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Nady [450]
3 years ago
9

From lowest energy to highest energy, which of the following correctly orders the different categories of electromagnetic radiat

ion?
a) infrared, visible light, ultraviolet, X rays, gamma rays, radio
b) visible light, infrared, X rays, ultraviolet, gamma rays, radio
c) radio, infrared, visible light, ultraviolet, X rays, gamma rays
d) gamma rays, X rays, visible light, ultraviolet, infrared, radio
e) radio, X rays, visible light, ultraviolet, infrared, gamma rays
Physics
2 answers:
Valentin [98]3 years ago
4 0

Answer: The answer is "C"

Explanation:

The electromagnetic waves are arranged in the increasing wavelength and energy by the following order;

Radio waves has the lowest energy and wavelength but of the highest frequency.

The Infra-red rays follows the visible light follows, the ultraviolet ray follows, the X-ray follows and then the Gamma-ray has the highest energy and wavelength but the lowest frequency.

Therefore the answer is C, radio, infra, visible, ultraviolet, x-ray, gamma.

Sliva [168]3 years ago
3 0

Answer:

C

Explanation:

Electromagnetic waves are categorized in terms of their wavelengths and frequency. This categorization is known as the Electromagnetic Spectrum.

When they are arranged in terms of increasing frequency, their wavelengths are decreasing. This is because wavelength and frequency are inversely proportional.

Since energy and frequency are directly proportional, increasing frequency would mean increasing energy.

In terms of increasing energy, the correct order is:

Radio waves

Infrared

Visible light

Ultraviolet

X rays

Gamma rays

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In terms of saturation (unsaturated, saturated, super-saturated). How would you classify the following?
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An electron moves at 0.130 c as shown in the figure (Figure 1). There are points: A, B, C, and D 2.10 μm from the electron.
Olegator [25]

Hi there!

We can use Biot-Savart's Law for a moving particle:
B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }

B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)

q = charge of particle (1.6 × 10⁻¹⁹ C)

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

r = distance from particle (2.10 μm)

There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:

B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }

Where 'θ' is the angle between the velocity and radius vectors.

a)
To find the angle between the velocity and radius vector, we find the complementary angle:

θ = 90° - 60° = 30°

Plugging 'θ' into the equation along with our other values:

B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }

B = \boxed{7.07 *10^{-10} T}

b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.

c)

In this instance, the radius vector and the velocity vector are perpendicular so

'θ' = 90°.

B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}

d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.

Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.

\boxed{B = 0 T}

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