A dipole is oriented along the x axis. The dipole moment is p (= qs). (Assume the center of the dipole is located at the origin with positive charge to the right and negative charge to the left.)
Calculate exactly the potential V (relative to infinity) at a location x, 0, 0 on the x axis and at a location 0, y, 0 on the y axis, by superposition of the individual 1/r contributions to the potential. (Use the following as necessary: q, ε0, x, s and y.)

Answer:

y = 20.99 V / A

there is no friction    y = 20.99 h

Explanation:

Let's solve this exercise in parts: first find the thrust on the block when it is submerged and then use the conservation of energy

when the block of ice is submerged it is subjected to two forces its weight  hydrostatic thrust

              F_net= ∑F = B-W

the expression stop pushing is

              B = ρ_water g V_ice

where rho_water is the density of pure water that we take as 1 10³ kg / m³ and V is the volume d of the submerged ice

We can write the weight of the body as a function of its density rho_hielo = 0.913 10³ kg / m³

             W = ρ-ice g V

              F_net = (ρ_water - ρ_ ice) g V

this is the net force directed upwards, we can find the potential energy with the expression

            F = -dU / dy

            ΔU = - ∫ F dy

            ΔU = - (ρ_water - ρ_ ice) g ∫ (A dy) dy

            ΔU = - (ρ_water - ρ_ ice) g A y² / 2

we evaluate between the limits y = 0,  U = 0, that is, the potential energy is zero at the surface

             U_ice = (ρ_water - ρ_ ice) g A y² / 2

now we can use the conservation of mechanical energy

starting point. Ice depth point

             Em₀ = U_ice = (ρ_water - ρ_ ice) g A y² / 2

final point. Highest point of the block

             [tex]Em_{f}[/tex] = U = m g y

as there is no friction, energy is conserved

            Em₀ = Em_{f}

            (ρ_water - ρ_ ice) g A y² / 2 = mg y

let's write the weight of the block as a function of its density

            ρ_ice = m / V

            m = ρ_ice V

we substitute

             (ρ_water - ρ_ ice) g A y² / 2 = ρ_ice V g y

              y = ρ_ice / (ρ_water - ρ_ ice) 2 V / A

let's substitute the values

             y = 0.913 / (1 - 0.913) 2 V / A

             y = 20.99 V / A

This is the height that the lower part of the block rises in the air, we see that it depends on the relationship between volume and area, which gives great influence if there is friction, as in this case it is indicated that there is no friction

                V / A = h

where h is the height of the block

                 y = 20.99 h

Answer:

Explanation:

Concave mirrors is otherwise known as converging mirrors: These are mirrors that are caved inwards (reflecting surface is on the outside curved part). It is called a converging mirror due to the fact that light converges to a point when it strikes and reflects from the surface of the mirror. This type of mirror is used to focus light; parallel rays that are directed towards it will be concentrated to a point.

For a concave mirror to reflect light with properties that are the same as a spotlight (directed light rays parallel to each other), one has to consider its property to gather light to a point after reflecting. Meaning that, we can achieve the spotlight by locatng the point where the rays will be parallel, this point is called the focal point.

Therefore, the light bulb should be placed at the focal point of the mirror.

Answer:

Explanation:

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

q is the charge on the electron

v is the velocity of the electron

B is the magnetic field strength

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

Given parameters

F =  1.40*10⁻¹⁶ N

q = 1.6*10⁻¹⁹C

v = 3.94*10³m/s

B = 1.23T

Required

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

Substituting the given parameters into the formula:

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

1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁹⁺³sinθ

1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁶sinθ

sinθ = 1.40*10⁻¹⁶/7.75392 * 10⁻¹⁶

sinθ = 1.40/7.75392

sinθ = 0.1806

θ = sin⁻¹0.1806

θ₁ = 10.4⁰

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

θ₂ = 180-10.4

θ₂ = 169.6⁰

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

Answer:

The number of turns of the inductor is 2000 turns.

Explanation:

Given;

emf of the inductor, E = 0.8 V

the rate of change of current with time, dI/dt = 10 A/s

steady current in the solenoid, I = 0.2 A

flux per turn, Ф = 8.0 μWb per

Determine the inductance of the solenoid, L

E = L(dI/dt)

L = E / (dI/dt)

L = 0.8 / (10)

L = 0.08 H

The inductance of the solenoid is given by;

[tex]L = \frac{\mu_o N^2 A}{l}[/tex]

Also, the magnetic field of the solenoid is given by;

[tex]B = \frac{\mu_o NI}{l}[/tex]

I is 0.2 A

[tex]B = \frac{\mu_oN(0.2)}{l} = \frac{0.2\mu_o N}{l}[/tex]

[tex]\frac{B}{0.2 } = \frac{\mu_o N}{l}[/tex]

[tex]L = \frac{\mu_o N^2 A}{l} \\\\L = \frac{\mu_o N }{l} (NA)\\\\L = \frac{B}{0.2} (NA)\\\\L = \frac{BA}{0.2} (N)[/tex]

But Ф = BA

[tex]L = \frac{\phi N}{0.2} \\\\\phi N = 0.2 L\\\\N = \frac{0.2 L}{\phi} \\\\N = \frac{0.2 *0.08}{8*10^{-6}}\\\\N = 2000 \ turns[/tex]

Therefore, the number of turns of the inductor is 2000 turns.

This question involves the concepts of magnetic flux, magnetic field, and inductance.

The inductor has "2000" turns.

The magnetic field due to an inductor coil is given as follows:

[tex]B=\frac{\mu_o NI}{L}\\\\[/tex]

where,

B = magnetic field

μ₀ = permeability of free space \

N = No. of turns

I = current = 0.2 A

L = length of inductor

Therefore,

[tex]\frac{\mu_oN}{L}=\frac{B}{0.2\ A}---------- eqn(1)[/tex]

Now, the inductance of a solenoid is given by the following formula:

[tex]E = L\frac{dI}{dt}\\\\L = \frac{E}{\frac{dI}{dt}}[/tex]

The inductance of solenoid can also be given using the following formula:

[tex]L = \frac{\mu_o N^2A}{L}[/tex]

comparing both the formulae, we get:

[tex]\frac{E}{\frac{dI}{dt}}= \frac{\mu_oN^2A}{L}\\\\E=\frac{dI}{dt}\frac{\mu_oN}{l}(NA)\\\\using\ eqn (1):\\\\E=\frac{dI}{dt}\frac{B}{0.2}(NA)\\\\[/tex]

where,

BA = magnetic flux = [tex]\phi[/tex] = 8 μWb/turn = 8 x 10⁻⁶ Wb/turn

N = No. of turns = ?

E = E.M.F = 0.8 volts

[tex]\frac{dI}{dt}[/tex] = rate of change in current = 10 A/s

Therefore,

[tex]0.8=(10)\frac{8\ x\ 10^{-6}}{0.2}N\\\\N=\frac{(0.8)(0.2)}{8\ x\ 10^{-5}}[/tex]

N = 2000 turns

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The attached picture shows the magnetic flux.

Answer:

9.88 milivolt

Explanation:

Given: diameter d = 5.2 cm

magnetic field B_1 = 1.35 T, final magnetic field B_2 =0 T

t = 0.29 sec.

we know emf =  - dΦ/dt

and flux Φ = BA

A= area

therefore emf ε = -A(B_2-B_1)/Δt

[tex]=-\pi(d/2)^2\frac{B_2-B_1}{\Delta t} \\=-\pi(0.052/2)^2\frac{0-1.35}{0.29} \\=98.8\times10^4\\=9.88 mV[/tex]

Answer:

c. 70 Ω

Explanation:

The R and R resistors are in parallel.  The 2R and 2R resistors are in parallel.  The 4R and 4R resistors are in parallel.  Each parallel combination is in series with each other.  Therefore, the equivalent resistance is:

Req = 1/(1/R + 1/R) + 1/(1/2R + 1/2R) + 1/(1/4R + 1/4R)

Req = R/2 + 2R/2 + 4R/2

Req = 3.5R

Req = 70Ω

Answer:

-0.73mA

Explanation:

Using amphere's Law

ε =−dΦB/ dt

=−(2.6T)·(7.30·10−4 m2)/ 1.00 s

=−1.9 mV

Using ohms law

ε=V =IR

I = ε/ R =−1.9mV/ 2.60Ω =−0.73mA

Answer:

1) a block going down a slope

2) a) W = ΔU + ΔK + ΔE, b) W = ΔE, c)  W = ΔK, d) ΔU = ΔK

Explanation:

In this exercise you are asked to give an example of various types of systems

1) a system where work is transformed into internal energy is a system with friction, for example a block going down a slope in this case work is done during the descent, which is transformed in part kinetic energy, in part power energy and partly internal energy that is represented by an increase in the temperature of the block.

2)

a) rolling a ball uphill

In this case we have an increase in potential energy, if there is a change in speed, the kinetic energy also increases, if the change in speed is zero, there is no change in kinetic energy and there is a change in internal energy due to the stationary rec in the point of contact

 W = ΔU + ΔK + ΔE

b) in this system work is transformed into internal energy

      W = ΔE

c) There is no friction here, therefore the work is transformed into kinetic energy

    W = ΔK

d) if you assume that there is no friction with the air, the potential energy is transformed into kinetic energy

      ΔU = ΔK

Answer:

A) 1.5 V

B) 4.5 V

Explanation:

A) Batteries in parallel have the same voltage as an individual battery.

V = 1.5 V

B) Batteries in series have a voltage equal to the sum of the individual batteries.

V = 1.5 V + 1.5 V + 1.5 V

V = 4.5 V

Answer:

A) 0.4 mA

B) 0.03 mA

Explanation:

Given that

voltage source, V = 120 V

to solve this question, we would be using the very basic Ohms Law, that voltage is proportional to the current and the resistance passing through the circuit, if temperature is constant.

mathematically, Ohms Law, V = IR

V = Voltage

I = Current

R = Resistance

from question a, we were given 300kΩ, substituting this value of resistance in the equation, we have

120 = I * 300*10^3 Ω

making I the subject of the formula,

I = 120 / 300000

I = 0.0004 A

I = 0.4 mA

Question said, currents above 10 mA causes involuntary muscle contraction, this current is way below 10 mA, so nothing happens.

B, we have Resistance, R = 4000kΩ

Substituting like in part A, we have

120 = I * 4000*10^3 Ω

I = 120 / 4000000

I = 0.00003 A

I = 0.03 mA

This also means nothing happens, because 0.03 mA is very much lesser compared to in the 10 mA

The current through a person will be:

a) 0.4 mA

b) 0.03 mA

Given:

Voltage, V = 120 V

It states that the voltage or potential difference between two points is directly proportional to the current or electricity passing through the resistance, and directly proportional to the resistance of the circuit.

Ohms Law, V = I*R

where,

V = Voltage

I = Current

R = Resistance

a)

Given: Resistance=  300kΩ

[tex]120 = I * 300*10^3 ohm\\\\I = 120 / 300000\\\\I = 0.0004 A[/tex]

Thus, current will be, I = 0.4 mA

b)

Given: R = 4000kΩ

[tex]120 = I * 4000*10^3 ohm\\\\I = 120 / 4000000\\\\I = 0.00003 A[/tex]

Thus, current will be, I = 0.03 mA

From calculations, we observe that nothing happens, because 0.03 mA is very much lesser compared to in the 10 mA.

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Answer:

The power of the mirror in diopters is 58.8 D

Explanation:

Given;

radius of curvature of the spoon, R = 3.4 cm = 0.034 m

The focal length of a mirror is given by;

[tex]f = \frac{R}{2} \\\\f = \frac{0.034}{2} \\\\f = 0.017 \ m[/tex]

The focal length of the mirror is 0.017 m

(b) The power of the mirror is given by;

[tex]P = \frac{1}{f}[/tex]

where;

P is the power of the mirror

f is the focal length

[tex]P = \frac{1}{f}\\\\P= \frac{1}{0.017}\\\\P = 58.8 \ D[/tex]

Thus, the power of the mirror in diopters is 58.8 D

Answer:

We can think the water stream as a solid object that is fired.

The distance between the fireperson and the building is 50m. (i consider that the position of the fireperson is our position = 0)

The angle is 30 above the horizontal. (yo wrote 300, but this has no sense because 300° implies that he is pointing to the ground).

The initial speed of the stream is 40m/s.

First, using the fact that:

x = R*cos(θ)

y = R*sin(θ)

in this case R = 40m/s and θ = 30°

We can use the above relation to find the components of the velocity:

Vx = 40m/s*cos(30°) = 34.64m/s

Vy = 20m/s.

First step:

We want to find the time needed to the stream to hit the buildin.

The horizontal speed is 34.64m/s and the distance to the wall is 50m

So we want that:

34.64m/s*t = 50m

t = 50m/(34.64m/s) = 1.44 seconds.

Now we need to calculate the height of the stream at t = 1.44s

Second step:

The only force acting on the water is the gravitational one, so the acceleration of the stream is:

a(t) = -g.

g = -9.8m/s^2

For the speed, we integrate over time and we get:

v(t) = -g*t + v0

where v0 is the initial speed: v0 = 20m/s.

The velocity equation is:

v(t) = -g*t + 20m/s.

For the position, we integrate again over time:

p(t) = -(1/2)*g*t^2 + 20m/s*t + p0

p0 is the initial height of the stream, this data is not known.

Now, the height at the time t = 1.44s is

p(1.44s) = -5.9m/s^2*(1.44s)^2 + 20m/s*1.44s + po

             = 16.57m + p0

So the height at wich the stream hits the building is 16.57 meters above the initial height of the fire hose.

Answer:

The values is  [tex]B = 3.2 *10^{-8} \ T[/tex]

The  direction is out of the plane

Explanation:

From the question we are told that

  The  magnitude of the electric field is  [tex]E = 9.6 \ V/m[/tex]

The  magnitude of the magnetic field is mathematically represented as

       [tex]B = \frac{E}{c}[/tex]

where c is the speed of light with value

      [tex]B = \frac{ 9.6}{3.0 *10^{8}}[/tex]

     [tex]B = 3.2 *10^{-8} \ T[/tex]

Given that the direction off the electromagnetic wave( c ) is  northward(y-plane ) and  the electric field(E) is eastward(x-plane ) then the magnetic field will be acting in the out of the page  (z-plane  )

Answer:

The wavelength of the emitted photons 532 nm, corresponds to a visible light having GREEN color.

Explanation:

Given;

energy of the emitted photons, E = 3.74 x 10⁻¹⁹ J

speed of light, c = 3 x 10⁸ m/s

Planck's constant, h = 6.63 x 10⁻³⁴ J.s

The wavelength of the emitted light will be calculated by applying energy of photons;

[tex]E = hf[/tex]

where;

E is the energy emitted light

h is Planck's constant

f is frequency of the emitted photon

But f = c / λ

where;

λ is the wavelength of the emitted photons

[tex]E = \frac{hc}{\lambda} \\\\\lambda = \frac{hc}{E} \\\\\lambda = \frac{6.63*10^{-34} *3*10^{8}}{3.74*10^{-19}} \\\\\lambda = 5.318 *10^{-7} \ m\\\\\lambda = 531.8 *10^{-9} \ m\\\\\lambda = 531.8 \ nm[/tex]

λ ≅ 532 nm

the wavelength of the emitted photons is 532 nm.

Therefore, the wavelength of the emitted photons 532 nm, corresponds to a visible light having GREEN color.

Answer:

The distance is [tex]D = 2.6 \ m[/tex]

Explanation:

From the question we are told that

    The wavelength of the light is  [tex]\lambda = 476.1 \ nm = 476.1 *10^{-9} \ m[/tex]

      The  distance between the slit is  [tex]d = 0.29 \ mm = 0.29 *10^{-3} \ m[/tex]

       The  between the first and second dark fringes is  [tex]y = 4.2 \ mm = 4.2 *10^{-3} \ m[/tex]

Generally  fringe width is mathematically represented as

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

Where D is the distance of the slit to the screen

   Hence

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

substituting values

       [tex]D = \frac{ 4.2 *10^{-3} * 0.29 *10^{-3}}{ 476.1 *10^{-9} }[/tex]

        [tex]D = 2.6 \ m[/tex]

Answer:

B

8.5 km/h east

Explanation:

Relative velocity= Va -Vb

=10-1.5

=8.5 km/h east

The concept relative speed is used when two or more bodies moving with some speed are considered. The relative speed of woman to the boat is 8.5 km/h east. The correct option is B.

The relative speed of two bodies is defined as the sum of their speeds if they are moving in the opposite direction and it is the difference of their speeds if they are moving in the same direction.

The speed of the moving body with respect to the stationary body is known as the relative speed. The term relative means in comparison to. The relative speed is a scalar quantity.

Here both the boat and women are travelling in the same direction. So the relative speed is given as:

Relative speed = 10 - 1.5 = 8.5 km / h

Therefore the relative speed is 8.5 km/h east.

Thus the correct option is B.

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Answer:

a. 0.342 kg-m² b. 2.0728 kg-m²

Explanation:

a. Since the skater is assumed to be a cylinder, the moment of inertia of a cylinder is I = 1/2MR² where M = mass of cylinder and r = radius of cylinder. Now, here, M = 56.5 kg and r = 0.11 m

I = 1/2MR²

= 1/2 × 56.5 kg × (0.11 m)²

= 0.342 kgm²

So the moment of inertia of the skater is

b. Let the moment of inertia of each arm be I'. So the moment of inertia of each arm relative to the axis through the center of mass is (since they are long rods)

I' = 1/12ml² + mh² where m = mass of arm = 0.05M, l = length of arm = 0.875 m and h = distance of center of mass of the arm from the center of mass of the cylindrical body = R/2 + l/2 = (R + l)/2 = (0.11 m + 0.875 m)/2 = 0.985 m/2 = 0.4925 m

I' = 1/12 × 0.05 × 56.5 kg × (0.875 m)² + 0.05 × 56.5 kg × (0.4925 m)²

= 0.1802 kg-m² + 0.6852 kg-m²

= 0.8654 kg-m²

The total moment of inertia from both arms is thus I'' = 2I' = 1.7308 kg-m².

So, the moment of inertia of the skater with the arms extended is thus I₀ = I + I'' = 0.342 kg-m² + 1.7308 kg-m² = 2.0728 kg-m²

a) The moment of inertia as the skater pulled his/her arm inward by assuming he/she is a cylinder is 0.3418kg-m².

b) If the body of the skater is assumed to be a cylinder of the same size, and the arms are rods extending straight out from his/her body being rotated at the ends, the moment of inertia is 1.7495kg-m².

Given the data in the question;

Mass of skater; [tex]M = 56.5kg[/tex]

a)

When the skater has his arms pulled inward by assuming they are cylinder of radius; [tex]R = 0.11 m[/tex]

Moment of inertia; [tex]I = \ ?[/tex]

From Parallel axis theorem; Moment of Inertia for a cylindrical body is expressed:

[tex]I = \frac{1}{2}MR^2[/tex]

Where M is the mass and R is the radius

We substitute our given values into the equation

[tex]I = \frac{1}{2}\ *\ 56.5kg\ *\ (0.11m)^2\\\\I = \frac{1}{2}\ *\ 56.5kg\ *\ 0.0121m^2\\\\I = 0.3418kg.m^2[/tex]

Therefore, the moment of inertia as the skater pulled his/her arm inward by assuming he/she is a cylinder is 0.3418kg-m²

b)

With the skater's arms extended by assuming that each arm is 5% of the mass of their body

Mass of each arm; [tex]M_a = \frac{5}{100} * M = \frac{5}{100} * 56.5kg = 2.825kg[/tex]

Remaining mass; [tex]M_b = M - 2M_a = 56.5kg - 2(2.825kg) = 50.85kg[/tex]

Assume the body is a cylinder of the same size and the arms are 0.875 m long rods extending straight out from their body being rotated at the ends.

Length of arm; [tex]L = 0.875 m[/tex]

From Parallel axis theorem; Moment of Inertia about vertical axis is expressed as:

[tex]I = \frac{1}{2}M_bR^2 + \frac{2}{3}M_aL^2[/tex]

We substitute in our values

[tex]I = \frac{1}{2}*50.85kg*(0.11m)^2 + \frac{2}{3}*2.825kg*(0.875m)^2\\\\I = [\frac{1}{2}*50.85kg * 0.0121m^2] + [\frac{2}{3}*2.825kg*0.765625m^2]\\\\I = 0.3076kg.m^2 + 1.4419kg.m^2\\\\I = 1.7495kg.m^2[/tex]

Therefore, if the body of the skater is assumed to be a cylinder of the same size, and the arms are rods extending straight out from his/her body being rotated at the ends, the moment of inertia is 1.7495kg-m².

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Answer:

the net work is zero

Explanation:

Work is defined by the expression

        W = F. ds

Bold type indicates vectors

In this problem, the friction force does not decrease, therefore it will be zero.

Consequently for work on a closed path it is zero.

The work in going from the initial point (0, 0, 0) to the end of each segment is positive and when it returns from the point of origin the angle is 180º, therefore the work is negative, consequently the net work is zero

Answer:

 I = 1.23 A

Explanation:

In an RL circuit current passing is described by

           I = E / R (1 - [tex]e^{-Rt/L}[/tex])

Let's reduce the magnitudes to the SI system

        L = 35 mH = 35 10⁻³ H

        t = 5.0 ms = 5.0 10⁻³ s

let's calculate

         I = 18/12 (1 - [tex]e^{-12 .. 5 {10}^{-3}/35 .. {10}^{-3} }[/tex]e (- 5 10-3 12/35 10-3))

         I = 1.5 (1- [tex]e^{-1.715}[/tex])

         I = 1.23 A

Explanation:

Pressure is the same for both plungers.

P = P

F / A = F / A

F / (¼ π d²) = F / (¼ π d²)

F / d² = F / d²

5 N / (0.05 m)² = F / (1 m)²

F = 2000 N

None of the options are correct.

Answer:

Change in kinetic energy (ΔKE) = 12.8 × 10⁴ J

Explanation:

Given:

Mass of truck(m) = 2,100 kg

Initial speed(v1) = 38 km/h = 38,000 / 3600 = 10.56 m/s

Final speed(v2) = 55 km/h = 55,000 / 3600 = 15.28 m/s

Find:

Change in kinetic energy (ΔKE)

Computation:

Change in kinetic energy (ΔKE) = 1/2(m)[v2² - v1²]

Change in kinetic energy (ΔKE) = 1/2(2100)[15.28² - 10.56²]

Change in kinetic energy (ΔKE) = 1,050[233.4784 - 111.5136]

Change in kinetic energy (ΔKE) = 1,050[121.9648]

Change in kinetic energy (ΔKE) = 128063.04

Change in kinetic energy (ΔKE) = 12.8 × 10⁴ J

Answer:

entropy

Explanation:

Answer:

382Hz

Explanation:

The question lacks the required option. Find the complete question in the attachment.

The long organ pipe open at both ends is called an open pipe. The fundamental frequency for an open pipe is expressed as F0 = V/2L

Harmonics are integral multiples of the fundamental frequency. For open pipes its harmonics are 2fo, 3fo, 4fo, 5fo...

Given fundamental frequency f0 to be 85 Hz, the following frequencies will be present as a standing wave;

First overtone f1 = 2fo = 2(85) = 170Hz

Second overtone f2 = 3fo = 3(85) = 255Hz

Third overtone = 4fo = 4(85) = 340Hz

Based on the option it can be seen that the only frequency that is not present as a standing wave is 382Hz

Answer:

your question answer is 22°

Answer:

True or False

Explanation:

Because.....

easy 50% chance you are right

Answer and Explanation:

The computation of the object distance related to two quantities is shown below:

It could find out by using the lens formula which is shown below:

[tex]\frac{1}{v} - \frac{1}{u} = \frac{1}{f}[/tex]

where,

v = image distance

u = object distance

f = focal length

It could be found by applying the above formula i.e considering the image distance, object distance and the focal length

Answer:

Explanation:

Relation between flux and inductance is as follows

φ = Li

where φ is flux associated with induction of inductance L when a current i flows through it

putting the values

3.25 x 10⁻³ x 800 = L x 2.9

L = .9 H

for induced emf in an induction , the relation is

emf induced = L di / dt

Putting the values

7.5 x 10⁻³ = .9 x di / dt

di / dt = 8.33 x 10⁻³ A / s

(a) The self inductance of the solenoid is 0.897 H.

(b) The magnitude of the rate of change of the current is 0.00836 A/s.

The given parameters;

The self inductance of the solenoid is calculated as follows;

[tex]emf = \frac{d\phi}{dt}\\\\emf = \frac{Ldi}{dt} \\\\d\phi = Ldi\\\\\phi = BA\\\\NBA = LI\\\\L = \frac{NBA}{I} \\\\L = \frac{N\phi}{I} \\\\L = \frac{800 \times 3.25\times 10^{-3}}{2.9} \\\\L = 0.897 \ H\\\\[/tex]

The magnitude of the rate of change of the current is calculated as follows;

[tex]emf = L \frac{di}{dt} \\\\\frac{di}{dt} \ = \frac{emf}{L} \\\\\frac{di}{dt} = \frac{7.5 \times 10^{-3}}{0.897} \\\\\frac{di}{dt} = 0.00836 \ A/s[/tex]

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Answer:

The magnetic field is 0.0857 T.

Explanation:

The electrons orbit the magnetic field with a centripetal force equal to

F = [tex]\frac{mv^{2} }{r}[/tex]

also, the force on an electron in a magnetic field is gotten as

F = Bqv

equating this two equations give

[tex]\frac{mv^{2} }{r}[/tex] = Bqv

mv/r = Bq

where m is the mass of the electron = 9.11 x 10^-31 kg

v is the the linear speed of the electron

B is the magnetic field on the electron

r is the radius of the orbital movement

q is the charge on an electron = 1.602 x 10^-19 C

but, the linear speed v = ωr

where ω is the angular speed of the electron

substituting into equation above, we have

mωr/r = Bq

which reduces to

mω = Bq

finally, w know that the angular speed is related to the frequency of the electron by

ω = 2πf

we then finally have

2mπf = Bq

where f is the frequency emitted by the electron = 2.4 GHz = 2.4 x 10^9 Hz

substituting values into the equation, we have

2 x 9.11 x 10^-31 x 3.142 x 2.4 x 10^9 = B x 1.602 x 10^-19

B = (1.3734 x 10^-20)/(1.602 x 10^-19) = 0.0857 T

= 85.7 mT

Answer:

787528.7 J

Explanation:

Work done: This can be defined as the product of force and distance along the direction of force. The S.I unit of work is Joules (J).

From the question,

W = Tcos∅(d)............. Equation 1

Where W = work done, T = tension in the rope, ∅ = the angle of the rope to the horizontal, d = distance.

But,

d = v(t)..................... equation 2

Where v = velocity, t = time

Substitute equation 2 into equation 1

W = Tcos∅(vt)............. Equation 3

Given: T = 210 N, ∅ = 23°, v = 7 km/h = 1.94 m/s, t = 35 min = 2100 s

Substitute into equation 3

W = 210(cos23°)(1.94×2100)

W = 787528.7 J