True or false, As the frequency of an energy wave increases , so does its wavelength

Has anyone done the science project: modeling waves/properties of light?

igor_vitrenko [27]

<h2>MARK BRAINLIEST</h2>

For this assignment, you will develop several models that show how light waves and mechanical waves are reflected, absorbed, or transmitted through various materials. For each model, you will write a brief description of the interaction between the wave and the material. You will also compose two <u><em>typewritten</em></u> paragraphs. The first will compare and contrast light waves interacting with different materials. The second will explain why materials with certain properties are well suited for particular functions.

<h2><u>Background Information</u></h2>

A wave is any disturbance that carries energy from one place to another. There are two different types of waves: mechanical and electromagnetic. A mechanical wave carries energy through matter. Energy is transferred through vibrating particles of matter. Examples of mechanical waves include ocean waves, sound waves, and seismic waves. Like a mechanical wave, an electromagnetic wave can also carry energy through matter. However, unlike a mechanical wave, an electromagnetic wave does not need particles of matter to carry energy. Examples of electromagnetic waves include microwaves, visible light, X-rays, and radiation from the Sun.

7 0

3 years ago

Read 2 more answers

A charge of 4.5 × 10-5 C is placed in an electric field with a strength of 2.0 × 104 . If the charge is 0.030 m from the source

snow_tiger [21]

Answer:

The electrical potential energy is 0.027 Joules.

Explanation:

The values from the question are

charge (q) = $4.5 \times 10^{-5} C$

Electric Field strength (E) = $2.0 \times 10^{4} N/C$

Distance from source (d) = 0.030 m

Now the formula for the electrical potential energy (U) is given by

$U = q \times E \times d$

So now insert the values to find the answer

$U = 4.5 \times 10^{-5} C \times 2.0 \times 10^{4} N/C \times 0.030 m$

On further solving

$U = 0.027 J$

8 0

3 years ago

The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, th

Inga [223]

According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.

To solve this problem we must apply the concept related to the conservation of energy theorem.

PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so

$E_i = E_f$

$0 = \frac{1}{2} (m_1+m_2)v_f^2-m_2gh+m_1gh$

$v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}$

PART B) Replacing the values given as,

$h= 1.7m\\m_1 = 3.5kg\\m_2 = 4.3kg \\g = 9.8m/s^2 \\$

$v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}$

$v_f = \sqrt{2(9.8)(1.7)(\frac{4.3-3.5}{3.5+4.3})}$

$v_f = 1.8486m/s$

Therefore the speed of the masses would be 1.8486m/s

6 0

3 years ago

PLZZ HELP WILL MARK BRAINLY The function of the respiratory system is to ___________.

pickupchik [31]

Breathe and now I’m just filling in more letters so it’ll go thru

3 0

3 years ago

Two identical cars, one on the moon and one on the earth, have the same speed and are rounding banked turns that have the same r

Naily [24]

Answer:

The value of the centripetal forces are same.

Explanation:

Given:

The masses of the cars are same. The radii of the banked paths are same. The weight of an object on the moon is about one sixth of its weight on earth.

The expression for centripetal force is given by,

$F_{c} = \dfrac{mv^{2}}{r}$

where, $m$ is the mass of the object, $v$ is the velocity of the object and $r$ is the radius of the path.

The value of the centripetal force depends on the mass of the object, not on its weight.

As both on moon and earth the velocity of the cars and the radii of the paths are same, so the centripetal forces are the same.

3 0

3 years ago