The capacitor is originally charged. How does the current I in the ammeter behave as a function of time after the switch is closed?

1. I = 0 always.

2. I = constant, not equal to 0.

3. I increases, then is constant.

4. I instantly jumps up, then slowly decreases.

5. None of the above.

Answer:

The number is  [tex]Z = 216 \ fringes[/tex]

Explanation:

From the question we are told that

      The wavelength is  [tex]\lambda = 520 \ nm = 520 *10^{-9} \ m[/tex]

       The length of the glass plates is [tex]y = 21.1cm = 0.211 \ m[/tex]

      The distance between the plates (radius of wire ) =  [tex]d = 0.028 mm = 2.8 *10^{-5} \ m[/tex]

   Generally the condition for constructive  interference in a film is mathematically represented as

            [tex]2 * t = [m + \frac{1}{2} ]\lambda[/tex]

Where  t is the thickness of the separation between the glass i.e  

    t  = 0 at the edge where the glasses are touching each other and  

     t =  2d at the edge where the glasses are separated by the wire  

   m is the order of the fringe it starts from  0, 1 , 2 ...

So  

       [tex]2 * 2 * d = [m + \frac{1}{2} ] 520 *10^{-9}[/tex]

=>   [tex]2 * 2 * (2.8 *10^{-5}) = [m + \frac{1}{2} ] 520 *10^{-9}[/tex]

=>    

       [tex]m = 215[/tex]

given that we start counting m from zero

   it means that the number of  bright fringes that would appear is

         [tex]Z = m + 1[/tex]

=>    [tex]Z = 215 +1[/tex]

=>     [tex]Z = 216 \ fringes[/tex]

Complete Question

An emf is induced in a conducting loop of wire 1.12m long as its shape is.

changed from square to circular. Find the average magnitude of the induced emf if the change in shape occurs in 0.125 ss and the local 0.504-TT magnetic field is perpendicular to the plane of the loop.

Answer:

The induced emf is  [tex]\epsilon = 0.0863 \ V[/tex]

Explanation:

From the question we are told that

      The  time taken is  [tex]\Delta t = 0.125 \ s[/tex]

       The magnitude of the magnetic field is  B =  0.504 T

        The length of the loop wire is  [tex]l = 1.12 \ m[/tex]

Generally the circumference of the wire when in circular form is  

          [tex]C = 2 \pi r[/tex]

=>        [tex]l = 2 \pi r[/tex]

=>         [tex]r =[/tex][tex]\frac{l}{2 \pi}[/tex]

=>          [tex]r =[/tex][tex]\frac{1.12}{2 * 3.142}[/tex]

=>        [tex]r =[/tex][tex]0.1782 \ m[/tex]

Now the area of the wire as a circle is

           [tex]A = \pi r^2[/tex]

    =>     [tex]A = 3.142 * (0.1782)^2[/tex]      

     =>    [tex]A = 0.0998 \ m^2[/tex]

The  length of one side of the square is

         [tex]b = \frac{l}{4}[/tex]

         [tex]b = \frac{1.12}{4}[/tex]

         [tex]b = 0.28 \ m[/tex]

Now the area of the wire as a square is

          [tex]A_s = b^2[/tex]

=>          [tex]A_s =(0.28 )^2[/tex]

             [tex]A_s = 0.0784 \ m^2[/tex]

Generally the induced emf is mathematically represented as

        [tex]\epsilon = \frac{B * [A - A_s ]}{\Delta t }[/tex]

=>      [tex]\epsilon = \frac{0.504 * [0.0998 - 0.0784 ]}{0.125 }[/tex]

=>      [tex]\epsilon = 0.0863 \ V[/tex]

Answer:

No.

Explanation:

Given the following :

Velocity (V) of ball = 5m/s

Radius = 1m

Can the ball reach the highest point of the circular track

of radius 1.0 m?

The highest point in the track could be considered as the diameter of the circle :

Radius = diameter / 2;

Diameter = (2 * Radius) = (2*1) = 2

Maximum height which the ball can reach :

Using the relation :

Kinetic Energy = Potential Energy

0.5mv^2 = mgh

0.5v^2 = gh

0.5(5^2) = 9.8h

0.5 * 25 = 9.8h

12.5 = 9.8h

h = 12.5 / 9.8

h = 1.2755

h = 1.26m

Therefore maximum height which can be reached is 1.26m.

Since h < Diameter

Answer:

The dispersion is [tex]D = 2.01220 *10^{5} \ rad/m[/tex]

Explanation:

From the question we are told that

    The wavelength of the light is  [tex]\lambda = 550 \ = 550 *10^{-9} \ n[/tex]

    The width of the grating is[tex]k = 1\ cm = 0.01 \ m[/tex]

    The  number of slit is  N =  1000 slits

    The order of the maxima is  m =  2

Generally the spacing between the slit is mathematically represented as

         [tex]d = \frac{k}{N}[/tex]

substituting values

        [tex]d = \frac{ 0.01}{1000}[/tex]

       [tex]d = 1.0 *10^{-5} \ m[/tex]

Generally the condition for constructive interference is

       [tex]d\ sin(\theta ) = m * \lambda[/tex]

substituting values

      [tex]1.0 *10^{-5} sin (\theta) = 2 * 550 *10^{-9}[/tex]

       [tex]\theta = sin^{-1} [\frac{ 2 * 550 *10^{-9}}{ 1.0 *10^{-5}} ][/tex]

      [tex]\theta = 6.315^o[/tex]

Generally the dispersion is mathematically represented as

           [tex]D = \frac{ m }{d cos(\theta )}[/tex]

substituting values

          [tex]D = \frac{ 2 }{ 1.0 *10^{-5} cos(6.315 )}[/tex]

           [tex]D = 2.01220 *10^{5} \ rad/m[/tex]

Answer:

factor that bug maximum KE change is 0.52284

Explanation:

given data

vertical distance = 6.5 cm

ripples decrease to =  4.7 cm

solution

We apply here formula for the KE of particle that executes the simple harmonic motion that is express as

KE = (0.5) × m × A² × ω²     .................1

and kinetic energy is  directly proportional to square of the amplitude.

so

[tex]\frac{KE2}{KE1} = \frac{A2^2}{A1^2}[/tex]      .............2

[tex]\frac{KE2}{KE1} = \frac{4.7^2}{6.5^2}[/tex]

[tex]\frac{KE2}{KE1}[/tex] = 0.52284

so factor that bug maximum KE change is 0.52284

The factor does the bug's maximum KE change should be considered as the 0.52284.

Since

vertical distance = 6.5 cm

ripples decrease to =  4.7 cm

So here we apply the given formula

KE = (0.5) × m × A² × ω²     .................1

here,

kinetic energy is directly proportional to square of the amplitude.

So,

= 4.7^2/ 6.5^2

= 0.52284

hence, The factor does the bug's maximum KE change should be considered as the 0.52284.

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Answer:

     e% = 0.99%   this value is within the 5% tolerance given by the manufacturer

Explanation:

Modern manufacturing methods establish a tolerance in order to guarantee homogeneous characteristics in their products, in the case of resistors the tolerance or error is given by

          e% = | R_nominal - R_measured | / R_nominal 100

where R_nominal is the one written in the resistance in your barcode, R_measured is the real value read with a multimeter and e% is the tolerance also written in the resistors

let's apply this formula to our case

R_nominal = 10 kΩ = 10000 Ω

R_measured = 100 99 Ω

        e% = | 10000 - 10099.1 | / 10000 100

        e% = 0.99%

this value is within the 5% tolerance given by the manufacturer

Answer:

a) 3.6 ft

b) 12.4 ft

Explanation:

Distance between mirrors = 6.2 ft

difference from from the mirror you face = 1.8 ft

a) you stand 1.8 ft in front of the mirror you face.

According to plane mirror rules, the image formed is the same distance inside the mirror surface as the distance of the object (you) from the mirror surface. From this,

your distance from your first "front" image = 1.8 ft + 1.8 ft = 3.6 ft

b) The mirror behind you is 6.2 - 1.8 = 4.4 ft behind you.

the back mirror will be reflected 3.6 + 4.4 = 8 ft into the front mirror,

the first image of your back will be 4.4 ft into the back mirror,

therefore your distance from your first "back" image = 8 + 4.4 = 12.4 ft

Answer:

The speed of light in air is 2.996x10⁸ m/s, and polystyrene is 1.873x10⁸ m/s.

Explanation:

To find the speed of light in air and in polystyrene we need to use the following equation:

[tex] c_{m} = \frac{c}{n} [/tex]

Where:

[tex]c_{m}[/tex]: is the speed of light in the medium

n: is the refractive index of the medium

In air:

[tex]c_{a} = \frac{c}{n_{a}} = \frac{2.997 \cdot 10^{8} m/s}{1.0003} = 2.996 \cdot 10^{8} m/s[/tex]

In polystyrene:

[tex]c_{p} = \frac{c}{n_{p}} = \frac{2.997 \cdot 10^{8} m/s}{1.6} = 1.873 \cdot 10^{8} m/s[/tex]  

Therefore, the speed of light in air is 2.996x10⁸ m/s, and polystyrene is 1.873x10⁸ m/s.

I hope it helps you!

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:

The reflected resistance in the primary winding is 6250 Ω

Explanation:

Given;

number of turns in the primary winding, [tex]N_P[/tex] = 50 turns

number of turns in the secondary winding, [tex]N_S[/tex] = 10 turns

the secondary load resistance, [tex]R_S[/tex] = 250 Ω

Determine the turns ratio;

[tex]K = \frac{N_P}{N_S} \\\\K = \frac{50}{10} \\\\K = 5[/tex]

Now, determine the reflected resistance in the primary winding;

[tex]\frac{R_P}{R_S} = K^2\\\\R_P = R_SK^2\\\\R_P = 250(5)^2\\\\R_P = 6250 \ Ohms[/tex]

Therefore, the reflected resistance in the primary winding is 6250 Ω

Answer:

chk photo

Explanation:

Answer:

a)step-down" transformer

b) 1440 turns

Explanation:

There are two types of transformers; step up transformers and step down transformers. A step down transformer converts a higher voltage to a lower voltage.

In a stepdown transformer, there are more turns in the primary coil than in the secondary coil, the turns ratio Ns/Np is less than 1 for a stepdown transformer.

If

Number of turns in primary coil Np= 2760

Number of turns in secondary coil Ns= unknown

Voltage in primary coil Vp= 230 V

Voltage in secondary coil Vs= 120 V

Ns/Np= Vs/VP

NsVp= NpVs

Ns= NpVs/VP = 2760 × 120/230

Ns= 1440 turns

Answer:

(a) The magnitude of induced emf in the loop while the magnetic field is increasing is 9.5 mV

(b) The magnitude of the induced voltage at a constant magnetic field is 124.7 mV

Explanation:

Given;

radius of the circular loop, r = 31.0 cm = 0.31 m

initial magnetic field, B₁ = 0.7 T

final magnetic field, B₂ = 2.3B₁ = 2.3 X 0.7 T = 1.61 T

duration of change in the field, t = 29

(a) The magnitude of induced emf in the loop while the magnetic field is increasing.

[tex]E = A*\frac{\delta B}{\delta t} \\\\[/tex]

[tex]E = A*\frac{B_2 -B_1}{\delta t}[/tex]

Where;

A is the area of the circular loop

A = πr²

A = π(0.31)² = 0.302 m²

[tex]E = A*\frac{B_2 -B_1}{\delta t} \\\\E = 0.302*\frac{1.61-0.7}{29} \\\\E = 0.0095 \ V\\\\E = 9.5 \ mV[/tex]

(b) the magnitude of the induced voltage at a constant magnetic field

E = A x B/t

E = (0.302 x 1.61) / 3.9

E = 0.1247 V

E = 124.7 mV

Therefore, the magnitude of the induced voltage at a constant magnetic field is 124.7 mV

Answer:

     λ = 52.5 nm

Explanation:

De Broglie's duality principle states that all matter has wave and particle characteristics, being related by the expression

          p = h / λ

where the moment

          p = mv

          λ = h / mv

let's calculate

        λ = 6.63 10⁻³⁴ / 1.675 10⁻²⁷ 7.54

        λ = 5.25 10⁻⁸ m

Let's reduce anm

       λ = 5.25 10⁻⁸ m (10⁹ nm / 1m)

       λ = 52.5 nm

Answer:

it is going to D. all of these are resistors

The researchers need to compare those who contracted the disease to those who did not.

Answer:

Explanation:

C. The temperature of the coffee will decrease while the temperature of the table increases.

The angle of the resultant vector is equal to the inverse tangent of the quotient of the y-component divided by the x-component of the resultant vector.

To find the angle of a resultant vector, the vector must be resolved into y-component and x-component.

The angle of this resultant vector is also known as the direction of the vector.

Mathematically, the direction of a resultant vector is given as;

[tex]\theta = tan^{-1} (\frac{R_y}{R_x} )\\\\where;\\\\\theta \ is \ the \ direction \ of \ the \ resultant \ vetcor\\\\R_y \ is \ the \ magnitude \ of \ the\ vector \ resolved \ in \ y - direction\\\\R_x \ is \ the \ magnitude \ of \ the\ vector \ resolved \ in \ x - direction[/tex]

Therefore, the angle of the resultant vector is equal to the inverse tangent of the quotient of the y-component divided by the x-component of the resultant vector.

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Answer:

Ultraviolet light from the sun.

Explanation:

This is an example of radiation.

Answer:

X-Ray

Explanation:

x-Ray is an example of radiation.

Answer: i dont do physics yet lol

Explanation:

Answer:

The magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

Explanation:

The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.

The radial acceleration is given by;

[tex]a_t = ar[/tex]

where;

a is the angular acceleration and

r is the radius of the circular path

[tex]a_t = ar\\\\a_t = 1.6 *0.13\\\\a_t = 0.208 \ m/s^2[/tex]

Determine time of the rotation;

[tex]\theta = \frac{1}{2} at^2\\\\0.4 = \frac{1}{2} (1.6)t^2\\\\t^2 = 0.5\\\\t = \sqrt{0.5} \\\\t = 0.707 \ s\\\\[/tex]

Determine angular velocity

ω = at

ω = 1.6 x 0.707

ω = 1.131 rad/s

Now, determine the radial acceleration

[tex]a_r = \omega ^2r\\\\a_r = 1.131^2 (0.13)\\\\a_r = 0.166 \ m/s^2[/tex]

The magnitude of total linear acceleration is given by;

[tex]a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{0.208^2 + 0.166^2} \\\\a = 0.266 \ m/s^2\\\\a = 0.27 \ m/s^2[/tex]

Therefore, the magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

It depends on what stage of the mission you're talking about.

==>  While it's sitting on the pad before launch, the forces on the rocket are balanced, so there's no net force on it.

==>  When the engines ignite, their thrust (d) is greater than the force of gravity.  So the net force on the rocket is upward, and the spacecraft accelerates upward.

==>  After the engines shut down, the net force acting on the rocket is due to Gravity (c).

. . . If the rocket has enough vertical speed, it escapes the Earth completely, and just keeps going.  

. . . If it has enough horizontal speed, it enters Earth orbit.  

. . . If it doesn't have enough vertical or horizontal speed, it falls back to Earth.    

A rocket will preserve to speed up so long as there's a resultant pressure upwards resulting from the thrust of the rocket engine.

A rocket launches whilst the pressure of thrust pushing it upwards is greater than the burden force because of gravity downwards. This unbalanced pressure reasons a rocket to accelerate upwards. A rocket will maintain to hurry up so long as there's a resultant force upwards resulting from the thrust of the rocket engine.

What's the net pressure of unbalanced?

If the forces on an item are balanced, the net pressure is zero. If the forces are unbalanced forces, the results do not cancel each difference. Any time the forces acting on an object are unbalanced, the net pressure is not 0, and the movement of the item modifications.

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Answer:

The velocity is 40 ft/sec.

Explanation:

Given that,

Force = 3200 lb

Angle = 30°

Speed = 64 ft/s

The resistive force with magnitude proportional to the square of the speed,

[tex]F_{r}=kv^2[/tex]

Where, k = 1 lb s²/ft²

We need to calculate the velocity

Using balance equation

[tex]F\sin\theta-F_{r}=m\dfrac{d^2v}{dt^2}[/tex]

Put the value into the formula

[tex]3200\sin 30-kv^2=m\dfrac{d^2v}{dt^2}[/tex]

Put the value of k

[tex]3200\times\dfrac{1}{2}-v^2=m\dfrac{d^2v}{dt^2}[/tex]

[tex]1600-v^2=m\dfrac{d^2v}{dt^2}[/tex]

At terminal velocity [tex]\dfrac{d^2v}{dt^2}=0[/tex]

So, [tex]1600-v^2=0[/tex]

[tex]v=\sqrt{1600}[/tex]

[tex]v=40\ ft/sec[/tex]

Hence, The velocity is 40 ft/sec.

Answer:

C if it is a negitive ion it has more electrons because protons determine what element it is

Answer:

1.37 ×10^-3 T

Explanation:

From;

B= μnI

μ = 4π x 10-7 N/A2

n= number of turns /length of wire = 1700/0.75 = 2266.67

I= 0.48 A

Hence;

B= 4π x 10^-7 × 2266.67 ×0.48

B= 1.37 ×10^-3 T

Answer:

The  radius of the earth is [tex]r = 6365.4 \ km[/tex]

Explanation:

From the question we are told that

     The distance at  Alexandria is  [tex]d_a = 800 \ km = 800 *10^{3} \ m[/tex]

      The angle of the sun is  [tex]\theta = 7.2 ^o[/tex]

So we want to first obtain the circumference of the earth

   So let assume that the earth is  circular ([tex]360 ^o[/tex])

  Now from question we know that the sun made an angle of [tex]7.2 ^o[/tex] so with this we will obtain how many  [tex](7.2 ^o)[/tex]  are in [tex]360^o[/tex]

 i.e    [tex]N = \frac{360}{7.2}[/tex]

=>      [tex]N = 50[/tex]

     With this  value we can evaluate the circumference as

             [tex]c = 50 * 800[/tex]

              [tex]c = 40000 \ km[/tex]

Generally circumference is mathematically represented as

        [tex]c = 2\pi r[/tex]

         [tex]40000 = 2 * 3.142 * r[/tex]

=>        [tex]r = 6365.4 \ km[/tex]

Answer:

The maximum temperature increase is [tex]\Delta T = 0.0497 \ ^oC[/tex]

Explanation:

From the question we are told that

    The mass of the bullet is [tex]m = 17.0 \ g =0.017 \ kg[/tex]

     The  speed is  [tex]v_1 = 785 \ m/s[/tex]

     The mass of the water is  [tex]m_w = 13.5 \ kg[/tex]

     The velocity it emerged with is  [tex]v_2 = 534 \ m/s[/tex]

Generally due to the fact that energy can nether be created nor destroyed but transferred from one form to another then  

the change in kinetic energy of the bullet =  the heat gained by the water

 So

 The change in kinetic energy of the water is  

          [tex]\Delta KE = \frac{1}{2} m (v_1^2 - v_2 ^2 )[/tex]

substituting values  

        [tex]\Delta KE =0.5 * 0.017 * (( 785)^2 - (534) ^2 )[/tex]

        [tex]\Delta KE = 2814.1 \ J[/tex]

Now the heat gained by the water is

     [tex]Q = m_w* c_w * \Delta T[/tex]

Here [tex]c_w[/tex] is the specific heat of water which has a value  [tex]c_w = 4190 J/kg \cdot K[/tex]

So  since   [tex]\Delta KE = Q[/tex]  

we have that

          [tex]2814.1 = 13.5 * 4190 * \Delta T[/tex]

          [tex]\Delta T = 0.0497 \ ^oC[/tex]

Answer:

The wavelength of the ligt beam in a material of refractive index 2.50 is 198 mm

d. 198 mm

Explanation:

Refractive index is given by;

[tex]\mu= \frac{\lambda_{vacuum}}{\lambda _{medium}}[/tex]

where;

[tex]\lambda_{vacuum}[/tex] is the wavelength of the light beam in vacuum

[tex]\lambda_{medium}[/tex] is the wavelength of the beam in a material

[tex]\mu= \frac{\lambda_{vacuum}}{\lambda _{medium}} \\\\\lambda_{vacuum} = \mu *\lambda _{medium}\\\\\ the \ wavelength \ of \ the \ light \ beam \ is \ constant \ in \ a \ vacuum\\\\ \mu_1 *\lambda _{medium}_1 = \mu_2 *\lambda _{medium}_2\\\\\lambda _{medium}_2 = \frac{ \mu_1 *\lambda _{medium}_1 }{ \mu_2} \\\\\lambda _{medium}_2 =\frac{1.5*330}{2.5} \\\\\lambda _{medium}_2 = 198 \ mm[/tex]

Therefore, the wavelength of the ligt beam in a material of refractive index 2.50 is 198 mm.

d. 198 mm

Answer:

Parte B : 31.18º , Parte C: 31.17º

Explanation:

Parte B: The angle between the glass and the water before it enters the water is going to be equal to the value of the angle when it enters the glass , 27.13º.

Using the formula n1 senθ1 = n2 senθ2 , where n1=1.51 , θ1=27.13º, n2=1.33 , it gives us θ2=31.18º.

Parte C: n1= 1 , θ1=43.5º, n2=1.33

Using the same formula : n1 senθ1 = n2 senθ2 , it gives us θ2= 31.17º.

Answer:

c = 3.00E108 m/s = 3.00E5 km/s

t = S / v = 3.84E5 / 3.00E5 = 1.28 sec