A toroidal solenoid has 590 turns, cross-sectional area 6.20 cm^2 , and mean radius 5.00 cm .Part A. Calcualte the coil's self-inductance.Part B. If the current decreases uniformly from 5.00 A to 2.00 A in 3.00 ms, calculate the self-induced emf in the coil.Part C. The current is directed from terminal a of the coil to terminal b. Is the direction of the induced emf froma to b or from b to a?

Answer:

Explanation:

The formula for calculating the voltage across an inductor is expressed as

[tex]V_l = IX_l\\\\Since\ X_l = 2\pi fL\\V_l = I(2\pi fL)[/tex]

Given parameters

current amplitude I = 1.50mA = 1.5*10⁻³A

inductance L = 0.450mH = 0.450*10⁻³H

Voltage across the inductor [tex]V_l[/tex] = 13.0V

Required

frequency f

Substituting the given parametres into the formula, we have;

[tex]V_l = I(2\pi fL)\\\\13 = 1.50*10^{-3}(2*3.14*f*0.450*10^{-3})\\\\13 = 4.239*10^{-6}f\\\\f = \frac{13}{4.239*10^{-6}} \\\\f = 3,066,761 Hertz\\\\f = 3.067MHz[/tex]

Hence, the frequency required is 3.067MHz

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

Explanation:

[tex]\theta=1.22 \frac{\lambda}{D}[/tex]

And, from equation ( 2 ), we get

[tex]\theta=\frac{x}{d}[/tex]

Thus,

[tex]\frac{x}{d} &=1.22 \frac{\lambda}{D}[/tex]

[tex]D &=1.22 \frac{\lambda d}{x}[/tex]

[tex]=1.22 \frac{550 \times 10^{-9} 3.84 \times 10^{8}}{5 \times 10^{3}}[/tex]

[tex]=0.0515 \mathrm{m}[/tex]

Thus, the diameter of the telescope's mirror that would allow us to see details as small as is

Answer:

The ability of a water strider to move along the surface of water without breaking the surface is due to:

SURFACE TENSION

Explanation:

this is because Water molecules are more attracted to each other than they are to other materials, so they generate a force to stay together called surface tension. Which allows the strider to move without breaking the surface

Explanation:

It is given that,

Distance covered by the cart to the right is 2 m

Distance covered by the cart to the left is 1 m

We need to find the total distance rolled by the cart. Total distance is equal to the sum of the distances covered by an object. It does depend on the direction.

Total distance = 2 m + 1 m

D = 3 m

The cart rolled to a total distance of 3 m.

....................

Answer:

T_object = 380.35 K

Explanation:

From Stefan–Boltzmann law, the power output is given by the formula:

P = σAT⁴

where;

σ is Stefan-Boltzmann constant

A is area of the radiating surface.

T is temperature of the body

Now, we are told that the power the object emitted is 3 times the power absorbed from the room.

Thus, we have;

P_e = 3P_a

Where P_e is power emitted and P_a is power absorbed.

So, we have;

σA(T_object)⁴ = 3σA (T_room)⁴

σA will cancel out to give;

(T_object)⁴ = 3(T_room)⁴

We are given T_room = 289 K

Thus;

(T_object)⁴ = 3 × 289⁴

(T_object) = ∜(3 × 289⁴)

T_object = 380.35 K

Answer:

Explanation:

As we've seen, humans have on average a heart rate of around 60 to 70 beats per minute, give or take. We live roughly 70 or so years, giving us just over 2 billion beats all up.Apr

Answer:

far away

Explanation:

There are different types of eye defect ranging from short sightedness, longsighted, astigmatism, presbyopia etc.

If someone is only able to see close ranged object clearly but not far distant object, then such person is suffering from short sightedness or myopia. This occurs when the light rays entering the eye does not converge on the retina. Instead of converging on the retina, the light ray is formed on a point in front of the retina. This causes the distance from the eye's lens to the retina shorter compared to that of a normal eye. This eye defect is usually corrected using concave lens in order to diverge the rays thereby allowing it to focus on the retina.

Hence, if the distance from your eye's lens to the retina is shorter than for a normal eye, you will struggle to see objects that are far away (at a far distant).

Answer:

1) Mass and acceleration

2) 4.0

3)When the net force applied to an object changes, the acceleration changes by the same factor.

4)acceleration

5)The acceleration is half of its original value

Explanation:

A wise man once said if you cheat together you succeed together! lol i am jk  

Answer:

An aircraft, flying in the vicinity of 18,000 ft altitude from west to east over the US at 12 Z today, will __LOSE___ altitude if the altimeter is not corrected

Answer:

The value is  [tex]\mu = 0.76[/tex]

Explanation:

From the question we are told that

    The  acceleration is [tex]a = 7.4 \ m /s^2[/tex]

Generally the force by which the truck bed (truck) is moving with is mathematically represented as

          [tex]F = ma[/tex]

Now for the truck cabinet to slip from the truck bed then the frictional force between the truck cabinet  is equal the force by which the the truck bed is moving with that is  

        [tex]F_f = F[/tex]

Here  [tex]F_f[/tex] is the frictional force which is mathematically represented as

         [tex]F_f = \mu * m * g[/tex]

substituting into above equation

         [tex]\mu * m * g = ma[/tex]

=>        [tex]\mu = \frac{a}{g}[/tex]

substituting values

           [tex]\mu = \frac{ 7.4 }{ 9.8}[/tex]

           [tex]\mu = 0.76[/tex]

Answer:

a) 1.05

b) 43.6°

Explanation:

a) The index refraction that describes the material can be found using Brewster's law:

[tex] \theta_{1} = arctan(\frac{n_{2}}{n_{1}}) [/tex]

where:

n₁ is the refractive index of the initial medium through which the light propagates (air) = 1

n₂ is the index of the material=?

θ₁ = 46.5°    

[tex] n_{2} = n_{1}tan(\theta_{1}) = tan(46.5) = 1.05 [/tex]

Hence, the material's index refraction is 1.05.

b) The angle of refraction can be found as follows:

[tex] n_{1}sin(\theta_{1}) = n_{2}sin(\theta_{2}) [/tex]

[tex]sin(\theta_{2}) = \frac{n_{1}sin(\theta_{1})}{n_{2}} = \frac{sin(46.5)}{1.05} = 0.69[/tex]

[tex] \theta_{2} = arcsin(0.69) = 43.6^{\circ} [/tex]

Therefore, the angle of refraction is 43.6°.

I hope it helps you!

Answer:

E = VdB

Explanation:

This is because canceling the electric and magnetic force means

q.vd. B= we

E= Vd. B

Answer: The correct answer for this question is letter (B) The electromagnetic waves reach Earth, while the mechanical waves do not. The sun generates both mechanical and electromagnetic waves. Space, between the sun and the earth is a nearly vacuum. So mechanical wave can not spread out in the vacuum.

Hope this helps!

Answer:

The electromagnetic waves reach Earth, while the mechanical waves do not

Answer:

Fluoroscopy

Explanation:

A Fluoroscopy is an imaging technique that uses X-rays to obtain real-time moving images of the interior of an object. In its primary application of medical imaging, a fluoroscope allows a physician to see the internal structure and function of a patient, so that the pumping action of the heart or the motion of swallowing, for example, can be watched.

Answer:

The average induced emf in the coil is 0.0286 V

Explanation:

Given;

diameter of the wire, d = 11.2 cm = 0.112 m

initial magnetic field, B₁ = 0.53 T

final magnetic field, B₂ = 0.24 T

time of change in magnetic field, t = 0.1 s

The induced emf in the coil is calculated as;

E = A(dB)/dt

where;

A is area of the coil = πr²

r is the radius of the wire coil = 0.112m / 2 = 0.056 m

A = π(0.056)²

A = 0.00985 m²

E = -0.00985(B₂-B₁)/t

E = 0.00985(B₁-B₂)/t

E = 0.00985(0.53 - 0.24)/0.1

E = 0.00985 (0.29)/ 0.1

E = 0.0286 V

Therefore, the average induced emf in the coil is 0.0286 V

We have that for the Question, it can be said that the average induced emf in the coil is

E=0.028565V

From the question we are told

A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to 0.24 T pointing down. What is the average induced emf in the coil?

Generally the equation for the Average emf induced   is mathematically given as

[tex]Emf_a=-NA\frac{dB}{dt}\\\\Where\\\\Area\\\\a=\pir^2\\\\a=\pi(0.056)^2\\\\a=0.00985\\\\[/tex]

Hence

[tex]dB=0.24-0.53\\\\dB=-0.29T[/tex]

Therefore

[tex]E=-\frac{1*0.00985*-0.29 }{0.10}[/tex]

E=0.028565V

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Answer:

Given that

capacitor being charged by a current i c has a displacementcurrent equal to i c between the plates∴

displacement current iD =i c=0.280 Aradius of the circular plate r = 4 cm=0.04 m

( A ) . displacement current density j D = iD / ( π r 2 )=0.28 / ( 3.14 * 0.04 2 )=55.73 A / m 2

( B ) . displacement current density j

D = ε( dE / dt )the rate at which the electric field between the plates is changing(

dE / dt ) = jD/ εdE/ dt ) = 55.73/ 8.85 * 10 -12=6.3*10 12 N / C - s( C ) . the induced magnetic field between theplates B = ( μ * r / 2π R 2) * i c ----( 1 )whereR= 2 cm=0.02 mr= 4 cm=0.04 mμ=permeability of free space=4π* 10 -7 H( D )substitute R = 1 cm = 0.01 m inequation ( 1 ),wegetanswer

(a) The car is undergoing an acceleration of

[tex]a=\dfrac{22.7\frac{\rm m}{\rm s}-0}{9.02\,\mathrm s}\approx2.52\dfrac{\rm m}{\mathrm s^2}[/tex]

so that in 9.02 s, it will have covered a distance of

[tex]x=\dfrac a2(9.02\,\mathrm s)^2\approx102\,\mathrm m[/tex]

The car has tires with diameter d = 58.5 cm = 0.585 m, and hence circumference π d ≈ 1.84 m. Divide the distance traveled by the tire circumference to determine how many revolutions it makes:

[tex]\dfrac{102\,\mathrm m}{1.84\,\mathrm m}\approx55.7\,\mathrm{rev}[/tex]

(b) The wheels have average angular velocity

[tex]\omega=\dfrac{\omega_f+\omega_i}2=\dfrac{\theta_f-\theta_i}{\Delta t}[/tex]

where [tex]\omega[/tex] is the average angular velocity, [tex]\omega_i[/tex] and [tex]\omega_f[/tex] are the initial and final angular velocities (rev/s), [tex]\theta_i[/tex] and [tex]\theta_f[/tex] are the initial and final angular displacements (rev), respectively, and [tex]\Delta t[/tex] is the duration of the time between initial and final measurements. The second equality holds because acceleration is constant.

The wheels start at rest, so

[tex]\dfrac{\omega_f}2=\dfrac{55.7\,\rm rev}{9.02\,\rm s}\implies\omega_f\approx12.4\dfrac{\rm rev}{\rm s}[/tex]

Answer:

Explanation:

Initial velocity , u = 30 m/s

final velocity , v = 10 m/s

time , t = 5 seconds

1. Acceleration = v - u / t

= 10 - 30 / 5

= -20 / 5

= - 4 m/s

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:

λ = 6.61 x 10⁻⁷ m = 661 nm

Explanation:

From the Young's Double Slit experiment, the the spacing between adjacent bright or dark fringes is given by the following formula:

Δx = λL/d

where,

Δx = fringe spacing = 2.86 mm = 2.86 x ⁻³ m

L = Distance between slits and screen = 2.12 m

d = slit separation = 0.49 mm = 0.49 x 10⁻³ m

λ = wavelength of light = ?

Therefore,

2.86 x 10⁻³ m = λ(2.12 m)/(0.49 x 10⁻³ m)

(2.86 x 10⁻³ m)(0.49 x 10⁻³ m)/(2.12 m) = λ

λ = 6.61 x 10⁻⁷ m = 661 nm

Answer:

Number of photons emit per second = 1.9 × 10²⁹  (Approx)

Explanation:

Given:

Frequency = 91.5 MHz

Power = 11.5 Kw = 11,500 J/s

Find:

Number of photons emit per second

Computation:

Total energy with frequency (E) = hf

Total energy with frequency (E) = 6.626×10⁻³⁴  × 91.5×10⁶

Total energy with frequency (E) = 6.06×10⁻²⁶ J

Number of photons emit per second = 11,500 / 6.06×10⁻²⁶

Number of photons emit per second = 1897.689 × 10²⁶

Number of photons emit per second = 1.9 × 10²⁹  (Approx)

Answer:

The frictional force the road applies to the rear tire is static friction and it acts opposite to the direction in which the car is traveling.

Explanation:

This question suggests that the car is accelerating forward. Thus, the easiest way for us to know what friction is doing is for us to know what happens when we turn friction off.

Now, if there is no friction and the car is stopped, if we push down on the accelerator, it will make the front wheels to spin in a clockwise manner. This spin occurs on the frictionless surface with the rear wheels doing nothing while the car doesn't move.

Now, if we apply friction to just the front wheels, the car will accelerate forward while the back wheels would be dragging along the road and not be spinning. Thus, friction opposes the motion and as such, it must act im a direction opposite to where the car is going. This must be static friction.

The frictional force the road applies to the rear tire is static friction and it acts opposite to the direction in which the car is traveling.

Answer:

A. 2.2*10^-2m

Explanation:

Using

Area = length x L/ uo xN²

So A = 0.7m * 25 x 10^-3H /( 4π x10^-7*

3000²)

A = 17.5*10^-3/ 1.13*10^-5

= 15.5*10^-2m²

Area= π r ²

15.5E-2/3.142 = r²

2.2*10^2m

Explanation:

Answer:

E. Coefficient of linear expansion

Answer:

a) the distance between the interference fringes is reduced by half

b) the distance between stripes is doubled

Explanation:

Interference experiments constructive interference is described by the expression

          d sin θ = m λ

let's use trigonometry to find the distance between the interference fringes

              tan θ=  y / L

dodne y is the distance from the central maximum, L the distance from the slit to the observation screen. In general these experiments are carried out at very small angles

            tan θ = sin θ / cos θ = sin θ

we substitute

             sin θ = y / L

            d y / L = m  λ

           y = m λ / d L

a) it asks us when the screen doubles its distance

           L ’= 2 L

subtitute in the equation

           y ’= m λ / (d 2L)

           y ’=( m λ / d L) /2

           y ’= y / 2

the distance between the interference fringes is reduced by half

b) the density of the network doubles

      if the density doubles in the same distance there are twice as many slits, so the distance between them is reduced by half

            d ’= d / 2

we substitute

          y ’= m λ (L d / 2)

          y ’= m λ / (L d) 2

          y ’= y 2

the distance between stripes is doubled

Explanation:

The frequency is given to be f = 8 Hz.

Period is the inverse of frequency.

T = 1/f = 0.125 s

Velocity is wavelength times frequency.

v = λf = (0.40 m) (8 Hz) = 3.2 m/s

The wave travels 3.2 meters every second.

Answer:

Explanation:

Formula for calculating the relationship between  the electromotive force (emf), current and number of turns of a coil in a transformer is expressed as shown:

[tex]\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]  where;

Vs and Vp are the emf in the secondary and primary coil respectively

Ns and Np are the number if turns in the secondary and primary coil respectively

Ip and Is are the currents in the secondary and primary coil respectively

Since the are all equal to each other, then we can equate any teo of the expression as shown;

[tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]

Given parameters

Np = 500-turns

Ns = 2000-turns

Is = 3.0Amp

Required

Current in the primary coil (Ip)

Using the relationship [tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]

[tex]I_p = \dfrac{N_sI_s}{N_p}[/tex]

[tex]I_p = \dfrac{2000*3}{500} \\\\I_p = \frac{6000}{500}\\ \\I_p = 12A\\[/tex]

Hence the current in the primary coil is 12Amp

Answer:

K.E = 5.53 eV = 8.85 x 10⁻¹⁹ J

Explanation:

First we calculate the energy of photon:

E = hc/λ

where,

E = Energy of Photon = ?

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

c = speed of light = 3 x 10⁸ m/s

λ = wavelength = 120 nm = 1.2 x 10⁻⁷ m

Therefore,

E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(1.2 x 10⁻⁷ m)

E = (16.565 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)

E = 10.35 eV

Now, from Einstein's Photoelectric equation we know that:

Energy of Photon = Work Function + K.E of Electron

10.35 eV = 4.82 eV + K.E

K.E = 10.35 eV - 4.82 eV

K.E = 5.53 eV = 8.85 x 10⁻¹⁹ J

The maximum kinetic energy of the ejected photoelectrons will be "8.85 × 10⁻¹⁹ J".

According to the question,

Speed of light, c = 3 × 10⁸ m/s

Wavelength, λ = 120 nm or,

                        = 1.2 × 10⁻⁷ m

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

Now,

The energy of photon will be:

→ E = [tex]\frac{hc}{\lambda}[/tex]

By substituting the values,

     = [tex]\frac{6.626\times 10^{-34}\times 3\times 20^8}{1.2\times 10^{-7}}[/tex]

     = [tex]\frac{16.565\times 10^{-19}}{\frac{1 \ eV}{1.6\times 10^{-19}} }[/tex]

     = 10.35 eV

By using Einstein's Photoelectric equation,  

Energy of Photon = Work function + K.E

                   10.35 = 4.82 + K.E

                       K.E = 10.35 - 4.82

                             = 5.53 eV or,

                             = 8.85 × 10⁻¹⁹ J

Thus the response above is correct.

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