17)

The y component of the electric field of an electromagnetic wave traveling in the +x direction through vacuum obeys the equation

Alexxx [7]

Answer: 8.6 µm

Explanation:

At a long distance from the source, the components (the electric and magnetic fields) of the electromagnetic waves, behave like plane waves, so the equation for the y component of the electric field obeys an equation like this one:

Ey =Emax cos (kx-ωt)

So, we can write the following equality:

ω= 2.2 1014 rad/sec

The angular frequency and the linear frequency are related as follows:

f = ω/ 2π= 2.2 1014 / 2π (rad/sec) / rad = 0.35 1014 1/sec

In an electromagnetic wave propagating through vacuum, the speed of the wave is just the speed of light, c.

The wavelength, speed and frequency, are related by this equation:

λ = c/f

λ = 3.108 m/s / 0.35. 1014 1/s = 8.6 µm.

7 0

3 years ago

A small sphere with mass mcarries a positive chargeqand is attached to one end of a silk fiber of lengthL. The other end of the

Aleksandr-060686 [28]

Answer:

(a): The magnitude of the electric force on the small sphere = $\dfrac{q\sigma}{2\epsilon_o}.$

(b): Shown below.

Explanation:

<u>Given:</u>

m = mass of the small sphere.
q = charge on the small sphere.
L = length of the silk fiber.
$\sigma$ = surface charge density of the large vertical insulating sheet.

When the dimensions of the sheet is much larger than the distance between the charge and the sheet, then, according to Gauss' law of electrostatics, the electric field experienced by the particle due to the sheet is given as:

$\rm E = \dfrac{\sigma}{2\epsilon_o}.$

<em>where,</em>

$\epsilon_o$ is the electrical permittivity of the free space.

The electric field at a point is defined as the amount of electric force experienced by a unit positive test charge, placed at that point. The magnitude electric field at a point and the magnitude of the electric force on a charge q placed at that point are related as:

$\rm F_e=qE.$

Thus, the magnitude of the electric force on the small sphere is given by

$\rm F_e = q\times \dfrac{\sigma }{2\epsilon_o}=\dfrac{q\sigma}{2\epsilon_o}.$

The sheet and the small sphere both are positively charged, therefore, the electric force between these two is repulsive, which means, the direction of the electric force on the sphere is away from the sheet along the line which is perepndicular to the sheet and joining the sphere.

When the sphere is in equilibrium, the tension in the fiber is given by the resultant of the weight of the sphere and the electric force experienced by it as shown in the figure attached below.

According to the fig.,

$\rm \tan \theta = \dfrac{F_e}{W}.$

<em>where,</em>

$\rm F_e$ = electric force on the sphere, acting along left.
$\rm W$ = weight of the sphere, acting vertically downwards.

<em />

$\rm F_e = \dfrac{q\sigma}{2\epsilon_o}\\\\W=mg\\\\Therefore,\\\\\tan\theta = \dfrac{\dfrac{q\sigma}{2\epsilon_o}}{mg}=\dfrac{q\sigma}{2mg\epsilon_o}.\\\Rightarrow \theta=\tan^{-1}\left ( \dfrac{q\sigma}{2mg\epsilon_o}\right ) .$

g is the acceleration due to gravity.

6 0

3 years ago

A 200 kg object is initially at rest on a horizontal frictionless surface. At time t = 0 , a horizontal force of 100 N applied t

s344n2d4d5 [400]

Answer:

b

Explanation:

5 0

3 years ago

In a shuffleboard game, the puck slides a total of 12 m before coming to rest. If the coefficient ofkinetic friction between the

Artist 52 [7]

Answer:

puck decelerates due to the kinetic frictional force μk mg

Explanation:

given data

total distance = 12 m

coefficient of kinetic friction = 0.28

solution

we will apply equation of motion that is

v² - u² = 2 × a × s ................1

we know acceleration will be

a = $\frac{-u^2}{2\times S}$

Then we have

Force = mass × acceleration .................2

m × $\frac{-u^2}{2\times S}$ = -μk mg

The puck decelerates due to the kinetic frictional force μk mg

and frictional force is negative as it opposes the motion.

so we get initial velocity of the puck which is strike.

4 0

3 years ago

Mechanical locks can accept a variety of inputs as keys, including magnetic strips on ID cards, radio signals from name badges,

lidiya [134]

Answer:

The answer is B) False

3 0

3 years ago