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An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is = 2.93 × 109 W/m2. What is the rms value of (a) the electric field and (b) the magnetic field in the electromagnetic wave emitted by the laser?

Answers

Answer 1

Answer:

A. The rms value of electric field be "1.05 × 10⁶ N/C".

B. The rms value of magnetic field will be "3.5 × 10⁻³ T".

Magnetic and Electric field

According to the question,

Intensity of the wave, S = 2.93 × 10⁹ W/m²

Free space permittivity, $\epsilon_0$ = 8.86 × 10⁻¹²

Speed of light, c = 3 × 10⁸

A. We know that,

The rms value of electric field,

→ $E_(rms)$ = $\sqrt{(S)/(\epsilon_0 c) }$

By substituting the values,

= $\sqrt{(2.93* 10^9)/((8.85* 10^(-12))(3* 10^8)) }$

= 1.05 × 10⁶ N/C

and,

B. We know that,

The rms value of magnetic field,

→ $B_(rms)$ = $(E_(rms))/(c)$

By substituting the values,

= $(1.05* 10^6)/(3* 10^8)$

= 3.5 × 10⁻³ T

Thus the above response is appropriate.

Find out more information about magnetic field here:

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Answer 2

Answer:

To solve this problem, it is necessary to apply the concepts related to the electric field according to the intensity of the wave, the permittivity constant in free space and the speed of light.

As well as the expression of the rms of the magnetic field as a function of the electric field and the speed of light.

PART A) The expression for the rms of electric field is

$E_(rms) = \sqrt{(S)/(\epsilon_0 c)}$

Where,

S= Intensity of the wave

$\epsilon_0$ = Permitivitty at free space

c = Light speed

Replacing we have that,

$E_(rms) = \sqrt{((2.93*10^9))/((8.85*10^(-12))(3*10^8))}$

$E_(rms) = 1.05*10^6N/C$

The RMS value of electric field is $1.05*10^6N/C$

PART B) The expression for the RMS of magnetic field is,

$B_(rms) = (E_(rms))/(c)\nB_(rms) = (1.05*10^6)/(3*10^8)\nB_(rms) =3.5*10^(-3)T$

The RMS of the magnetic field is $3.5*10^(-3)T$

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An effect analogous to two-slit interference can occur with sound waves, instead of light. In an open field, two speakers placed 1.30 m apart are powered by a single function generator producing sine waves at 1200-Hz frequency. A student walks along a line 12.5 m away and parallel to the line between the speakers. She hears an alternating pattern of loud and quiet, due to constructive and destructive interference. What is : (a) the wavelength of this sound and (b) the distance between the central maximum and the first maximum (loud) position along this line

Answers

Answer:

2.72 m

Explanation:

wavelength of sound λ = velocity / frequency

= 340 / 1200

= .2833 m

Distance of point of first constructive interference

= λ D / d ( D is distance of the screen and d is distance between source of sound.

Here D = 12.5 m

d = 1.3 m

λ D / d= ( .2833 x 12.5) / 1.3

= 2.72 m

Distance of point of first constructive interference = 2.72 m

Final answer:

The wavelength of the produced sound is approximately 0.29 m. Constructive interference occurs when the path difference between the two waves is a multiple of this wavelength, allowing you to calculate the distance between the central maximum and first maximum loud position.

Explanation:

For part (a) of the question, we need to calculate the wavelength of the sound wave. The wave speed (v) is given by the multiplication of frequency (f) and wavelength (λ). The speed of sound in air is approximately 343 m/s and given that the frequency produced by the function generator is 1200 Hz, the wavelength can be calculated using the formula λ = v / f = 343 / 1200 ≈ 0.29 m.

For part (b) the distance between the central maximum (loud) position and the first maximum along this line requires understanding of sound wave interference and constructive interference. For constructive interference to occur, the path difference between the two waves needs to be a multiple of the wavelength. Thus, in the first constructive interference position (first maximum loud position), the path difference equals one wavelength (0.29m). Since the student is walking 12.5 m away and parallel to the line between the speakers (which is the hypotenuse of a right triangle stakeout, with one side being 0.65m), we can use Pythagorean theorem to find out the distance.

Learn more about Sound Waves Interference here:

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A 39 kg block of ice slides down a frictionless incline 2.8 m along the diagonal and 0.74 m high. A worker pushes up against the ice, parallel to the incline, so that the block slides down at constant speed. (a) Find the magnitude of the worker's force. How much work is done on the block by (b) the worker's force, (c) the gravitational force on the block, (d) the normal force on the block from the surface of the incline, and (e) the net force on the block?

Answers

Answer:

(a) Fw = 101.01 N

(b) W = 282.82 J

(d) N = 368.61 N

(e) Net force = 0 N

Explanation:

(a) In order to calculate the magnitude of the worker's force, you take into account that if the ice block slides down with a constant speed, the sum of forces, gravitational force and work's force, must be equal to zero, as follow:

$F_g-F_w=0$ (1)

Fg: gravitational force over the object

Fw: worker's force

However, in an incline you have that the gravitational force on the object, due to its weight, is given by:

$F_g=Wsin\theta=Mg sin\theta$ (2)

M: mass of the ice block = 39 kg

g: gravitational constant = 9.8m/s^2

θ: angle of the incline

You calculate the angle by using the information about the distance of the incline and its height, as follow:

$sin\theta=(0.74m)/(2.8m)=0.264\n\n\theta=sin^(-1)(0.264)=15.32\°$

Finally, you solve the equation (1) for Fw and replace the values of all parameters:

$F_w=F_g=Mgsin\theta\n\nF_w=(39kg)(9.8m/s^2)sin(15.32\°)=101.01N$

The worker's force is 101.01N

(b) The work done by the worker is given by:

$W=F_wd=(101.01N)(2.8m)=282.82J$

(c) The gravitational force on the block is, without taking into account the rotated system for the incline, only the weight of the ice block:

$F_g=Mg=(39kg)(9.8m/s^2)=382.2N$

The gravitational force is 382.2N

(d) The normal force is:

$N=Mgcos\theta=(39kg)(9.8m/s^2)cos(15.32\°)=368.61N$

(e) The speed of the block when it slides down the incle is constant, then, by the Newton second law you can conclude that the net force is zero.

A parallel-plate capacitor is charged and then disconnected from the battery. By what factor does the stored energy change when the plate separation is then doubled?

Answers

Answer:

U/U₀ = 2

(factor of 2 i.e U = 2U₀)

Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected

Explanation:

Energy stored in a capacitor can be expressed as;

U = 0.5CV^2 = Q^2/2C

And

C = ε₀ A/d

Where

C = capacitance

V = potential difference

Q = charge

A = Area of plates

d = distance between plates

U = Q^2/2C = dQ^2/2ε₀ A

The initial energy of the capacitor at d = d₀ is

U₀ = Q^2/2C = d₀Q^2/2ε₀ A ....1

When the plate separation is increased after the capacitor has been disconnected, the charge Q of the capacitor remain constant.

The final energy stored in the capacitor at d = 2d₀ is

U = 2d₀Q^2/2ε₀ A ...2

The factor U/U₀ can be derived by substituting equation 1 and 2

U/U₀ = (2d₀Q^2/2ε₀ A)/( d₀Q^2/2ε₀ A )

Simplifying we have;

U/U₀ = 2

U = 2U₀

Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected.

PLEASE HELP!!! Sophia says that if we add two electrons to oxygen in order to fill its valence shell, it’s expected charge would be +2. Is she correct? If not, explain the error in her thinking.

Answers

The error in her thinking is that oxygen has has six electrons and a negative charge is acquired by nitrogen when it gains two electrons.

Oxygen is a member of group 16. The elements in group 16 has six valence electrons. This means that they need an extra two electrons to complete their octet.

If an atom gains two electrons, it will have a charge of -2 and not +2, a positive charge means that the atom lost electrons. Nonmetals like oxygen do not loose electrons rather they gain electrons.

Learn more: brainly.com/question/14156701

A 3.1 kg ball is thrown straight upward with a speed of 18.2 m/s. Use conservation of energy to calculate the maximum height the ball can reach.

Answers

Answer:

h = 16.9 m

Explanation:

When a ball is thrown upward, its velocity gradually decreases, until it stops for a moment, when it reaches the maximum height, while its height increases. Thus, the law conservation of energy states in this case, that:

Kinetic Energy Lost by Ball = Potential Energy Gained by Ball

(0.5)m(Vf² - Vi²) = mgh

h = (0.5)(Vf² - Vi²)/g

where,

Vf = Final Speed of Ball = 0 m/s (Since, ball stops for a moment at highest point)

Vi = Initial Speed of Ball = 18.2 m/s

g = acceleration due to gravity = - 9.8 m/s² ( negative for upward motion)

h = maximum height the ball can reach = ?

Therefore, using values in the equation, we get:

h = (0.5)[(0 m/s)² - (18.2 m/s)²]/(-9.8 m/s²)

h = 16.9 m

Lightning can sometimes occur on hot and humid summer evenings when there are no thunderstorms.(A) True
(B) False