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3 years ago

The energy of an electromagnetic wave is related to

Physics

1 answer:

Sedbober [7]3 years ago

6 0

Answer:

The energy carried by an electromagnetic wave is proportional to the frequency of the wave. The wavelength and frequency of the wave are connected via the speed of light: Electromagnetic waves are split into different categories based on their frequency (or, equivalently, on their wavelength).

Explanation:

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A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.

iris [78.8K]

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

$F = \frac{kq_1q_2}{d^2}$

Here

k = Coulomb's Constant

$q_{1,2} =$ Charge of each object

d = Distance

Our values are given as,

$q_1 = 1 \mu C$

$q_2 = 6 \mu C$

d = 1 m

$k = 9*10^9 Nm^2/C^2$

a) The electric force on charge $q_2$ is

$F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}$

$F_{12} = 54 mN$

Force is positive i.e. repulsive

b) As the force exerted on $q_2$ will be equal to that act on $q_1$ ,

$F_{21} = F_{12}$

$F_{21} = 54 mN$

Force is positive i.e. repulsive

c) If $q_2 = -6 \mu C$ , a negative sign will be introduced into the expression above i.e.

$F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}$

$F_{12} = F_{21} = -54 mN$

Force is negative i.e. attractive

6 0

2 years ago

What is the intensity of 60dB sound?

polet [3.4K]

Answer:

The intensity of the sound in W/m² is 1 x 10⁻⁶ W/m².

Explanation:

Given;

intensity of the sound level, dB = 60 dB

The intensity of the sound in W/m² is calculated as;

$dB = 10 Log[\frac{I}{I_o} ]\\\\$

where;

I₀ is threshold of hearing = 1 x 10⁻¹² W/m²

I is intensity of the sound in W/m²

Substitute the given values and for I;

$dB = 10 Log[\frac{I}{I_o} ]\\\\60 = 10 Log[\frac{I}{I_o} ]\\\\6 = Log[\frac{I}{I_o} ]\\\\10^6 = \frac{I}{I_o} \\\\I = 10^6 \ \times \ I_o\\\\I = 10^6 \ \times \ 1^{-12} \ W/m^2 \\\\I = 1\ \times \ 10^{-6} \ W/m^2$

Therefore, the intensity of the sound in W/m² is 1 x 10⁻⁶ W/m².

5 0

2 years ago

What is the frequency of orange light is it has a wavelength of 6.2 x 10^-6<br> meters?

Tanya [424]

λ=v/f

λ-wavelength

v-speed

f-frequency

we have the wavelength(6.2 x 10^-6meters) and we use the speed of light which is equal to 3*10^8m/s

6.2*10^-6m=3*10^8m/s/f

f=(3*10^8m/s)/(6.2*10^-6)≈0.48*10^14Hz

6 0

2 years ago

Read 2 more answers

A 460 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exert

ollegr [7]

Answer:

The rocket should be launched at a horizontal distance of <u>6.45 m</u> left of the loop.

Explanation:

Given:

Mass of the rocket model (m) = 460 g = 0.460 kg [1 g = 0.001 kg]

Speed of the cart (v) = 3.0 m/s

Thrust force by the rocket engine (F) = 8.5 N

Vertical height of the loop (y) = 20 m

Let the horizontal distance left to the loop for launch be 'x'. Also, let 't' be the time taken by the rocket to reach the loop.

Now, there are two types of motion associated with the rocket- one is horizontal and the other vertical.

So, we will apply kinematics of motion in the two directions separately.

Vertical motion:

Given:

Force acting in the vertical direction is given as:

$F_y=F-mg=8.5-0.46\times 9.8=3.992\ N$

So, acceleration in the vertical direction is given as:

Acceleration = Force ÷ mass

$a_y=\frac{F_y}{m}=\frac{3.992\ N}{0.46\ kg}=8.678\ m/s^2$

Vertical displacement of rocket is same as the height of loop. So, $y=20\ m$

There is no initial velocity in the vertical direction. So, $u_y=0\ m/s$

Now, applying equation of motion in vertical direction. we have:

$y=u_yt+\frac{1}{2}a_yt^2\\\\20=0+\frac{1}{2}\times 8.678t^2\\\\20=4.339t^2\\\\t^2=\frac{20}{4.339}\\\\t=\sqrt{\frac{20}{4.339}}=2.15\ s$

Now, time taken to reach the loop is 2.15 s.

Horizontal motion:

There is no acceleration in the horizontal motion. So, displacement in the horizontal direction is equal to the product of horizontal speed and time.

Also, displacement of the rocket in the horizontal direction is nothing but the horizontal distance of its launch left of the loop. So,

$x=vt\\\\x=3.0\ m/s\times 2.15\ s\\\\x=6.45\ m$

Therefore, the rocket should be launched at a horizontal distance of 6.45 m left of the loop.

5 0

3 years ago

07. How do scientists use spectroscopy (the study of light frequencies released

Vera_Pavlovna [14]

Every element is able to be recognized individually in many different ways. A very easy and common way is using light absorption also known as spectroscopy. Every atom has electrons, and these electrons like to stay in their lowest-energy configuration. However, when photons collide with an electron it can increase it to a higher energy level.. This is absorption, and each element’s electrons absorb light at specific wavelengths related to the difference between energy levels in that atom. But the electrons want to return to their original levels, so they don’t hold onto the energy for long. When they emit the energy, they release photons with exactly the same wavelengths of light that were absorbed in the first place. An electron can release this light in any direction, so most of the light is emitted in directions away from our line of sight. Therefore, a dark line appears in the spectrum at that particular wavelength.

Because the wavelengths at which absorption lines occur are unique for each element, astronomers can measure the position of the lines to determine which elements are present in a target. The amount of light that is absorbed can also provide information about how much of each element is present.

8 0

3 years ago