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2 years ago

What vibrates in an EM wave

Physics

2 answers:

AlladinOne [14]2 years ago

7 0

Answer:magnetic and electric fields

Explanation:

Tema [17]2 years ago

3 0

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Electromagnetic waves are waves that consist of vibrating electric and magnetic fields. ... The fields can exert force over objects at a distance. An electromagnetic wave begins when an electrically charged particle vibrates. This causes a vibrating electric field, which in turn creates a vibrating magnetic field.

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PLEASE HELP ASAP

aleksandrvk [35]

Answer:

changing from a solid to a gas without changing into a liquid. :)

6 0

2 years ago

A mass free to vibrate on a level, frictionless surface at the end of a horizontal spring is pulled 35 cm from its equilibrium p

saul85 [17]

Answer:

0.67 s

Explanation:

This is a simple harmonic motion (SHM).

The displacement, $x$ , of an SHM is given by

$x = A\cos(\omega t)$

A is the amplitude and $\omega$ is the angular frequency.

We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or $\frac{\pi}{4}$ radian.

From trigonometry, $\sin A =\cos B$ if A and B are complementary.

At $t = 0$ , $x = 3.5$

$3.5 = A\cos(\omega \times0)$

$A =3.5$

$x = 3.5\cos(\omega t)$

At $t = 0.12$ , $x = 1.5$

$1.5 = 3.5\cos(0.12\omega)$

$\cos(0.12\omega)=\dfrac{1.5}{3.5}=0.4286$

$0.12\omega =\cos^{-1}0.4286$

$0.12\omega = 1.13$

$\omega = 9.4$

The period, $T$ , is related to $\omega$ by

$T = \dfrac{2\pi}{\omega} = \dfrac{2\times3.14}{9.4}=0.67$

5 0

2 years ago

Light with a wavelength of 400 nm strikes the surface of cesium in a photocell, and the maximum kinetic energy of the electrons

Firdavs [7]

Answer:

The longest wavelength of light that is capable of ejecting electrons from that metal is 1292 nm.

Explanation:

Given that,

Wavelength = 400 nm

Energy $E=1.54\times10^{-19}\ J$

We need to calculate the longest wavelength of light that is capable of ejecting electrons from that metal

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$\lambda=\dfrac{hc}{E}$

Put the value into the formula

$\lambda=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{1.54\times10^{-19}}$

$\lambda=1292\times10^{-9}\ m$

$\lambda=1292\ nm$

Hence, The longest wavelength of light that is capable of ejecting electrons from that metal is 1292 nm.

8 0

3 years ago

A standing wave is created on a string of length 1.2 m. If the speed of the wave on the string is 60.0 m/s, what is the fundamen

coldgirl [10]

Answer:

Fundamental frequency in the string will be 25 Hz

Explanation:

We have given length of the string L = 1.2 m

Speed of the wave on the string v = 60 m/sec

We have to find the fundamental frequency

Fundamental frequency in the string is equal to $f=\frac{v}{2L}$ , here v is velocity on the string and L is the length of the string

So frequency will be equal to $f=\frac{v}{2L}=\frac{60}{2\times 1.2}=25Hz$

So fundamental frequency will be 25 Hz

6 0

3 years ago

Read 2 more answers

A parallel-plate capacitor with plates of area 600 cm^2 is charged to a potential difference V and is then disconnected from the

Julli [10]

Answer:

Explanation:

a) The charge Q on the positive charge capacitor can be gotten using the formula Q = CV

C = capacitance of the capacitor (in Farads )

V = voltage (in volts) = 100V

C = ∈A/d

∈ = permittivity of free space = 8.85 × 10^-12 F/m

A = cross sectional area = 600 cm²

d= distance between the plates = 0.7cm

C = 8.85 × 10^-12 * 600/0.7

C = 7.59*10^-9Farads

Q = 7.59*10^-9 * 100

Q = 7.59*10^-7Coulombs

Q = 0.759*10^-6C

Q = 0.759µC

b) Energy stored in a capacitor is expressed as E = 1/2CV²

E = 1/2 * 7.59*10^-9 * 100²

E = 0.0000395Joules

E = 39.5*10^-6Joules

E = 39.5µJ

7 0

2 years ago