cos θ1 sin θ2)

      =  1100 * sin 45 / ( cos 45 sin 45 + cos 45 sin 45 )

     =  1100 * 0.7071 /( 0.7071 * 0.7071 +0 .7071 * 0.7071 )

     = 777.82 /( 0.5 + 0.5)

    = 777.82 m/s

(b) 

 the speed of projectile , V1 (f) = sinθ2 * V2(f) / sinθ1

                                                              = sin 45 * 777.82 / sin 45

                                                              = 777.82 m/s

So, the target proton's speed is  777.82 m/s

And, the projectile proton's speed is  777.82 m/s

Answer:

The answer to this questions are, (a) 0.685 * 10^-4 T/sec (b) 0.946 * 10^-6V

Explanation:

Solution

Recall

radius r = 0.02m

N = 11 turns

Instant I = 3 amperes

Velocity v =3.3m/s

x = 0.17 m

(a) What is the magnitude of the rate of change of the magnetic field inside the coil

db/dt =μ₀T/2π x²

Thus,

= 4π * 10^-7* 3* 3.3/2π * 0.17²

=685.12* 10^-7

which is now,

0.685 * 10^-4 T/sec

(b) What is the magnitude of the voltmeter reading.

μ =  N (db/dt)πr²

Note: this includes all 11 turns of the coil

Thus,

= 11 * 0.6585* 10^-4* 3.14 * (0.02)²

= 946 * 10^-7

which is = 0.946 * 10^-6V

Note: Kindly find an attached copy or document of the complete question of this exercise

The rate of change of a magnetic field in a coil can be determined using Faraday's law and the chain rule. Starting with the formula for the magnetic field created by current in a straight wire, we differentiate with respect to time to get the rate of change.

The magnitude of the rate of change of the magnetic field inside a coil can be represented algebraically using Faraday's law of induction. Starting with the equation B = μ * I / 2λr (where I is the current, r is the distance to the wire, and μ is the permeability of free space), we can find the rate of change of the magnetic field using the chain rule.

Faraday's law is represented as E = - dΦ/dt, where E is the induced emf, Φ is the magnetic flux, and t is time. Since the magnetic flux Φ is the product of the magnetic field (B) and the area enclosed by the loop (A), we can express this as d/dt (B * A). By applying the chain rule, we can find the rate of change of the magnetic field in the coil.

Last, magnetic field B is determined by Ampère's law, often used in calculations involving magnetic fields around conductors and coils. For a long straight wire, field lines form concentric circles around the wire, following the right-hand rule.

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Answer: S = 4.67 W/m², U = 1.29 J

Explanation:

Given

Time of flow, t = 12.3 s

Area of flow, a = 0.0225 s

Amplitude, E = 59.3 V/m

Intensity, S = ?

I = E² / cμ, where

μ = permeability of free space

c = speed of light

E = E(max) / √2

E = 59.3 / √2

E = 41.93 V/m

I = 41.93² / (2.99*10^8 * 1.26*10^-6)

I = 1758.125 / 376.74

I = 4.67 W/m²

Energy that flows through

U = Iat

U = 4.67 * 0.0225 * 12.3

U = 1.29 J

Therefore, the intensity is 4.67 W/m² and the energy is 1.29J

Answer:

A) Intensity = 4.664 W/m²

B) U = 1.29J

Explanation:

A) The intensity of the wave is related to a time-averaged version of a quantity called the Poynting vector, and is given by the formula:

I = (E_rms/cμo)

Where;

c = speed of light which has a value of 3 x 10^(8) m/s

μo = permeability of free space which has a constant value of 4π x 10^(-7) N/A²

E_rms is root mean square value of electric field

In the question, we are given maximum amplitude of the electric field. In this case, we would have to calculate the "root-mean-square" or "rms" value through the relationship:

E_rms = E_max/√2

Thus, E_rms = 59.3/√2 = 41.93 V/m

Thus, Intensity, I = (E_rms/cμo)= [41.93²/(3 x 10^(8) x 4π x 10^(-7))]

I = 4.664 W/m²

B) The formula for the energy flowing is given by the formula ;

U = IAt

Where;

I is intensity

A is area

t is time in seconds

Thus, U = 4.664 x 0.0225 x 12.3 = 1.29J

Answer:

216.59 s.

Explanation:

Using,

F = ma................. Equation 1

Where F = force exerted by the tugboat, m = mass of the ocean liner, a = acceleration of the ocean liner

make a the subject of the equation

a = F/m.................. Equation 2

Given: F = 195 kN = 195000 N, m = 38000 Mg = 38000000 kg.

Substitute into equation 2

a = 195000/38000000

a = 5.13×10⁻³ m/s²

Also using,

Assuming the liner is decelerating

a = (u-v)/t............ Equation 3

Where v = final velocity, u = initial velocity, t = time

make t the subject of the equation

t = (u-v)/a............. Equation 4

Given: u = 4 km/h = 4(1000/3600) = 1.111 m/s, v = 0 m/s, a = 0.00513 m/s²

Substitute into equation 4

t = (1.111-0)/0.00513

t = 216.59 s.

Answer:h=19.4 m

Explanation:

Given

mass of automobile [tex]m=750\ kg[/tex]

Initial height of automobile [tex]h_o=5\ m[/tex]

Velocity at this instant [tex]v=16.8\ m/s[/tex]

If the car stops somewhere at a height [tex]h[/tex]

Thus conserving total energy we get

[tex]K_i+U_i=K_f+U_f[/tex]

[tex]\frac{1}{2}mv^2+mgh_o=\frac{1}{2}m(0)^2+mgh[/tex]

[tex]\frac{v^2}{2g}+h_o=h[/tex]

[tex]h=5+\frac{16.8^}{2\times 9.8}[/tex]

[tex]h=5+14.4[/tex]

[tex]h=19.4\ m[/tex]

Answer:Calculate displacement of an object that is not acceleration, given initial position and velocity.

Calculate final velocity of an accelerating object, given initial velocity, acceleration, and time.

Calculate displacement and final position of an accelerating object, given initial position, initial velocity, time, and acceleration.

Four men racing up a river in their kayaks.

Figure 1. Kinematic equations can help us describe and predict the motion of moving objects such as these kayaks racing in Newbury, England. (credit: Barry Skeates, Flickr).

We might know that the greater the acceleration of, say, a car moving away from a stop sign, the greater the displacement in a given time. But we have not developed a specific equation that relates acceleration and displacement. In this section, we develop some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration already covered.

Notation: t, x, v, a

First, let us make some simplifications in notation. Taking the initial time to be zero, as if time is measured with a stopwatch, is a great simplification. Since elapsed time is

Δ

t

=

t

f

−

t

0

, taking

t

0

=

0

means that

Δ

t

=

t

f

, the final time on the stopwatch. When initial time is taken to be zero, we use the subscript 0 to denote initial values of position and velocity. That is,

x

0

is the initial position and

v

0

is the initial velocity. We put no subscripts on the final values. That is,

t

is the final time,

x

is the final position, and

v

is the final velocity. This gives a simpler expression for elapsed time—now,

Δ

t

=

t

. It also simplifies the expression for displacement, which is now

Δ

x

=

x

−

x

0

. Also, it simplifies the expression for change in velocity, which is now

Δ

v

=

v

−

v

0

. To summarize, using the simplified notation, with the initial time taken to be zero,

Explanation: