Estimate the peak electric field inside a 1.2-kW microwave oven under the simplifying approximation that the microwaves propagate as a plane wave through the oven's 700-cm2 cross-sectional area.

Answer:

9.42*10^-4 m^3/s

Explanation:

d=3/100 m

=0.03 m

A=3.14*0.03^2/6

=4.71*10^-4 m^2

Volume flow rate V = A * s

= 4.71*10^-4 * 2

= 9.42*10^-4

So, the volume flw rate of fluid is 9.42*10^-4 m^3/s

Answer:

Explanation:

a )

This type of spectrum is called line emission spectrum . Because it consists of lines . It is emission spectrum because it is due to emission of radiation from a source .

b ) The wavelength of a photon  is inversely proportional to its energy .  Photon  due to transition between n = 1 and n = 3 will have higher energy than

that due to transition between n = 2 and n = 5 . So the later photon ( B)  will have greater wavelength or photon  due to transition between n = 2 and n = 5 will have greater wavelength .

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, [tex]q_1=-3\ nC[/tex]

It is placed at a distance of 9 cm at x axis

Charge, [tex]q_2=+4\ nC[/tex]

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

[tex]\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0[/tex]

Here,

[tex]r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}[/tex]

So,

[tex]\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}[/tex]

Squaring both sides,

[tex]3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m[/tex]

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

According to the question,

Charge,

Now,

→ [tex]\frac{kq_1}{r_1} +\frac{kq_2}{r_2} =0[/tex]

or,

→ [tex]\frac{kq_1}{r_1} =-\frac{kq_2}{r_2}[/tex]

→  [tex]\frac{q_1}{r_1} = \frac{q_2}{r_2}[/tex]

here,

[tex]r_1 = \sqrt{y^2+81}[/tex]

[tex]r^2 = \sqrt{y^2+225}[/tex]

By substituting the values,

→      [tex]\frac{-3 }{\sqrt{y^2+225} } = -\frac{4}{\sqrt{y^2+225} }[/tex]

By applying cross-multiplication,

[tex]3\times \sqrt{y^2+225} = 4\times \sqrt{y^2+81}[/tex]

By squaring both sides, we get

→  [tex]9(y^2+225) = 16(y^2+81)[/tex]

    [tex]9y^2+2025 = 16 y^2+1296[/tex]

   [tex]2025-1296=7y^2[/tex]

               [tex]7y^2=729[/tex]

                  [tex]y = 10.2 \ m[/tex]

Thus the solution above is correct.

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Answer:

Explanation:

d

Answer:

352 m/s

Explanation:

The velocity of sound in air is approximated as:

v ≈ 331 + 0.6 T

where v is the velocity in m/s and T is the temperature in Celsius.

At T = 35:

v = 331 + 0.6 (35)

v = 352 m/s

Answer:

2

Explanation:

since the second object was in it stationary, it velocity is 0 m/s

Answer:

The time rate of change of the electric field between the plates is  [tex]\frac{E }{t} = 2.29 *10^{14} \ N \cdot C \cdot s^{-1}[/tex]  

Explanation:

From the question we are told that

    The  radius is  [tex]r = 2.8 \ cm = 0.028 \ m[/tex]

     The distance of separation is  [tex]d = 1.1 \ mm = 0.0011 \ m[/tex]

      The  current is  [tex]I = 5 \ A[/tex]

Generally the electric field generated is mathematically represented as

         [tex]E = \frac{q }{ \pi * r^2 \epsilon_o }[/tex]

Where [tex]\epsilon_o[/tex] is the permitivity of free space with a value

          [tex]\epsilon_o = 8.85*10^{-12 }\ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

So the time rate of change of the electric field between the plates is mathematically represented as

        [tex]\frac{E }{t} = \frac{q}{t} * \frac{1 }{ \pi * r^2 \epsilon_o }[/tex]

But [tex]\frac{q}{t } = I[/tex]

So  

       [tex]\frac{E }{t} = * \frac{I }{ \pi * r^2 \epsilon_o }[/tex]

substituting values  

        [tex]\frac{E }{t} = * \frac{5 }{3.142 * (0.028)^2 * 8.85 *10^{-12} }[/tex]

        [tex]\frac{E }{t} = 2.29 *10^{14} \ N \cdot C \cdot s^{-1}[/tex]

Answer:

0.0081T

Explanation:

The magnetic field B in the toroid is proportional to the applied current I and the number of turns N per unit length L of the toroid. i.e

B ∝ I [tex]\frac{N}{L}[/tex]

B = μ₀ I [tex]\frac{N}{L}[/tex]                   ----------------(i)

Where;

μ₀ = constant of proportionality called the magnetic constant = 4π x 10⁻⁷N/A²

Since the radius (r = 4.2cm = 0.042m) of the toroid is given, the length L is the circumference of the toroid given by

L = 2π r

L = 2π (0.042)

L = 0.084π

The number of turns N = 1000

The current in the toroid = 1.7A

Substitute these values into equation (i) to get the magnetic field as follows;

B = 4π x 10⁻⁷ x  1.7 x  [tex]\frac{1000}{0.084\pi }[/tex]        [cancel out the πs and solve]

B = 0.0081T

The magnetic field along the central radius is 0.0081T

Answer:

Electric potential at position, x = -20 m, = -100 V

Electric potential at position, x = 0 m, = 0

Electric potential at position, x = 10 m, = 50 V

Electric potential at position, x = 11 m, = 55 V

Electric potential at position, x = 99 m, 495 V

Explanation:

Given;

magnitude of electric field, E = 5.0 V/m

at position x = 10 m, electric potential = 50 V

Electric potential at position, x = -20 m

V = Ex

V = 5 (-20)

V = -100 V

Electric potential at position, x = 0 m

V = Ex

V = 5(0)

V = 0

Electric potential at position, x = 10 m

V = Ex

V = 5(10)

V = 50 V

Electric potential at position, x = 11 m

V = Ex

V = 5(11)

V = 55 V

Electric potential at position, x = 99 m

V = Ex

V = 5(99)

V = 495 V

Answer:

B. only when the current in the first coil changes

Explanation:

This is because for current to be induced in the second coil they must be an change in current inyhe first coil in line with Faraday's first law of electromagnetic induction. Which state that whenever their is a change in magnetic lines of flux an emf is induced

Answer:

1000j

Explanation:

work done = force x distance

w = 5 x 10 x 20 = 1000joules

Answer:

B. over the symbol.

Explanation:

vectors are represented with a symbol carrying an arrow head with also indicates direction

Answer:

(a) 0.59 ms

(b) 0.15 ms

(c) The significance is that the speed of sound in different media determines the time interval of perception by the ears, which are at constant distance apart.

Explanation:

(a) distance between ears = 20 cm = 0.2 m

speed of sound generated = 340 m/s

time = ?

speed = [tex]\frac{distance covered}{time taken}[/tex]

⇒ time taken, t = [tex]\frac{distance covered}{speed}[/tex]

                        = [tex]\frac{0.2}{340}[/tex]

                        = 5.8824 × [tex]10^{-4}[/tex]

                        = 0.59 ms

The time interval of the arrival of the sound at the right ear to the left ear is 0.59 ms.

(b) distance between ears = 20 cm = 0.2 m

speed of sound in water = 1530 m/s

time = ?

speed = [tex]\frac{distance covered}{time taken}[/tex]

⇒ time taken, t = [tex]\frac{distance covered}{speed}[/tex]

                         = [tex]\frac{0.2}{1530}[/tex]

                         = 1.4815 × [tex]10^{-4}[/tex]

                         = 0.15 ms

The sound heard by the right ear of the diver would arrive at the left 0.15 ms latter.

(c) The significance is that the speed of sound in different media, determines the time interval of perception by the ears, which are at constant distance apart.

A) The time interval between when the sound arrives at the right ear and the sound arrives at the left ear is; t = 0.588 × 10⁻³ seconds

B) The time interval between when the sound arrives at the right ear and the sound arrives at the left ear if the speed of sound in water is 1,530 m/s is; t = 0.131 × 10⁻⁵ seconds

C) The significance about the time interval of the two situations is that;

Transmission of sound varies with different mediums.

A) We are given;

Distance between the two ears; d = 20 cm = 0.2 m

Speed of sound; v = 340 m/s

Since the sound source is placed at the right ear, the time interval for it to get to the left ear is;

t = d/v

t = 0.2/340

t = 0.588 × 10⁻³ seconds

B) We are now told that the speed of sound in water is 1530 m/s. Thus;

t = 0.2/1530

t = 0.131 × 10⁻⁵ seconds

C) We can see that in answer A and B, the time interval is different even when the distance remained the same. This means that, the time interval of hearing a sound changes with respect to the medium of transmission.

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Answer:

Explanation:

The whole wave front may be divided into two halves , the upper half and the lower half . Waves coming from top of the slit or top of upper half and top of lower half or from the mid point of slit can form minima at given point only when there is phase difference of π radian or path difference of λ or one wavelength. Every other point in upper half and corresponding point in lower half will interfere destructively at that point and will form dark spot at the given point . In this way minima will be formed at that point

Hence the phase difference between the Huygens wavelet from the top of the slit and the wavelet from the midpoint of the slit at first minima  is π radian .

This is note the complete question, the complete question is:

One of the lousy things about getting old (prepare yourself!) is that you can be both near-sighted and farsighted at once. Some original defect in the lens of your eye may cause you to only be able to focus on some objects a limited distance away (near-sighted). At the same time, as you age, the lens of your eye becomes more rigid and less able to change its shape. This will stop you from being able to focus on objects that are too close to your eye (far-sighted). Correcting both of these problems at once can be done by using bi-focals, or by placing two lenses in the same set of frames. An old physicist instructor can only focus on objects that lie at distance between 0.47 meters and 5.4 meters.

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest 2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness?

Answer:  3.04 D

Explanation:

when an object is held 21 cm away from the instructor's eyes, the spectacle lens must produce 0.47m ( the near point) away.

An image of 0.47m from the eye will be ( 47 - 2 )

i.e 45 cm from the spectacle lens since the spectacle lens is 2cm away from the eye.

Also, the image distance will become negative

gap between lense and eye = 2cm

Therefore;

image distance d₁ = - 45cm = - 0.45m

object distance  d₀ = 21 - 2 = 19cm = 0.19m

P = 1/f = 1/ d = 1/d₀ + 1/d₁ = 1/0.19 + (-1/0.45)

P = 1/f =  5.26315789 - 2.22222222

P = 1/f = 3.04093567 ≈ 3.04 D

Answer:

1) 14 kW

2) 4.67 x 10^-5 N

Explanation:

Area of solar panel = 10 m^2

Intensity of sun's radiation incident on earth = 1.4 kW/m^2

Solar power absorbed = ?

We know that the intensity of radiation on a given area is

[tex]I[/tex] = [tex]\frac{P}{A}[/tex]

where I is the intensity of the radiation

P is the power absorbed due to this intensity on a given area

A is the area on which this radiation is incident

From the equation, we have

P = IA

P = 1.4 kW/m^2  x  10 m^2 = 14 kW

b) For a perfect absorbing surface, the radiation pressure is given as

p = I/c

where p is the radiation pressure

I is the incident light intensity = 1.4 kW/m^2 = 1.4 x 10^3 kW/m^2

c is the speed of light = 3 x 10^8 m/s

substituting values, we have

p = (1.4 x 10^3)/(3 x 10^8) = 4.67 x 10^-6 Pa

we know that Force = pressure x area

therefore force on the solar panels is

F = 4.67 x 10^-6 x 10 = 4.67 x 10^-5 N

Answer:

1. λ = 0.48 cm = 4800 μm

2. υ = 6.25 x 10¹⁰ Hz

3. E = 4.14 x 10⁻²³ J

Explanation:

1.

Since, the wavelength is defined as the distance between two consecutive or successive crests or troughs. Therefore, in this case the wavelength will be equal to:

Wavelength = λ = Distance between 5 successive crests/5

λ = 2.4 cm/5

λ = 0.48 cm = 4800 μm

2.

The frequency of photon can be given as:

υ = c/λ

where,

υ = frequency of photon = ?

c = speed of light = 3 x 10⁸ m/s

λ = wavelength = 0.48 cm = 0.0048 m

Therefore,

υ = (3 x 10⁸ m/s)/(0.0048 m)

υ = 6.25 x 10¹⁰ Hz

3.

Now, the energy of photon is given as:

E = hυ

where,

E = Energy = ?

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

Therefore,

E = (6.626 x 10⁻³⁴ J.s)(6.25 x 10¹⁰ Hz)

E = 4.14 x 10⁻²³ J

Answer: d = 33 cm or 0.33 m

Explanation: In physics, Work is the amount of energy transferred to an object to make it move. It can be expressed by:

W = F.d.cosθ

F is the force applied to the object, d is the displacement and θ is the angle formed between the force and the displacement.

For the ice block, the angle is 0, i.e., force and distance are at the same direction, so:

W = F.d.cos(0)

W = F.d

To determine d:

d = [tex]\frac{W}{F}[/tex]

d = [tex]\frac{3}{9}[/tex]

d = 0.33 m

The distance d the block ice moved is 33 cm.

The correct option is option (B)

The orbit of the planets is elliptical with the sun at one of the foci.

According to Kepler's Laws:

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Answer:

The frequency is    [tex]f = 0.221 \ Hz[/tex]

Explanation:

From the question we are told that  

     The  time taken for it to decay to half its original size is [tex]t = 3.40 \ ms = 3.40 *10^{-3} \ s[/tex]

Let the voltage of the capacitor when it is fully charged be  [tex]V_o[/tex]

Then the voltage of the capacitor at time t is  said to be  [tex]V = \frac{V_o}{2}[/tex]

   Now  this voltage can be  mathematical represented as

      [tex]V = V_o * e ^{-\frac{t}{RC} }[/tex]

Where  RC  is the time constant

   substituting values  

    [tex]\frac{V_o}{2} = V_o * e ^{-\frac{3.40 *10^{-3}}{RC} }[/tex]

    [tex]0.5 = e^{-\frac{3.40 *10^{-3}}{RC} }[/tex]

    [tex]- \frac{0.5}{RC} = ln (0.5)[/tex]

     [tex]-\frac{0.5}{RC} = -0.6931[/tex]

     [tex]RC = 0.721[/tex]

Generally the cross-over frequency for a low pass filter is mathematically represented as

          [tex]f = \frac{1}{2 \pi * RC }[/tex]

substituting values  

           [tex]f = \frac{1}{2* 3.142 * 0.72 }[/tex]

           [tex]f = 0.221 \ Hz[/tex]

Answer:

A. Attractive

B. ( μ₀I² ) / ( 2πd )

Explanation:

A. We know that currents in the same direction attract, and currents in the opposite direction repel, according to ampere's law. In this case the current in the two wires are flowing in the same direction, and hence the force between the two wires are attractive.

B. Suppose that two wires of length [tex]l_1[/tex] and [tex]l_2[/tex] both carry the current [tex]I[/tex] in the same direction ( given ). In the presence of a magnetic field produced by wire 1, a force of magnitude m say, is experienced by wire 2. The magnitude of the magnetic field produced by wire 1 at distance say d, from it's axis, should thus be the following -

[tex]B_1[/tex] = μ₀I / 2πd

The force experienced by wire 2 should thus be -

[tex]F_2[/tex] = I( [tex]l_2[/tex] [tex]*[/tex] [tex]B_1[/tex] )

= I [tex]*[/tex] [tex]l_2 * B_1 *[/tex] Sin( 90 )

= I [tex]*[/tex] [tex]l_2[/tex] ( μ₀I / 2πd )

Therefore the force per unit length experienced by wire 2 toward wire 1 should be ...

( [tex]F_2[/tex] / [tex]l_2[/tex] ) = ( μ₀I² ) / ( 2πd ) ... which is our solution

Answer:

A. the inertia of shelly

Explanation:

Answer:

A.  the inertia of Shelly

Explanation:

newtons first law states that an object at rest stays at rest unless on opposite force act against it, so when the buss started to move it acted on shelly forcing her to move back. Shelly's inertia was at a halt and was forced back by the motion of the bus.

Answer:

The rejected by the air conditioning system is 1.75 kilowatts.

Explanation:

A air conditioning system is a refrigeration cycle, whose receives heat from cold reservoir with the help of power input before releasing it to hot reservoir. The First Law of Thermodynamics describes the model:

[tex]\dot Q_{L} + \dot W - \dot Q_{H} = 0[/tex]

Where:

[tex]\dot Q_{L}[/tex] - Heat rate from cold reservoir, measured in kilowatts.

[tex]\dot Q_{H}[/tex] - Heat rate liberated to the hot reservoir, measured in kilowatts.

[tex]\dot W[/tex] - Power input, measured in kilowatts.

The heat rejected is now cleared:

[tex]\dot Q_{H} = \dot Q_{L} + \dot W[/tex]

If [tex]\dot Q_{L} = 1\,kW[/tex] and [tex]\dot W = 0.75\,kW[/tex], then:

[tex]\dot Q_{H} = 1\,kW + 0.75\,kW[/tex]

[tex]\dot Q_{H} = 1.75\,kW[/tex]

The rejected by the air conditioning system is 1.75 kilowatts.

Answer:

Three halves of a wavelength I.e 7lambda/2

Explanation:

See attached file pls

Answer:

(b) Increases

Explanation:

The potential energy between two point charges is given as;

[tex]U = F*r = \frac{kq_1q_2}{r}[/tex]

Where;

k is the coulomb's constant

q₁  ans q₂ are the two point charges

r is the distance between the two point charges

Since the two charges have opposite sign;

let q₁ be negative and q₂ be positive

Substitute in these charges we will have

[tex]U = \frac{k(-q_1)(q_2)}{r} \\\\U = - \frac{kq_1q_2}{r}[/tex]

The negative sign in the above equation shows that as the distance between the two charges increases, the potential energy increases as well.

Therefore, as you move the point charges farther and farther apart, the potential energy of this system relative to infinity Increases.

Answer:

Explanation:

Let the charge on proton be q .

energy gain by proton in a field having potential difference of V₀

= V₀ q

Due to gain of energy , its kinetic energy becomes 1/2 m v₀²

where m is mass and v₀ is velocity of proton

V₀ q = 1/2 m v₀²

In the second case , gain of energy in electrical field

= 2 V₀q , if v be the velocity gained in the second case

2 V₀q = 1/2 m v²

1/2 m v² = 2 V₀q = 2 x 1/2 m v₀²

mv² = 2  m v₀²

v = √2 v₀

Answer:

sphere #1 carries positive charge and #2 carries negative charge

This is because from the laws of static electricity, disconnecting the copper wire makes #1 to be positively charged and #2 to be negatively charged

Answer:

408N at 89.89°

Explanation:

This problem requires that we resolve the force vectors into

x- and y

-componentsOnce this is done, we can add the components easily, as the one 2-dimensional problem will be two 1-dimensional problems.

Finally, we will convert the resultant force into standard form and find the equilibrant.

Resolve into components:

F1x =F1cos 180°= 232(−1)=−232N

F1y=F1sin180°=0N

F2x=F2cos(−140°)=194(−0.766)=−148.6N

F1y=F1sin(−140°)=232(−0.643)=−149.17N

Note the change of the angle used to give the direction of

F2. Standard angles (rotation from thex

-axis; counterclockwise is +) should be used to avoid sign errors in the results.

Now, we add the components:

Fx=F1x+F2x=−380.6N

Fy=F1y+F1y=−148.17N

Technically, this is the resultant force. However, it should be changed back into standard form. Here's how:

F=√(Fx)2(Fy)2=√(−380.6)^2(−148.17)^2=408N

θ=tan−1(−148.17−380.6)

=89.89°

Answer:

Explanation:

Using the law of gravitation of force to solve this question. The law states that the Force of attraction between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distances between them.

Mathematically, F = GMaMb/R² ... 1

G is the gravitational constant

Ma and Mb are the masses of the balls

R is the distance between the balls

If the mass of ball B is tripled and the magnitude of the separation of the balls is increased to 3R, the force between them will be;

F₂ = GMa(3Mb)/(3R)²

F₂ = 3GMaMb/9R² ... 2

Dividing equation 1 by 2 we will have;

F₂/F = (3GMaMb/9R²)/GMaMb/R²

F₂/F =  3GMaMb/9R² * GMaMb/R²

F₂/F = 3/9

F₂/F = 1/3

F₂ = 1/3 F

This shows that the magnitude of the new attractive force is one-third that of the initial attractive force

Answer:

Explanation:

Using E= hc/wavelength

6.63 x10^-34 x3x10^8/ 236nm

19.86*10^-26/236*10^-9

=0.08*10^-35Joules