If the magnetic field of an electromagnetic wave is in the +x-direction and the electric field of the wave is in the +y direction, the wave is traveling in what direction? Explain your answer.

Answers

Answer 1

Answer:

Explanation:

The direction of propagation of electromagnetic wave

is given by the direction of vector E x B where E is electrical field , B is magnetic field .

Given Electric field  = E i because it is along x axis

Magnetic field = Bj because it is along y axis

E x B = Ei x Bj

= EB k .

so direction of E  x B is along k direction or z  - axis so wave is propagating along z - axis .

Answer 2

The direction of motion of electromagnetic wave will be +z-direction.

Electromagnetic waves are waves that consist of the electric field and magnetic field.

The electric and magnetic fields are perpendicular to each other and the wave propagates in the direction perpendicular to both the fields.

Now, the direction of wave motion can be estimated by taking the cross-product of directional unit vectors of the electric and magnetic fields.

The electric field is in the +y direction and the magnetic field is in the +x-direction.

So, the direction of the wave will be,

[tex]i\times j=k[/tex]

Therefore, the direction of motion of electromagnetic wave will be +z-direction.

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Related Questions

At what rate must Uranium 235 undergo fission by neutron bombardment to generate energy at a rate of 100 W (1 W

Answers

Complete Question

At what rate must Uranium 235 undergo fission by neutron bombardment to generate energy at a rate of 100 W (1 W = 1 J/s)? Assume each fission reaction releases 200 MeV of energy.

Answer

a. Approximately [tex]5*10^{10}[/tex] fissions per second.

b. Approximately [tex]6*10^{12 }[/tex]fissions per second.

c. Approximately [tex]4*10^{11}[/tex] fissions per second.

d. Approximately [tex]3*10^{12}[/tex] fissions per second.

e. Approximately[tex]3*10^{14}[/tex] fissions per second.

Answer:

The correct option is  d

Explanation:

From the question we are told that

       The energy released by each fission reaction [tex]E = 200 \ MeV = 200 *10^{6} * 1.60 *10^{-19} =3.2*10^{-11} \ J /fission[/tex]

Thus to generated  [tex]100 \ J/s[/tex] i.e  (100 W  ) the rate of fission is  

              [tex]k = \frac{100}{3.2 *10^{-11} }[/tex]

              [tex]k =3*10^{12} fission\ per \ second[/tex]

) Calculate current passing in an electrical circuit if you know that the voltage is 8 volts and the resistance is 10 ohms

Answers

Explanation:

Hey, there!

Here, In question given that,

potential difference (V)= 8V

resistance (R)= 10 ohm

Now,

According to the Ohm's law,

V= R×I { where I = current}

or, I = V/R

or, I = 8/10

Therefore, current is 4/5 A or 0.8 A.

(A= ampere = unit of current).

Hope it helps...

Can abnormality exist outside of a cultural context

Answers

you should search this up and put your own thoughts into it, it’s always good to learn something new!!

A thermos bottle works well because:

a. its glass walls are thin
b. silvering reduces convection
c. vacuum reduces heat radiation
d. silver coating is a poor heat conductor
e. none of the above

Answers

Answer:

A thermos bottle works well because:

A) Its glass walls are thin

Answer:

A thermos bottle works well because:

C

Vacuum reduces heat radiation

The velocity function (in meters per second) is given for a particle moving along a line. Find the total distance traveled by the particle during the given interval

Answers

Answer:

s=((vf+vi)/2)t vf is final velocity and vi is initial velocity

Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visible light with no greater than 1.0 mW total power. They're relatively safe because the eye's blink reflex limits exposure time to 250 ms.

Requried:
a. Find the intensity of a 1-mW class 2 laser with beam diameter 2.0 mm .
b. Find the total energy delivered before the blink reflex shuts the eye.
c. Find the peak electric field in the laser beam.

Answers

Answer:

a) 318.2 W/m^2

b) 2.5 x 10^-4 J

c) 1.55 x 10^-8 v/m

Explanation:

Power of laser P = 1 mW = 1 x 10^-3 W

exposure time t = 250 ms = 250 x 10^-3 s

If beam diameter = 2 mm = 2 x 10^-3 m

then

cross-sectional area of beam A = [tex]\pi d^{2} /4[/tex] = (3.142 x [tex](2*10^{-3} )^{2}[/tex])/4

A = 3.142 x 10^-6 m^2

a) Intensity I = P/A

where P is the power of the laser

A is the cros-sectional area of the beam

I = ( 1 x 10^-3)/(3.142 x 10^-6) = 318.2 W/m^2

b) Total energy delivered E = Pt

where P is the power of the beam

t is the exposure time

E = 1 x 10^-3 x 250 x 10^-3 = 2.5 x 10^-4 J

c) The peak electric field is given as

E = [tex]\sqrt{2I/ce_{0} }[/tex]

where I is the intensity of the beam

E is the electric field

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

[tex]e_{0}[/tex] = 8.85 x 10^9 m kg s^-2 A^-2

E = [tex]\sqrt{2*318.2/3*10^8*8.85*10^9}[/tex]  = 1.55 x 10^-8 v/m

(a)  The intensity of laser beam is  [tex]318.2 \;\rm W/m^{2}[/tex].

(b)  The total energy delivered before the blink reflex shuts the eye is [tex]2.5 \times 10^{-4} \;\rm J[/tex].

(c)  The required value of peak electric field in the laser beam is [tex]1.55 \times 10^{-8} \;\rm V/m[/tex].

Given data:

The power of laser is, [tex]P=1 \;\rm mW = 1 \times 10^{-3} \;\rm W[/tex].

The exposure time is, [tex]t = 250\;\rm ms = 250 \times 10^{-3} \;\rm s[/tex].

The beam diameter is, [tex]d = 2 \;\rm mm = 2 \times 10^{-3} \;\rm m[/tex].

a)

The standard expression for the intensity of beam is given as,

I = P/A

Here, P is the power of the laser  and A is the cross-sectional area of the beam. And its value is,

[tex]A =\pi /4 \times d^{2}\\\\A =\pi /4 \times (2 \times 10^{-3})^{2}\\\\A =3.142 \times 10^{-6} \;\rm m^{2}[/tex]

Then intensity is,

[tex]I = (1 \times 10^{-3})/(3.142 \times 10^{-6})\\\\I =318.2 \;\rm W/m^{2}[/tex]

Thus, the intensity of laser beam is [tex]318.2 \;\rm W/m^{2}[/tex].

(b)

The expression for the total energy delivered is given as,

E = Pt

Solving as,

[tex]E = 1 \times 10^{-3} \times (250 \times 10^{-3})\\\\E = 2.5 \times 10^{-4} \;\rm J[/tex]

Thus, the total energy delivered before the blink reflex shuts the eye is [tex]2.5 \times 10^{-4} \;\rm J[/tex].

(c)

The expression for the peak electric field is given as,

[tex]E = \sqrt{\dfrac{2I}{c \times \epsilon_{0}}}[/tex]

Solving as,

[tex]E = \sqrt{\dfrac{2 \times 318.2}{(3 \times 10^{8}) \times (8.85 \times 10^{9})}}\\\\E =1.55 \times 10^{-8} \;\rm V/m[/tex]

Thus, the required value of peak electric field in the laser beam is [tex]1.55 \times 10^{-8} \;\rm V/m[/tex].

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5. The speed of a transverse wave on a string is 170 m/s when the string tension is 120 ????. To what value must the tension be changed to raise the wave speed to 180 m/s?

Answers

Answer:

The tension on string when the speed was raised is 134.53 N

Explanation:

Given;

Tension on the string, T = 120 N

initial speed of the transverse wave, v₁ = 170 m/s

final speed of the transverse wave, v₂ = 180 m/s

The speed of the wave is given as;

[tex]v = \sqrt{\frac{T}{\mu} }[/tex]

where;

μ is mass per unit length

[tex]v^2 = \frac{T}{\mu} \\\\\mu = \frac{T}{v^2} \\\\\frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}[/tex]

The final tension T₂ will be calculated as;

[tex]T_2 = \frac{T_1 v_2^2}{v_1^2} \\\\T_2 = \frac{120*180^2}{170^2} \\\\T_2 = 134.53 \ N[/tex]

Therefore, the tension on string when the speed was raised is 134.53 N

1. (I) If the magnetic field in a traveling EM wave has a peak magnitude of 17.5 nT at a given point, what is the peak magnitude of the electric field

Answers

Answer:

The electric field is [tex]E = 5.25 V/m[/tex]

Explanation:

From the question we are told that

    The peak magnitude of the magnetic field is  [tex]B = 17.5 nT = 17.5 *10^{-9}\ T[/tex]

Generally the peak magnitude of the electric field is mathematically represented as

         [tex]E = c * B[/tex]

Where c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]

So

       [tex]E = 3.0 *10^{8} * 17.5 *10^{-9}[/tex]

       [tex]E = 5.25 V/m[/tex]

The peak magnitude of the electric field will be "5.25 V/m".

Magnetic field

According to the question,

Magnetic field's peak magnitude, B = 17.5 nT or,

                                                           = 17.5 × 10⁻⁹ T

Speed of light, c = 3.0 × 10⁸ m/s

We know the relation,

→ E = c × B

By substituting the values, we get

      = 3.0 × 10⁸ × 17.5 × 10⁻⁹

      = 5.25 V/m

Thus the above approach is appropriate.

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A solid block is attached to a spring scale. When the block is suspended in air, the scale reads 21.2 N; when it is completely immersed in water, the scale reads 18. 2 N. What are the volume and density of the block?

Answers

Answer:

7066kg/m³

Explanation:

The forces in these cases (air and water) are: Fa =mg =ρbVg Fw =(ρb −ρw)Vg where ρw = 1000 kg/m3 is density of water and ρb is density of the block and V is its density. We can find it from this two equations:

Fa /Fw = ρb / (ρb −ρw) ρb = ρw (Fa /Fa −Fw) =1000·(1* 21.2 /21.2 − 18.2)

= 7066kg/m³

Explanation:

Answer:

The volume of the block is 306 cm³

The density of the block is 7.07 g/cm³

Explanation:

Given;

weight of block in air, [tex]W_a[/tex] = 21.2 N

Weight of block in water, [tex]W_w[/tex] = 18.2 N

Mass of the block in air;

[tex]W_a = mg[/tex]

21.2 = m x 9.8

m = 21.2 / 9.8

m = 2.163 kg

mass of the block in water;

[tex]W_w = mg[/tex]

18.2 = m x 9.8

m = 18.2 / 9.8

m = 1.857 kg

Apply Archimedes principle

Mass of object in air  - mass of object in water = density of water   x  volume                  of object

2.163 kg - 1.857 kg = 1000 kg/m³ x Volume of block

0.306 kg = 1000 kg/m³ x Volume of block

Volume of the block = [tex]\frac{0.306 \ kg}{1000 \ kg/m^3}[/tex]

Volume of the block = 3.06 x 10⁻⁴ m³

Volume of the block = 306 cm³

Determine the density of the block

[tex]Density = \frac{mass}{volume} \\\\Density =\frac{2163 \ g}{306 \ cm^3} \\\\Density = 7.07 \ g/cm^3[/tex]

What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with a wavelength of 580 nm

Answers

Answer:

Explanation:

In case of soap film , light gets reflected from denser medium , hence interference takes place between two waves , one reflected from upper and second from lower surface . For destructive interference the condition is

2μt = nλ where μ is refractive index of water , t is thickness , λ is wavelength of light and n is an integer .

2 x 1.34 x t = 1  x 580

t = 216.42 nm .

Thickness must be 216.42 nm .

Scouts at a camp shake the rope bridge they have just crossed and observe the wave crests to be 9.70 m apart. If they shake the bridge twice per second, what is the propagation speed of the waves (in m/s)?

Answers

Answer:

The speed of the wave is 19.4 m/s

Explanation:

The wave's crest to crest distance (the wavelength of this rope's wave) λ= 9.70 m

The bridge is shaken twice, meaning that two wavelengths passed a given point on the rope per sec. The frequency of a wave is the amount of that wave that passes a given point in a second.

this means that the frequency f = 2 Hz

The speed of a wave = fλ = 9.70 x 2 = 19.4 m/s

Structures on a bird feather act like a diffraction grating having 8500 lines per centimeter. What is the angle of the first-order maximum for 577 nm light shone through a feather?

Answers

Answer:

29.5°

Explanation:

To find the distance d

d = 1E10^-2/8500lines

= 1.17x 10-6m

But wavelength in first order maximum is 577nm

and M = 1

So

dsin theta= m. Wavelength

Theta= sin^-1 (m wavelength/d)

= Sin^-1 ( 1* 577 x10^-8m)/1.17*10^-6

= 493*10^-3= sin^-1 0.493

Theta = 29.5°

A rectangular conducting loop of wire is approximately half-way into a magnetic field B (out of the page) and is free to move. Suppose the magnetic field B begins to decrease rapidly in strength

Requried:
What happens to the loop?

1. The loop is pushed to the left, toward the magnetic field.
2. The loop doesn’t move.
3. The loop is pushed downward, towards the bottom of the page.
4. The loop will rotate.
5. The loop is pushed upward, towards the top of the page.
6. The loop is pushed to the right, away from the magnetic field

Answers

Answer:

. The loop is pushed to the right, away from the magnetic field

Explanation

This decrease in magnetic strength causes an opposing force that pushes the loop away from the field

iven a 36.0 V battery and 14.0 Ω and 84.0 Ω resistors, find the current (in A) and power (in W) for each when connected in series.

Answers

Answer:

0.367A = Current of both resistors

For resistor 1: 1.89W; For resistor 2: 11.3W

Explanation:

When the resistors are connected in series, the equivalent resistance is the sum of both resistors, that is:

R = 14.0Ω + 84.0Ω = 98.0Ω

Using Ohm's law, we can find the current of the circuit (Is the same for both resistors):

V = RI

V / R = I

36.0V / 98.0Ω = I

0.367A = Current of both resistors

Power is defined as:

P = I²*R

For resistor 1:

P = 0.367A²*14.0Ω = 1.89W

For resistor 1:

P = 0.367A²*84.0Ω = 11.3W

A straight wire that is 0.56 m long is carrying a current of 2.6 A. It is placed in a uniform magnetic field, where it experiences a force of 0.24 N. The wire makes an angle of 900 with the magnetic field. What is the magnitude of the magnetic field

Answers

Answer:

0.165Tesla

Explanation:

The Force experienced by the wire in the uniform magnetic field is expressed as F = BILsin∝ where;

B is the magnetic field (in Tesla)

I is the current (in amperes)

L is the length of the wire (in meters)

∝ is the angle that the conductor makes with the magnetic field.

Given parameters

L = 0.56 m

I = 2.6A

F = 0.24N

∝  = 90°

Required

magnitude of the magnetic field (B)

Substituting the given values into the formula given above we will have;

F = BILsin∝

0.24 = B * 2.6 * 0.56 sin90°

0.24 =  B * 2.6 * 0.56 (1)

0.24 = 1.456B

1.456B = 0.24

Dividing both sides by 1.456 will give;

1.456B/1.456 = 0.24/1.456

B ≈ 0.165Tesla

Hence the magnitude of the magnetic field is approximately 0.165Tesla

A circular loop of wire of area 25 cm2 lies in the plane of the paper. A decreasing magnetic field B is coming out of the paper. What is the direction of the induced current in the loop?

Answers

Answer:

counterclockwise

Explanation:

given data

area = 25 cm²

solution

We know that a changing magnetic field induces the current and induced emf is express as

[tex]\epsilon = -N \frac{d \phi }{dt}[/tex]     ..................................1

and we will get here direction of the induced current in the loop that is express by the Lens law that state that the direction of induces current is such that the magnetic flux due to the induced current opposes the change in magnetic flux due to the change in magnetic field

so when magnetic field decrease and point coming out of the paper.

so induced current in the loop will be counterclockwise

A resistor and an inductor are connected in series to an ideal battery of constant terminal voltage. At the moment contact is made with the battery, the voltage across the inductor is

Answers

Answer:

The voltage is equal to the batteries terminal voltage

Explanation:

Explanation:

Grocery store managers contend that there is less total energy consumption in the summer if the store is kept at a low temperature. Make arguments to support or refute this claim, taking into account that there are numerous refrigerators and freezers in the store.

Answers

Answer:

Argument in favor of less total energy consumption if the store is kept at a low temperature

Explanation:

Have in mind that if the store has numerous refrigerators and freezers, the energy consumption of those machines have to be included into the analysis.

Recall that the efficiency (or Coefficient Of Performance - COP) of a frezzer or refrigerator is inversely proportional to the temperature difference between the inside of th machine and the environment where it is operation, therefore the smaller the difference, the highest their efficiency. Therefore, the cooler the environment (the temperature at which the store is kept) the better performance of the running refrigerators and freezers.

Exercise 2.4.5: Suppose we add possible friction to Exercise 2.4.4. Further, suppose you do not know the spring constant, but you have two reference weights 1 kg and 2 kg to calibrate your setup. You put each in motion on your spring and measure the frequency. For the 1 kg weight you measured 1.1 Hz, for the 2 kg weight you measured 0.8 Hz. a) Find k (spring constant) and c (damping constant). Find a formula for the mass in terms of the frequency in Hz. Note that there may be more than one possible mass for a given frequency. b) For an unknown object you measured 0.2 Hz, what is the mass of the object? Suppose that you know that the mass of the unknown object is more than a kilogram.

Answers

Answer:

a) k = 95.54 N / m,   c =   19.55 , b)      m₃ = 0.9078 kg

Explanation:

In a simple harmonic movement with friction, we can assume that this is provided by the speed

          fr = -c v

when solving the system the angular value remains

          w² = w₀² + (c / 2m)²

They give two conditions

1) m₁ = 1 kg

     f₁ = 1.1 Hz

the angular velocity is related to frequency

         w = 2π f₁

Let's find the angular velocity without friction is

         w₂ = k / m₁

we substitute

        (2π f₁)² = k / m₁ + (c / 2m₁)²

2) m₂ = 2 kg

    f₂ = 0.8 Hz

        (2π f₂)² = k / m₂ + (c / 2m₂)²

we have a system of two equations with two unknowns, so we can solve it

we solve (c / 2m)² is we equalize the expression

           (2π f₁)² - k / m₁ = (2π f₂²) 2 - k / m₁

           k (1 / m₂ - 1 / m₁) = 4π² (f₂² - f₁²)

           k = 4π² (f₂² -f₁²) / (1 / m₂ - 1 / m₁)

a) Let's calculate

           k = 4 π² (0.8² -1.1²) / (½ -1/1)

           k = 39.4784 (1.21) / (-0.5)

           k = 95.54 N / m

now we can find the constant of friction

              (2π f₁) 2 = k / m₁ + (c / 2m₁)²

           c2 = ((2π f₁)² - k / m₁) 4m₁²

           c2 = (4ππ² f₁² - k / m₁) 4 m₁²

let's calculate

           c² = (4π² 1,1² - 95,54 / 1) 4 1²

           c² = (47.768885 - 95.54) 8

           c² = -382.1689

           c =   19.55    

b) f₃ = 0.2 Hz

   m₃ =?

              (2πf₃)² = k / m₃ + (c / 2m₃) 2

we substitute the values

              (4π² 0.2²) = 95.54 / m₃ + 382.1689 2/4 m₃²

              1.579 = 95.54 / m₃ + 95.542225 / m₃²

let's call

              x = 1 / m₃

              x² = 1 / m₃²

- 1.579 + 95.54 x + 95.542225 x² = 0

              60.5080 x² + 60.5080 x -1 = 0

                x² + x - 1.65 10⁻² = 0

                  x = [1 ±√ (1- 4 (-1.65 10⁻²)] / 2

                  x = [1 ± 1.03] / 2

                  x₁ = 1.015 kg

                  x₂ = -0.015 kg

Since the mass must be positive we eliminate the second results

                  x₁ = 1 / m₃

                 m₃ = 1 / x₁

                  m₃ = 1 / 1.1015

             

An organ pipe open at both ends is 1.5 m long. A second organ pipe that is closed at one end and open at the other is 0.75 m long. The speed of sound in the room is 330 m/s. Which of the following sets of frequencies consists of frequencies which can be produced by both pipes?

a. 110Hz,220Hz, 330 Hz
b. 220Hz 440Hz 66 Hz
c. 110Hz, 330Hz, 550Hz
d. 330 Hz, 550Hz, 440Hz
e. 660Hz, 1100Hz, 220Hz

Answers

Answer:

A. 110Hz,220Hz, 330 Hz

Explanation:

for organ open at open both ends;

the length of the organ for the fundamental frequency, L = A---->N + N----->A

A---->N  = λ /4 and N----->A = λ /4

L = λ /4 + λ /4 = λ /2

[tex]L = \frac{\lambda}{2} \\\\\lambda = 2L[/tex]

λ  = 2 x 1.5m = 3.0 m

Wave equation is given by;

V = Fλ

Where;

V is the speed of sound

F is the frequency of the wave

F = V/ λ

F₀ = V / 2L

Where;

F₀  is the fundamental frequency

F₀ = 330 / 2(1.5)

F₀ = 330 / 3

F₀ = 110 Hz

the length of the organ for the first overtone, L = A---->N + N----->A + A----->N +  N----->A

L = 4λ /4

L = λ

λ = 1.5 m

F₁ = 330 / 1.5

F₁ = 220 Hz

Thus, F₁ = 2F₀

For open organ at one end

the length of the organ for the fundamental frequency, L = N------A

L = λ /4

λ = 4L

F₀ = V/4L

F₀ = 330 / (4 x 0.75)

F₀ = 110 Hz

the length of the organ for the first overtone, L = N-----N + N-----A

L = λ/2 + λ / 4

L = 3λ /4

F₁ = 3F₀

F₁ = 3 x 110

F₁ = 330 Hz

Thus the fundamental frequency for both organs is 110 Hz,

The first overtone for the organ open at both ends is 220 Hz

The first overtone for the organ open at one end is 330 Hz

The correct option is "A. 110Hz,220Hz, 330 Hz"

The correct option is option (A)

the frequencies produced by the pipes are (A) 110Hz,220Hz, 330 Hz

Frequencies and overtones:

(I) For an organ pipe open at open both ends the frequency of different modes is given by:

F =  nv/2L

where

F is the frequency

L is the length of the organ pipe

v is the speed of the wave

and, n is the mode of frequency

the fundamental frequency corresponds to n = 1, given by:

F₀ = v/2L

F₀ = 330 / 2(1.5)

F₀ = 330 / 3

F₀ = 110 Hz

The first overtone corresponds to n = 2, the second overtone corresponds to n = 3, and so on...

F₁ =2v/2L

F₁ = 330 / 1.5

F₁ = 220 Hz

Thus, F₁ = 2F₀

The difference between successive overtones is F₀

(II) For an organ pipe open at one end the frequency of different modes is given by:

F =  nv/4L

where

F is the frequency

L is the length of the organ pipe

v is the speed of the wave

and, n is the mode of frequency

the fundamental frequency corresponds to n = 1, given by:

F₀ = V/4L

F₀ = 330 / (4 x 0.75)

F₀ = 110 Hz

For an organ pipe open at one end, only those overtones are present which correspond to odd n, that is n = 3,5,...so:

F₁ = 3F₀

F₁ = 3 x 110

F₁ = 330 Hz

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How wide is the central diffraction peak on a screen 2.20 mm behind a 0.0328-mmmm-wide slit illuminated by 588-nmnm light?

Answers

Answer:

[tex]y = 0.0394 \ m[/tex]

Explanation:

From the question we are told that

        The  distance of the screen is  [tex]D = 2.20 \ m[/tex]

       The distance of separation of the slit is  [tex]d = 0.0328 \ mm = 0.0328*10^{-3} \ m[/tex]

        The  wavelength of light is  [tex]\lambda = 588 \ nm = 588 *10^{-9} \ m[/tex]

Generally the condition for constructive interference is

            [tex]dsin\theta = n * \lambda[/tex]

=>        [tex]\theta = sin^{-1} [ \frac{ n * \lambda }{d } ][/tex]

here n = 1 because we are considering the central diffraction peak

=>        [tex]\theta = sin^{-1} [ \frac{ 1 * 588*10^{-9} }{0.0328*10^{-3} } ][/tex]

=>       [tex]\theta = 1.0274 ^o[/tex]

Generally the width of central diffraction peak on a screen is mathematically evaluated as

           [tex]y = D tan (\theta )[/tex]

substituting values

        [tex]y = 2.20 * tan (1.0274)[/tex]

        [tex]y = 0.0394 \ m[/tex]

Find the total electric potential due to these charges at the point P, whose coordinates are (4.00, 0) m. SOLUTION

Answers

Answer:

Some parts of your question is missing attached below is the missing parts and the answer provided is pertaining to your question alone

answer : -6661.59 volts

Explanation:

The total electric potential can be calculated using this relation

V = k [tex](\frac{q1}{r1} + \frac{q2}{r2})[/tex]

q 1 = 1.62 uc

r1 = 4.00 m

q2 = -5.73 uc

r2 = 5.00 m  

k = 8.99 * 10^9 N.m^2/c^2

insert the given values into the above equation

V = ( 8.99 * 10^9 ) * [tex](\frac{1.62*10^{-6} }{4} + \frac{-5.73*10^{-6} }{5})[/tex]  =  -6661.59 volts

A solenoid inductor has an emf of 0.80 V when the current through it changes at the rate 10.0 A/s. A steady current of 0.20 A produces a flux of 8.0 μWb per turn.

Required:
How many turns does the inductor have?

Answers

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

Learn more about magnetic flux here:

brainly.com/question/24615998?referrer=searchResults

The attached picture shows the magnetic flux.

If the
refractive index of benzere is 2.419,
what is the speed of light in benzene?

Answers

Answer:

[tex]v=1.24\times 10^8\ m/s[/tex]

Explanation:

Given that,

The refractive index of benzene is 2.419

We need to find the speed of light in benzene. The ratio of speed of light in vacuum to the speed of light in the medium equals the refractive index. So,

[tex]n=\dfrac{c}{v}\\\\v=\dfrac{c}{n}\\\\v=\dfrac{3\times 10^8}{2.419}\\\\v=1.24\times 10^8\ m/s[/tex]

So, the speed of light in bezene is [tex]1.24\times 10^8\ m/s[/tex].

If a negatively charged rod is held near a neutral metal ball, the ball is attracted to the rod. This happens:_______

a. because of magnetic effects
b. because the ball tries to pull the rod's electrons over to it
c. because the rod polarizes the metal
d. because the rod and the ball have opposite charges

Answers

Answer:

c. because the rod polarizes the metal.

Explanation:

Bringing the negatively charged rod close to the neutral metal ball causes the neutral metal ball to be polarized with induced positive charge on it. The polarizing of the formally neutral metal ball is due to the negative charge on the metal rod (bodies induce a charge opposite of their own charge on a nearby neutral body). The ball and rod then attract themselves because bodies with opposite charges attract each other, unlike bodies with same charges that repel each other.

Which statement belongs to Dalton’s atomic theory? Atoms have a massive, positively charged center. Atoms cannot be created or destroyed. Atoms can be broken down into smaller pieces. Electrons are located in energy levels outside of the nucleus.

Answers

Answer:

the correct statement is

* atoms cannot be created or destroyed

Explanation:

The Datlon atomic model was proposed in 1808 and represents atoms as the smallest indivisible particle of matter, they were the building blocks of matter and are represented by solid spheres.

Based on the previous descriptive, the correct statement is

* atoms cannot be created or destroyed

Answer:

the Answer is b hope it help

Explanation:

The ceiling of your lecture hall is probably covered with acoustic tile, which has small holes separated by about 6.1 mm. Using light with a wavelength of 578 nm, how far could you be from this tile and still resolve these holes

Answers

Answer:

8.65x10^3m

Explanation:

See attached file

A hydraulic system is being used to lift a 1500-kg car. If the large piston under the car has a diameter of 50 cm, the small piston has a diameter of 4.0 cm, and the car is lifted a distance of 1.3 m, how much work is done on the car

Answers

Answer:

W = 122.3 J

Explanation:

First, we need to find out the force applied to the smaller piston. We know that the pressure applied to smaller piston must be equally transmitted to the larger piston. Therefore,

P₁ = P₂

F₁/A₁ = F₂/A₂

F₂ = F₁(A₂/A₁)

where,

F₁ = Force of Larger Piston = Weight of car = mg = (1500 kg)(9.8 m/s²)

F₁ = 14700 N

F₂ = Force applied to smaller piston = ?

A₁ = Area of larger piston = πd₁²/4

A₂ = Area of smaller piston = πd₂²/4

Therefore,

F₂ = (14700 N)[(πd₂²/4)/(πd₁²/4)]

F₂ = (14700 N)(d₂²/d₁²)

where,

d₁ = diameter of large piston = 50 cm

d₂ = diameter of small piston = 4 cm

Therefore,

F₂ = (14700 N)[(4 cm)²/(50 cm)²]

F₂ = 94.08 N

Now, for the work done on the car:

Work Done = W = F₂ d

where,

d = displacement of car = 1.3 m

Therefore,

W = (94.08 N)(1.3 m)

W = 122.3 J

Water flows at speed v in a pipe of radius R. At what speed does the water flow through a constriction in which the radius of the pipe is R/3

Answers

Answer:

   v₂ = 9 v

Explanation:

For this exercise in fluid mechanics, let's use the continuity equation

           v₁ A₁ = v₂ A₂

where v is the velocity of the fluid, A the area of ​​the pipe and the subscripts correspond to two places of interest.

The area of ​​a circle is

           A = π R²

let's use the subscript 1 for the starting point and the subscript 2 for the part with the constraint

     

In this case v₁ = v and the area is

            A₁ = π R²

in the second point

           A₂= π (R / 3)²

we substitute in the continuity equation

           v π R² = v₂ π R² / 9

            v = v₂ / 9

           

            v₂ = 9 v

A donkey is attached by a rope to a wooden cart at an angle of 23° to the horizontal. the tension in the rope is 210 n. if the cart is dragged horizontally along the floor with a constant speed of 7 km/h, calculate how much work the donkey does in 35 minutes.

Answers

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

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