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3 years ago

An electric generator uses __________ to convert mechanical energy into electrical energy.

Physics

1 answer:

Dafna11 [192]3 years ago

3 0

Answer:

Devices that convert mechanical energy to electric energy. ... An electric generator is a device that changes kinetic energy to electrical energy through electromagnetic induction. ... [Figure1]. Q: What might happen to the current produced by an electric generator if the poles of the magnet kept reversing?

Explanation:

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Place the items in order from the largest wavelength to the shortest wavelength.

Butoxors [25]

From largest to shortest wavelength:

Radio waves, Microwaves, Infrared radiation, Red visible, Orange visible, Yellow visible, Green visible, Blue visible, Violet visible, Ultraviolet, X-rays, Gamma rays

Explanation:

Electromagnetic waves are oscillations of the electric and the magnetic field in a plane perpendicular to the direction of motion the wave.

Electromagnetic waves are the only type of waves able to travel in a vacuum, and in a vacuum they always at the same speed, the speed of light, equal to:

$c=3.0\cdot 10^8 m/s$

Electromagnetic waves are classified into 7 different types, according to their wavelength/frequency. From slongest to shortest wavelength, they are ranked as follows:

Radio waves

Microwaves

Infrared radiation

Visible light

Ultraviolet

X-rays

Gamma rays

Visible light is the only part of the spectrum that the human eye is able to see. Depending on the wavelength of the visible light, we perceive the radiation as a different color. In order from longest to shortest wavelength, colors are:

Red

Orange

Yellow

Green

Blue

Indigo

Violet

Therefore, the correct order from largest to shortest wavelength in the given list is:

Radio waves

Microwaves

Infrared radiation

Red visible

Orange visible

Yellow visible

Green visible

Blue visible

Violet visible

Ultraviolet

X-rays

Gamma rays

Learn more about electromagnetic waves:

brainly.com/question/9184100

brainly.com/question/12450147

#LearnwithBrainly

4 0

3 years ago

1. boiling point of water

melomori [17]

Answer:

what is the question. . .

5 0

3 years ago

To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(

DerKrebs [107]

Answer:

0.0549 m

Explanation:

Given that

equation y(x,t)=Acos(kx−ωt)

speed v = 8.5 m/s

amplitude A = 5.5*10^−2 m

wavelength λ = 0.5 m

transverse displacement = ?

v = angular frequency / wave number

and

wave number = 2π/ λ

wave number = 2 * 3.142 / 0.5

wave number = 12.568

angular frequency = v k

angular frequency = 8.5 * 12.568

angular frequency = 106.828 rad/sec ~= 107 rad/sec

equation y(x,t)=Acos(kx−ωt)

y(x,t)= 5.5*10^−2 cos(12.568 x−107t)

when x =0 and and t = 0

maximum y(x,t)= 5.5*10^−2 cos(12.568 (0) − 107 (0))

maximum y(x,t)= 5.5*10^−2 m

and when x = x = 1.52 m and t = 0.150 s

y(x,t)= 5.5*10^−2 cos(12.568 (1.52) −107(0.150) )

y(x,t)= 5.5*10^−2 × (0.9986)

y(x,t) = 0.0549 m

so the transverse displacement is 0.0549 m

5 0

4 years ago

A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium wit

Masja [62]

Answer:

Wavelength of the incident wave in air = 1 m

Wavelength of the incident wave in medium 2 = 0.33 m

Intrinsic impedance of media 1 = 377 ohms

Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

a) Wavelength of the incident wave in air

The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz

Speed of light in air, c = 3 * 10⁸ Hz

Wavelength of the incident wave in air:

$\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m$

Wavelength of the incident wave in medium 2

The refractive index of air in the lossless dielectric medium:

$n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3$

$\lambda_{2} = \frac{c}{nf}\\\lambda_{2} = \frac{3 * 10^{6} }{3 * 3 * 10^{6}}\\\lambda_{2} = 1/3\\\lambda_{2} = 0.33 m$

b) Intrinsic impedances of media 1 and media 2

The intrinsic impedance of media 1 is given as:

$n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

$n_1 = \sqrt{\frac{4\pi * 10^{-7} }{8.84 * 10^{-12} } }$

$n_1 = 377 \Omega$

The intrinsic impedance of media 2 is given as:

$n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

ϵr = 9

$n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }$

$n_2 = 125.68 \Omega$

c) The reflection coefficient,r and the transmission coefficient,t at the boundary.

Reflection coefficient, $r = \frac{n - n_{0} }{n + n_{0} }$

You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

$r = \frac{3 - n_{0} }{3 + n_{0} }$

Transmission coefficient at the boundary, t = r -1

d) The amplitude of the incident electric field is $E_{0} = 10 V/m$

Maximum amplitudes in the total field is given by:

$E = tE_{0}$ and $E = r E_{0}$

E = 10r, E = 10t

3 0

3 years ago

10 turns of wire are closely wound around a pencil as shown in the figure. when measured using a scale as shown, the length of t

Mila [183]

Answer:

a. The thickness of the wire is 2.5 mm.

b. The wire is 0.25 cm thick.

Explanation:

Number of turns of the wire = 10

The length of total turns = 25 mm

a. The thickness of the wire can be determined by;

thickness of the wire = $\frac{length of total turns}{number of turns}$

= $\frac{25}{10}$

= 2.5 mm

Therefore, the wire is 2.5 mm thick.

b. To determine the thickness of the wire in centimetre;

10 mm = 1 cm

So that,

2.5 mm = x

x = $\frac{2.5}{10}$

= 0.25 cm

The wire is 0.25 cm thick.

8 0

3 years ago