A person with mass mp = 74 kg stands on a spinning platform disk with a radius of R = 2.31 m and mass md = 183 kg. The disk is initially spinning at ω = 1.8 rad/s. The person then walks 2/3 of the way toward the center of the disk (ending 0.77 m from the center). 1)What is the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk? kg-m2 2)What is the total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk? kg-m2 3)What is the final angular velocity of the disk? rad/s 4)What is the change in the total kinetic energy of the person and disk? (A positive value means the energy increased.) J 5)What is the centripetal acceleration of the person when she is at R/3? m/s2 6)If the person now walks back to the rim of the disk, what is the final angular speed of the disk? rad/s

To solve the problem it is necessary to apply Ohm's law. From this it is established that the voltage is the equivalent to the product between the current and the resistance, therefore we have to,

[tex]V = IR[/tex]

Here,

V = Voltage

I = Current

R = Resistance

Rearranging to find the resistance,

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

Replacing,

[tex]R = \frac{100V}{20A}[/tex]

[tex]R = 5\Omega[/tex]

Therefore the resistance should be [tex]5\Omega[/tex]

The resistance should be [tex]\bold { 5\Omega}[/tex] in order to limit the emf to 100 V.

Ohm's law:

It states that the voltage is the equival to the product of the current and the resistance.

[tex]\bold{V =I\times R}[/tex]

Where,

V = Voltage = 100 Volts

I = Current  = 20 Ampere

R = Resistance

Put the values and solve it for R

[tex]\bold {R =\dfrac{100}{20}}\\\\\bold {R = 5\Omega}[/tex]

Therefore, the resistance should be [tex]\bold { 5\Omega}[/tex] in order to limit the emf to 100 V.

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Final answer:

To hold the projectile in position, the force required is 20 Newtons.

Explanation:

To calculate the force required to hold the projectile in position, we need to consider the acceleration of the projectile in the pipe.

From the given information, we can calculate the acceleration using the formula:

Acceleration = Change in velocity / Time taken

Since the mean velocity is given as 6 m/s, we can use this as the final velocity and assume the initial velocity is 0 m/s.

Substituting the values in the formula, we have:

Acceleration = (6 m/s - 0 m/s) / 0.3 m = 20 m/s^2

Now we can calculate the force using Newton's second law:

Force = Mass * Acceleration

Assuming the mass of the projectile is 1 kg, we have:

Force = 1 kg * 20 m/s^2 = 20 N

The electric potential at point A in the electric field= 0.099 x 10 ⁻¹v

Explanation:

Given data,

charge = 5.5 x 10¹² C

k =9.00 x 10⁹

The electric potential V of a point charge can found by,

V= kQ / r

Assuming, r=5.00×10⁻² m

V= 5.5 x 10⁻¹²C x  9.00 x 10⁹ / 5.00×10⁻² m

V=  49.5 x 10⁻³/ 5.00×10⁻²

Electric potential V=  0.099 x 10⁻¹v

The electric potential at a point in the electric field created by a point charge is calculated by the formula V = kQ/r. The provided constants are [tex]Q = 5.5 \times 10^{-12}\, \text{C}[/tex] and [tex]k = 9.00 \times 10^9[/tex] The distance r from the point charge is necessary to calculate the electric potential.

The electric potential at point A in the electric field created by a point charge can be found using the formula V = kQ/r, where V is the electric potential, k is Coulomb's constant, Q is the value of the point charge, and r is the distance from the point charge.

Given, the charge [tex]Q = 5.5 \times 10^{-12}\, \text{C}[/tex], and Coulomb's constant [tex]k = 9.00 \times 10^9[/tex]

The distance r is not given in the question. Therefore, we can't calculate the electric potential at point A. However, if distance r from the point charge were provided (in meters), we could substitute k, Q, and r into the formula and solve for V.

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Answer:

(a)  277.05 Ω

(b) 15.39 Ω

(c) 3.76 W

Explanation:

(a)

Applying,

P = V²/R.......................... Equation 1

Where P = Power dissipated by the string. V = Voltage source, R = equivalent resistance of the light string

Make R the subject of the equation

R = V²/P................... Equation 2

Given: V = 130, P = 61  W

Substitute into equation 2

R = 130²/61

R = 277.05 Ω

(b) The resistance of a single light is given as

R' = R/18 (since the light are connected in series and the are identical)

Where R' = Resistance of the single light.

R' = 277.05/18

R' = 15.39 Ω

(c)

Heat dissipated in a single light is given as

P' = I²R'..................... Equation 3

Where P' = heat dissipated in a single light, I = current flowing through each light.

We can calculate for I using

P = VI

make I the subject of the equation

I = P/V

I = 61/130

I = 0.469 A.

Also given: R' = 15.39 Ω

Substitute into equation 3

P' = 0.496²(15.39)

P' = 3.76 W

(a)The equivalent resistance of the light string is 277.05 Ω

(b)The power is dissipated in a single ligh15.39 Ω

(c)The resistance of the light string now3.76 W

(a) P is = V²/R.......................... Equation 1

Where P is = Power dissipated by the string.

Then V = Voltage source,

After that R is = the equivalent resistance of the light string

Now Make R the subject of the equation

R is = V²/P................... Equation 2

Then Given: V = 130,

P = 61 W

After that Substitute into equation 2

Then R = 130²/61

Therefore, R = 277.05 Ω

(b) When The resistance of a single light is given as

R' is = R/18 (since the light are connected in series and are identical)

Now Where R' is = Resistance of the single light.

R' is = 277.05/18

Therefore, R' is = 15.39 Ω

(c) When Heat dissipated in a single light is given as

P' is = I²R'..................... Equation 3

Where P' is = heat dissipated in a single light,

Then I = current flowing through each light.

Now We can calculate for I using

P is = VI

Now we make I the subject of the equation

After that I = P/V

Then I = 61/130

I is = 0.469 A.

Also given: R' is = 15.39 Ω

Then Substitute into equation 3

P' is = 0.496²(15.39)

Therefore, P' is = 3.76 W

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Answer:

Explanation:

Given that,

Initial speed of the girl is

u = 1.4m/s

Height she is going is

H = 2.45m

Incline plane she will pass to that height

L = 12.4m

Mass of girl and bicycle is

M=60kg

Frictional force that oppose motion is

Fr = 41N

Speed at lower end of inclined plane

V2 = 6.7m/s

Work done by the girl when the car travel downward

Using conservation of energy

K.E(top) + P.E(top) + work = K.E(bottom) + P.E(bottom) + Wfr

Where Wfr is work done by friction

Wfr = Fr × d

P.E(bottom) is zero, sicne the height is zero at the ground

K.E is given as ½mv²

Then,

½M•u² + MgH + W = ½M•V2² + 0 + Fr×d

½ × 60 × 1.4² + 60×9.8 × 2.45 + W = ½ × 60 × 6.7² + 41 × 12.4

58.8 + 1440.5 + W = 1855.1

W = 1885.1 —58.8 —1440.5

W = 355.8 J

Answer:

a)  the geometric optics is adequate (Ray)

b) wave optics should be used for the second case

Explanation:

In general, the approximation of the geometric optics is adequate when the dimensions of the system are much greater than the wavelengths and the wave optics should be used for cases in which the size of the system is from the length of the wave.

Let's apply this to our case

a) in this case the size of the system is d = 26 cm=0.26 m and the wavelength is alm = 530 10⁺⁹ m

   in this case

                    d >>> λ

therefore the geometric optics is adequate (Ray)

b) in this case the system has a size of d = 114 10⁻⁹ m with a wavelength of λ= 722 nm

for this case d of the order of lam

therefore wave optics should be used for the second case

To solve this problem we will apply the concepts related to angular resolution based on wavelength and the slit width at this case of the cat's pupils. This relationship is given as,

[tex]\theta = \frac{\lambda}{a}[/tex]

Here,

[tex]\lambda[/tex] =  Wavelength

a = Slit width

[tex]\theta = \frac{519*10^{-9}m}{0.55*10^{-3}m}[/tex]

[tex]\theta = 9.43*10^{-4} Rad[/tex]

Therefore the angular resolution is [tex]9.43*10^{-4}rad[/tex]

Final answer:

The angular resolution for a cat observing birds can be estimated using the formula θ = 1.22 λ / D, with a pupil width of 0.550 mm and a wavelength of 519 nm, yields an angular resolution of approximately 1.152 × 10⁻⁶ radians.

Explanation:

The question refers to angular resolution in optics, specifically related to the diffraction limit of a cat's eyes observing birds. In physics, the angular resolution for a circular aperture, like the pupil of a cat's eye, can be estimated using the formula θ = 1.22 λ / D, where θ is the angular resolution in radians, λ is the wavelength of the light, and D is the diameter of the aperture (the pupil in this case). Given that the cat's pupil has a slit width of a = 0.550 mm (which we will use in place of the diameter for this rough estimate) and the light has an average wavelength of λ = 519 nm, the angular resolution θ can be calculated as follows:

θ = 1.22 × 519 x 10⁻⁹ m / 0.550 x 10^-3 m

θ = 1.22 × 519 / 550 × 10⁻⁶

θ = 1.22 × 0.944 × 10⁻⁶

θ = 1.152 × 10⁻⁶ radians

Given the specified conditions, this calculation estimates the angular resolution for the cat's eyes. Note, however, that this is a simplification and doesn't take into account the complexities of a vertical slit pupil versus a circular aperture.

The correct question is;

An object of mass m attached to a spring of force constant K oscillates with simple harmonic motion. The maximum displacement from equilibrium is A and the total mechanical energy of the system is E.

What is the system's potential energy when its kinetic energy is equal to ¾E?

Answer:

P.E = ⅛KA²

Explanation:

From conservation of energy, the total energy in the system is given as the sum of potential and kinetic energy.

Thus,

Total Energy; E = K.E.+P.E.

In simple harmonic motion, the total energy is given by;

E = ½KA²

We are told that kinetic energy is ¾E.

Thus, ½KA² = ¾(½KA²) + P.E

P.E = ½KA² - ⅜KA²

P.E = ⅛KA²

The maximum possible wavelength for constructive interference from two shifted loudspeakers at given interference maxima positions is calculated to be 0.5738 m.

The question is about the phenomenon of constructive interference of sound waves emitted from two loudspeakers that are shifted apart. The problem requires identifying the maximum possible wavelength of the sound waves produced by the speakers.

Constructive interference occurs when two waves meet and their crests (high points) and troughs (low points) align. This occurs effectively when the path difference is either zero or a multiple of the wavelength. When the path difference is specifically an integral multiple of the wavelength, the interference is constructive and has a maximum value.

Given that the loudspeakers are moved to a distance of 3.0 m sideways and 7.0 m forward, the distance separating the two is calculated using Pythagoras's theorem as √((7)^2+(3)^2) = √58 m. The maxima of the constructive interference are at positions 14, 12, and 34 the distance between the speakers, thus the maximum possible wavelength is 2 * √58 / 34 = 0.5738 m.

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Answer:

The linear mass  density is of the string  [tex]\mu= 5.51*10^{-3} kg / m[/tex]

Explanation:

    From the question we are told that

            The distance between wall and pulley is [tex]d = 0.405m[/tex]

            The mass on the hook is [tex]m = 25.4\ kg[/tex]

            The frequency of oscillation is  [tex]f = 261.6 Hz[/tex]

    Generally, the frequency of oscillation is mathematically  represented as

            [tex]f = \frac{1}{2d} \sqrt{\frac{T}{\mu} }[/tex]

Where T is the tension mathematically represented as

                T = mg

Substituting values

             [tex]T = 25.4 *9.8[/tex]

                [tex]=248.92N[/tex]

   [tex]\mu[/tex] is the mass linear density

Making  [tex]\mu[/tex] the subject of the formula above

            [tex]\mu = \frac{T}{(2df)^2}[/tex]

Substituting values

         [tex]\mu = \frac{248.92}{(2 * 0.405 * 261.6)^2}[/tex]

           [tex]\mu= 5.51*10^{-3} kg / m[/tex]

Answer:

0.005550 Kg/m

Explanation:

The picture attached below shows the full explanation

Answer:

R

Explanation:

Given that,

Two stage rocket traveling at

V = 1210 m/s with respect to earth

First stage

When fuel Is run out , explosive bolts releases and push rocket backward at speed is

V1 = 40m/s relative to second stage

Therefore

V1 = 40 - V2

The first stage is 3 times as massive as the second stage

I.e Mass of first stage is 3 times the second stage

Let Mass of second stage be

M2 = M

Then, M1 = 3M

Velocity of second stage V2?

Applying conservation of linear momentum

Momentum before explosion = momentum after explosion

Momentum p=mv

Then,

(M1+M2)V = —M1•V1 + M2•V2

(3M+M)•1210 = —3M•(40-V2) +M•V2

4M × 1210 = —120M + 3M•V2 + MV2

4840M = —120M + 4M•V2

4840M + 120M = 4M•V2

4960M = 4M•V2

Then, V2 = 4960M / 4M

V2 = 1240 m/s

Answer:

(a)  change in the internal energy of the gas is zero

(b) the work done by the gas is 16.93 kJ

(c) the heat flow is 16.93 kJ, which is into the gas

Explanation:

Given;

number of moles of gas, n = 6.35 moles

temperature of the gas, T = 320 K

initial volume of the gas, V₁ = 1.45 L

final volume of the gas, V₂ = 3.95 L

Part (a)

For isothermal expansion, temperature is constant and internal energy will also be constant.

Therefore, change in the internal energy of the gas is zero since the gas expanded isothermally (constant temperature).

ΔU = Q - W

where;

ΔU is change in internal energy

Q is heat transferred to the system

W is the work done by the system

Thus, Q = W

ΔU = 0

Part (b)

the work done by the gas

[tex]W = nRTln{[\frac{V_2}{V_1}][/tex]

where;

R is gas constant = 8.314 J/mol.K

[tex]W = (6.35)(8.314)(320)ln{[\frac{3.95}{1.45}]}\\\\W =16930.4\ J\\\\W = 16.93\ kJ[/tex]

Part (c)

the heat flow into or out of the gas

Q = ΔU + W

Q = 0 + 16.93 kJ

Q = 16.93 kJ

Since the heat flow is positive, then it is heat flow into the gas.

Answer:

Explanation:

In a scenario where a particle of charge overrightarrow{B} enters a magnetic field with a velocity overrightarrow{V}, it experiences a force overrightarrow{F} given by: overrightarrow{F}=q(\overrightarrow{V}\times \overrightarrow{B}).

implies F=BqVSin\theta.

Where theta is the angle between the velocity vector of the particle and the magnetic field vector.

When a particle enters the magnetic field at an angle 90, it moves in a circular path as it experiences a centripetal force, given by F=\frac{mV^2}{R}.

Where R is the radius of the circle, V is its velocity and m is its mass

Thus, magnetic force becomes F=BqVSin90^o=\frac{mV^2{R}\implies R=\frac{mV}{Bq}.

The equation changes as below, when velocity is doubled, let us assume that the radius is given by R_1.

R_1=\frac{2mV}{Bq}=2R.

Therefore, it is obvious that the velocity of a charged particle in a circular arc is directly proportional to the radius of the arc. The radius of the circular arc doubles when the velocity of the charged particle in the circular orbit doubles only if the mass, charge and magnetic field of the particle remains constant

Hence when velocity is doubled radius of the circle also gets doubled.

When the speed of the particle doubles, the radius of the arc also doubles.

The magnetic force on the particle is calculated as follows;

[tex]F = qvB[/tex]

The centripetal force on the particle is calculated as follows;

[tex]F_c = \frac{mv^2}{R}[/tex]

The speed of the particle is calculated as follows;

[tex]\frac{mv^2}{R} = qvB\\\\mv = qBR\\\\v = \frac{qBR}{m} \\\\\frac{v_1}{R_1} = \frac{v_2}{R_2}[/tex]

when the speed of the particle doubles;

[tex]\frac{v_1}{R_1} = \frac{2v_1}{R_2} \\\\R_2v_1 = 2R_1v_1\\\\R_2 = 2R_1[/tex]

Thus, we can conclude that when the speed of the particle doubles, the radius of the arc also doubles.

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Answer:

Catastrophic events of weather related outcomes.

Explanation:

The mental map of a person from a certain place may change due to long periods of time outside, since this gives the mind time to forget certain important details, they may also change according to people's experience and perception, places, regions and environments since these places are also changing and the perception we had is no longer the same. A mental map is a first person perspective of an area that an individual possesses. This type of subconscious map shows a person how a place looks and how to interact with it.

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Answer:

1. downward

2. to the left

3. downward and to the left

Explanation:

This is gotten by using vector law of triangle Addition which states that If 2 vectors acting simultaneously on a body are represented both in magnitude and direction by 2 sides of a triangle taken in an order then the resultant(both magnitude and direction) of these vectors is given by 3rd side of that triangle taken in opposite order.

Answer:

Δx = 3.477 x 10⁻³ m = 3.477 mm

Explanation:

The distance between two consecutive dark fringes is given by the following formula, in Young's Double Slit experiment:

Δx = λL/d

where,

Δx = distance between two consecutive dark fringes = ?

λ = wavelength of light = 4.9 x 10² nm = 4.9 x 10⁻⁷ m

L = Distance between slits and screen = 2.2 m

d = slit separation = 0.31 mm = 0.31 x 10⁻³ m

Therefore,

Δx = (4.9 x 10⁻⁷ m)(2.2 m)/(0.31 x 10⁻³ m)

Δx = 3.477 x 10⁻³ m = 3.477 mm

The distance (in mm) between the first and second dark fringes of the

interference pattern is 3.477 mm

This is calculated by using the formula in Young's Double Slit experiment:

Δx = λL/d

where,

Δx = distance between two consecutive dark fringes which is unknown

λ which is wavelength of light = 4.9 x 10² nm = 4.9 x 10⁻⁷ m

L which is distance between slits and screen = 2.2 m

d which is slit separation = 0.31 mm = 0.31 x 10⁻³ m

We then substitute them into the equation

Δx = (4.9 x 10⁻⁷ m ×2.2 m)/(0.31 × 10⁻³ m)

Δx = 3.477 x 10⁻³ m = 3.477 mm

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To solve this problem we will apply the concepts related to the principle of destructive and constructive interference. Mathematically this expression can be given as

[tex]2nt = m\lambda[/tex]

Here,

n = Index of refraction

t = Thickness

m = Order of the reflection

[tex]\lambda[/tex] = Wavelength

We have all of this values, therefore replacing,

[tex]2(1.28)t = (1)(600nm)[/tex]

[tex]t = 233nm[/tex]    

Therefore the thickness of the oil slick is 233nm

Answer:

the density of a blue block is ___twice _____ the density of a red block.

Explanation:

If both blocks have the same mass, then the block with the lesser volume will be more densely packed when compared to the block with the larger volume. This is because the molecules are more closely packed.

Check the image below for detailed calculations for proof

If the blue block and red block have the same mass but different volumes - specifically, the blue block's volume is half that of the red block - then the density of the blue block is twice that of the red block.

Given the blocks weigh the same, their masses are equal. Density is defined as mass per unit volume (mass/volume), meaning the density is influenced by both the mass and the volume of an object. As the red and blue blocks have different volumes, their densities must be different if their masses are the same.

In the case of the blue block and the red block, the blue block has half the volume of the red block but the same mass. Thus, because the blue block's volume is half that of the red block but the mass is the same, the density of the blue block is twice the density of a red block, not half. This is because the proportion of mass to volume in the blue block is larger than in the red block. In other words: Density of blue block = 2 * Density of red block.

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Answer:

The total length of the spring would be 0.65 m

Explanation:

The Concept

Hooke's law evaluates the increment of  spring in relation to the force acting on the body. Hooke's law states that for a spring undergoing deformation, the  force applied is directly proportional to the deformation experienced by the spring. Hooke's law is represented thus;

F = k x ..................1

where F is the force applied to the spring

k is the spring constant

x is the spring stretch or extension

Step by Step Calculations

We have to obtain x before adding it to the nominal length, We make x the subject formula in equation 1;

x = F/k

but F = m x g

so, x = (m x g)/k

given that, the mass of the person m =150 kg

g is the acceleration due to gravity = 9.81 m/[tex]s^{2}[/tex]

k is the spring constant = 10000 N/m

then x = (9.81 m/[tex]s^{2}[/tex] x 150 kg)/10000 N/m

x = 0.14715 m

the extension experienced by the spring after the compression is 0.14715 m

The total length of the spring would be;

L = 0.14715 m + 0.5 m = 0.64715

L ≈  0.65 m

Therefore the total length of the spring would be 0.65 m

Answer:

how large a magnetic field would you experience = 8.16 x 10∧-4T

Explanation:

I = 20KA = 20,000A

r = 4.9 m

how large a magnetic field would you experience =  u.I/2πr

how large a magnetic field would you experience = (4π x10∧-7) × 20000/2π × 4.9

how large a magnetic field would you experience = 8.16 x 10∧-4T

Answer: 8.16*10^-4 T

Explanation:

Given

Current of the lightening bolt, I = 20 kA

Distance from the strike of the lightening bolt, r = 4.9 m

To solve, we use the formula

B = [μ(0) * I] / 2πr, where

B = magnetic field of the lightening

μ = permeability constant = 4π*10^-7 N/A²

I = current of the lightening

r = distance from the lightening strike

B = [(4 * 3.142*10^-7) * 20*10^3] / (2 * 3.142 * 4.9)

B = (12.568*10^-7 * 20*10^3) / 6.284 * 4.9

B = 0.025 / 30.79

B = 8.16*10^-4 T

The magnetic field to be experienced would be 8.16*10^-4 T large

Answer:

Look up attached file

Explanation:

To find the time it takes for the current to drop to 12% of its value and the energy dissipated by the circuit, we can use the equations for the decay of current in an RL circuit and the energy in an inductor respectively.

To determine the time it takes for the current in the circuit to drop to 12% of its value at t = 0, we need to use the equation for the decay of current in an RL circuit. The equation is given by I(t) = I(0) * exp(-t/τ), where I(t) is the current at time t, I(0) is the initial current, t is the time, and τ is the time constant. In this case, we can use I(t) = I(0) * exp(-t/τ) = 0.12 * I(0). To find the time, we rearrange the equation as t = -τ * ln(0.12).

The energy dissipated by the circuit can be calculated using the equation for the energy in an inductor, which is given by E = 1/2 * L * I(0)^2, where E is the energy, L is the inductance, and I(0) is the initial current. In this case, we can substitute the values given to find the energy dissipated.

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The time needed for the current in the circuit to drop to 12% of its value at t = 0 is approximately 20.63 ms. The energy dissipated in the circuit during that time is approximately 7.22 mJ.

Given, initial current I0 can be obtained from Ohm's law, I0 = V / R = 108 / 1100 = 0.09818 A. The circuit value diminishes exponentially over time and can be expressed as I = I0 * e-Rt/2L. Therefore, to find out when the current drops to 12% of its initial value we set I / I0 = 0.12 = e-Rt/2L, and solve for t. After calculation, we find t ≈ 20.63 ms.

As for part (b), the energy dissipated in an RL circuit over time is given by W = 1/2 * L * I2, where I is the current at time t, which is given by the relation: I = I0 * e-Rt/2L. Performing the integration over the time period t = 0 to 20.63 ms, we find that the energy dissipated is approximately 7.22 mJ.

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Answer:

Explanation: check the attached document for step by step solution.

Answer:

[tex]6.75\mu C/m^2[/tex]

Explanation:

We are given that

Diameter,d=[tex]1\mu m=1\time 10^{-6} m[/tex]

[tex]1\mu m=10^{-6} m[/tex]

Radius,r=[tex]\frac{d}{2}=\frac{1}{2}\times 10^{-6}=0.5 \times 10^{-6} m[/tex]

Density,[tex]\rho=900kg/m^3[/tex]

Total number of electrons,n=39

Charge on electron =[tex]1.6\times 10^{-19} C[/tex]

Total charge=[tex]q=ne=39\times 1.6\times 10^{-19}=62.4\times 10^{-19} C[/tex]

Distance,s=2mm=[tex]2\times 10^{-3} m[/tex]

Mass =[tex]density\times volume=900\times \frac{4}{3}\pi r^3=900\times \frac{4}{3}\pi(0.5\times 10^{-6})^3=4.7\times 10^{-16} kg[/tex]

Initial velocity,u=0

Final speed,v=4.5 m/s

[tex]v^2-u^2=2as[/tex]

[tex](4.5)^2-0=2a(2\times 10^{-3})[/tex]

[tex]20.25=4a\times 10^{-3}[/tex]

[tex]a=\frac{20.25}{4\times 10^{-3}}=5062.5m/s^2[/tex]

Force,F=ma

[tex]qE=ma[/tex]

[tex]q(\frac{\sigma}{2\epsilon_0})=ma[/tex]

[tex]\sigma=\frac{2\epsilon_0ma}{q}=\frac{2\times 8.85\times 10^{-12}\times 4.7\times 10^{-16}\times 5062.5}{62.4\times 10^{-19}}[/tex]

[tex]\epsilon_0=8.85\times 10^{-12}[/tex]

[tex]\sigma=6.75\times 10^{-6}C/m^2=6.75\mu C/m^2[/tex]

To solve this problem we must find the values of the equivalent resistances in both section 1 and section 2. Later we will calculate the total current and the total voltage. With the established values we can find the values of the currents in the 3 Ohms resistance and the power there.

The equivalent resistance in section 1 would be

[tex]R_{eq1} = \frac{(3\Omega)(6\Omega)}{(3+6)\Omega}[/tex]

[tex]R_{eq1} = 2\Omega[/tex]

The equivalent resistance in section 2 would be

[tex]R_{eq2} = R_{eq1} +4\Omega[/tex]

[tex]R_{eq2} = 6\Omega[/tex]

Now the total current will be,

[tex]I_t = \frac{V_t}{R_{eq2}}[/tex]

[tex]I_t = \frac{12V}{6\Omega}[/tex]

[tex]I_t = 2.0A[/tex]

Finally the total Voltage will be,

[tex]V = IR_{eq1}[/tex]

[tex]V = (2.0A)(2.0\Omega)[/tex]

[tex]V = 4V[/tex]

Since the voltage across the 3 and 6 Ohms resistor is the same, because they are in parallel, the current in section 3 would be

[tex]I_{3.0\Omega} = \frac{V}{R}[/tex]

[tex]I_{3.0\Omega} = \frac{4.0V}{3.0\Omega}[/tex]

[tex]I_{3.0\Omega} = 1.3A[/tex]

Finally the power ratio is the product between the current and the voltage then,

[tex]P_{3.0\Omega} = I_{3.0\Omega} V[/tex]

[tex]P_{3.0\Omega} = (1.3A)(4.0V)[/tex]

[tex]P_{3.0\Omega} = 5.3W[/tex]

Therefore the correct answer is D.

Final answer:

To find the power dissipated in the 3.0-Ω resistor, calculate the total resistance and current, then apply Ohm's law to find the voltage across and current through the resistor. Finally, use the power formula P = V²/R, resulting in a power dissipation of (a) 12 W for the 3.0-Ω resistor.

Explanation:

The question deals with finding the power dissipated in the 3.0-Ω resistor. To answer this, we first need to find the total resistance of the circuit and the current through the circuit. We then apply this current to the parallel resistors to find the voltage across and the current through the 3.0-Ω resistor before calculating its power dissipation.

First, calculate the resistance of the resistors in parallel. Using the formula for resistors in parallel:

1/Rparallel = 1/R₁ + 1/R₂

1/Rparallel = 1/3.0 + 1/6.0

1/Rparallel = 1/3 + 1/6 = 2/6 + 1/6 = 3/6

1/Rparallel = 1/2

Rparallel = 2.0 Ω

Now, add the series resistor to find the total resistance:

Rtotal = Rparallel + Rseries

Rtotal = 2.0 + 4.0 = 6.0 Ω

Then, using Ohm's law, calculate the total current from the battery:

I = V/Rtotal

I = 12V/6.0Ω

I = 2.0 A

The current through the 3.0-Ω resistor in parallel is the same as the total current, so we use Ohm's law V = IR to find the voltage across the 3.0-Ω resistor:

V3.0-Ω = 2.0 A × 3.0 Ω = 6.0 V

Finally, calculate the power dissipated by the 3.0-Ω resistor:

P = V2/R

P = (6.0 V)2/3.0 Ω

P = 36 W/3.0 Ω

P = 12 W

Therefore, the power dissipated in the 3.0-Ω resistor is 12 W, which corresponds to choice (a).

Complete Question

 The complete question is shown on the first uploaded image

Answer:

The magnetic field is [tex]B_{net} = \frac{1}{4} * mT[/tex]

And the direction is  [tex]-\r k[/tex]

Explanation:

      From the question we are told that

                 The magnetic field at the center is [tex]B = 1mT[/tex]

Generally magnetic field is mathematically represented as

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

We are told that it is equal to 1mT

So

                [tex]B = \frac{\mu_o I}{2R} = 1mT[/tex]

From the first diagram we see that the effect of the current flowing in the circular loop is  (i.e the magnetic field generated)

                         [tex]\frac{\mu_o I}{2R} = 1mT[/tex]

 This implies that the effect of a current flowing in the smaller semi-circular loop is (i.e the magnetic field generated)

                   [tex]B_1 = \frac{1}{2} \frac{\mu_o I}{2R}[/tex]

and  for the larger semi-circular loop  is

                 [tex]B_2 = \frac{1}{2} \frac{\mu_o I}{2 * (2R)}[/tex]

Now a closer look at the second diagram will show us that the current in the semi-circular loop are moving in the opposite direction

    So the net magnetic field would be

                   [tex]B_{net} = B_1 - B_2[/tex]

                        [tex]= \frac{1}{2} \frac{\mu_o I}{2R} -\frac{1}{2} \frac{\mu_o I}{2 * (2R)}[/tex]

                        [tex]=\frac{\mu_o I}{4R} -\frac{\mu_o I }{8R}[/tex]

                        [tex]=\frac{\mu_o I}{8R}[/tex]

                        [tex]= \frac{1}{4} \frac{\mu_o I}{2R}[/tex]

Recall  [tex]\frac{\mu_o I}{2R} = 1mT[/tex]

    So  

             [tex]B_{net} = \frac{1}{4} * mT[/tex]

Using the Right-hand rule we see that the direction is into the page which is [tex]-k[/tex]

Answer:

option (b)

Explanation:

mass of proton, mp = m

mass of deuteron, md = 2m

charge on proton, qp = q

charge on deuteron, qd = q

The magnetic force on the charged particle when it is moving is given by

F = q v B Sinθ

where, θ is the angle between the velocity and magnetic field.

Here, θ = 90°

Let v is the velocity of both the particle when they enters in the magnetic field.

The force on proton is given by

Fp = q x v x B ...... (1)

The force on deuteron is

Fd = q x v x B .... (2)

Divide equation (1) by equation (2)

Fp / Fd = 1

Thus, the ratio of force on proton to the force on deuteron is 1 : 1.

Thus, option (b) is correct.

Answer:

[tex]\frac{g_{2}}{g_{1}} = \frac{1}{4}[/tex]

Explanation:

The period of the simple pendulum is:

[tex]T = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex]

Where:

[tex]l[/tex] - Cord length, in m.

[tex]g[/tex] - Gravity constant, in [tex]\frac{m}{s^{2}}[/tex].

Given that the same pendulum is test on each planet, the following relation is formed:

[tex]T_{1}^{2}\cdot g_{1} = T_{2}^{2}\cdot g_{2}[/tex]

The ratio of the gravitational constant on planet CornTeen to the gravitational constant on planet Earth is:

[tex]\frac{g_{2}}{g_{1}} = \left(\frac{T_{1}}{T_{2}} \right)^{2}[/tex]

[tex]\frac{g_{2}}{g_{1}} = \left(\frac{2\,s}{4\,s} \right)^{2}[/tex]

[tex]\frac{g_{2}}{g_{1}} = \frac{1}{4}[/tex]

Answer:

larger, because her angular speed is larger.

Explanation:

The rotational kinetic energy is proportional to the square of the angular velocity while it is linearly proportional to the moment of inertia. So the increase of angular speed will have a larger effect of the kinetic energy than the decrease of the moment of inertia.

Answer:

Rotational kinetic energy will Increase

Explanation:

Rotational kinetic energy KE is

KE = 1/2 x I x w^2

Where I is moment of inertia,

w is angular velocity.

It can be seen that increasing angular velocity increases rotational kinetic energy.