The intensity of the electromagnetic wave which travels in an insulating magnetic material in a medium is 5.766×10⁻⁸ W/m².
The intensity of a wave is the total power delivered per unit area. It can be given as,
It can also be given as,
Here, () is relative permeability, () is physical constant, (k) is dielectric constant, (E) is the amplitude of electric field, and is the permittivity of free space.
Here, the electromagnetic wave with frequency 65.0hz travels in an insulating magnetic material that has dielectric constant 3.64 and relative permeability 5.18 at this frequency.
As the electric field has amplitude 7.20×10−3v/m. Thus, put the values in the above formula to find the intensity as,
Hence, the intensity of the electromagnetic wave which travels in an insulating magnetic material in a medium is 5.766×10⁻⁸ W/m².
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Answer:
(a) 1.87×10⁶ V/m
(b) 1.12 mm closer
Explanation:
(a)
Electric Field = Electric potential/distance.
E = V/d ................... Equation 1
Where E = Electric Field, V = Electric potential, d = distance.
Given: V = 5575 V, d = 2.98 mm = 0.00298 m.
Substitute into equation 1
E = 5575/0.00298
E = 1.87×10⁶ V/m
(b)
without exceeding the breakdown strength,
make d the subject of equation 1
d = V/E.............. Equation 2
Given: E = 3×10⁶ V/m, V = 5575 V
Substitute into equation 2
d = 5575/3000000
d = 1.86 mm.
the plate will be = 2.98-1.86 = 1.12 mm closer
Light has wavelength 600 nm in a vacuum ,the frequency of the light is 2 × Hz.
The separation between such a wave motion's crests and troughs would be known as the wavelength of photons.
The total number of waves that pass a specific location in a predetermined amount of time is known as frequency.
Calculation of frequency
Given data:
wavelength = 600 nm = 600 × m
index of refraction = 1.5.
Frequency can be calculated by using the formula:
v = f × wavelength
f = wavelength / v
Where, f = Frequency , v is velocity.
put the given data in above equation.
f = wavelength / v
f = 600 × m / 3 ×
f = 200 × .
f = 2 ×
Therefore, the frequency of the light is 2 × Hz.
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v = f lambda
in vac ... 3X10^8 = 600x10^-9xf
in glass speed slower, poss 2/3 that of vacuum
Answer:
The flux through the surface of the cube is
Solution:
As per the question:
Edge of the cube, a = 8.0 cm =
Volume Charge density,
Now,
To calculate the electric flux:
(1)
where
= electric flux
= permittivity of free space
Volume Charge density for the given case is given by the formula:
(2)
Volume of cube,
Thus
Thus from eqn (2), the total charge is given by:
Now, substitute the value of 'q' in eqn (1):
Taking into account the definition of wavelength, frecuency and propagation speed, the frequency of light waves with wavelength of 5×10⁻⁷ m is 6×10¹⁴ Hz.
First of all, wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. It is expressed in units of length (m).
On the other side, frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).
Finally, the propagation speed is the speed with which the wave propagates in the medium, that is, it is the magnitude that measures the speed at which the wave disturbance propagates along its displacement.
The propagation speed relate the wavelength (λ) and the frequency (f) inversely proportional using the following equation:
v = f× λ
All electromagnetic waves propagate in a vacuum at a constant speed of 3×10⁸ m/s, the speed of light.
In this case, you know:
Replacing in the definition of propagation speed:
3×10⁸ m/s = f× 5×10⁻⁷ m
Solving:
3×10⁸ m/s ÷ 5×10⁻⁷ m= f
f= 6×10¹⁴ Hz
In summary, the frequency of light waves with wavelength of 5×10⁻⁷ m is 6×10¹⁴ Hz.
Learn more about wavelength, frecuency and propagation speed:
brainly.com/question/2232652?referrer=searchResults
Answer:
Speed of light =m/s
wavelength = m
frequency = ?
we have
Speed = frequency × wavelength
= frequency ×
Frequency = hz
What is the kinetic energy of an electron in a circular orbit around the gold nucleus?
Answer:
a) F = -1.82 10⁻¹⁵ N, b) K = 9.1 10⁻¹⁶ J
Explanation:
a) To calculate the force between the nucleus and the electrons, let's use the Coulomb equation
F = k q Q / r²
as the nucleus occupies a very small volume compared to electrons, we can suppose it as punctual
let's calculate
F = 9 10⁹ (-1.6 10⁻¹⁹) (79 1.6 10⁻¹⁹) / (10⁻¹⁰)²
F = -1.82 10⁻¹⁵ N
b) they ask us for kinetic energy
let's use Newton's second law
F = m a
acceleration is centripetal
a = v² / r
we substitute
F = m v² / r
v = √ (F r / m)
v = √ (1.82 10⁻¹⁵ 10⁻¹⁰ / 9.1 10⁻³¹)
v = √ (0.2 10⁻¹⁶)
v = 0.447 10⁸ m / s
kinetic energy is
K = ½ m v²
K = ½ 9.1 10⁻³¹ (0.447 10⁸)²
K = 0.91 10⁻¹⁵ J
K = 9.1 10⁻¹⁶ J