Ask question

Ask question

All categories

borishaifa [10]

3 years ago

10

Electromagnetic waves are sometimes demonstrated by rippling waves in water. However, electromagnetic waves are different from w

ater waves. Which is a
characteristic that makes electromagnetic waves and water waves different?
O A. All water waves have higher frequencies.
OB. All water waves have longer wavelengths.
OC. All electromagnetic waves are transverse waves.
O D. All electromagnetic waves can travel in a vacuum.

2 answers:

uranmaximum [27]3 years ago

8 0

D my teacher had gone over it with us before

algol [13]3 years ago

3 0

The answer is D: all electromagnetic waves can travel in a vacuum

You might be interested in

What is quantum programing

Komok [63]

Answer: Quantum programming is the process of assembling sequences of instructions, called quantum programs, that are capable of running on a quantum computer. Quantum programming languages help express quantum algorithms using high-level constructs.

Explanation:

7 0

4 years ago

Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude?

FromTheMoon [43]

Given: Please see the attached image below.

To be able to subtract vectors, we can either use the parallelogram method or the triangle method. Take note that the only difference is that alternatively adding vectors A and B, we will instead be adding A and – B. When we ponder of vector subtraction, we must anticipate about it in terms of adding a negative vector. A negative vector has the same magnitude as the original vector, however, it has an opposite direction.

So in this problem, the two vectors that will have the largest magnitude are A & F when subtracted (i.e., when one vector is subtracted from the other).

7 0

4 years ago

A branch of mechanics dealing with the mathematical methods of describing motion

alexdok [17]

Kinematics is the branch of Physics that deals with
the mathematical methods of describing motion.

3 0

4 years ago

Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degree

kvv77 [185]

Answer:

1

When second polarizer is removed the intensity after it passes through the stack is

$I_f_3 = 27.57 W/cm^2$

2 When third polarizer is removed the intensity after it passes through the stack is

$I_f_2 = 102.24 W/cm^2$

Explanation:

From the question we are told that

The angle of the second polarizing to the first is $\theta_2 = 21^o$

The angle of the third polarizing to the first is $\theta_3 = 61^o$

The unpolarized light after it pass through the polarizing stack $I_u = 60 W/cm^2$

Let the initial intensity of the beam of light before polarization be $I_p$

Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as

$I_1 = \frac{I_p}{2}$

Now according to Malus’ law the intensity of light that would emerge from the second polarizing filter is mathematically represented as

$I_2 = I_1 cos^2 \theta_1$

$= \frac{I_p}{2} cos ^2 \theta_1$

The intensity of light that will emerge from the third filter is mathematically represented as

$I_3 = I_2 cos^2(\theta_2 - \theta_1 )$

$I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)]$

making $I_p$ the subject of the formula

$I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}$

Note that $I_u = I_3$ as $I_3$ is the last emerging intensity of light after it has pass through the polarizing stack

Substituting values

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}$

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}$

$=234.622W/cm^2$

When the second is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as

$I_f_3 = \frac{I_p}{2} cos ^2 \theta_2$

$I_f_3$ is the intensity of the light emerging from the stack

substituting values

$I_f_3 = \frac{234.622}{2} * cos^2(61)$

$I_f_3 = 27.57 W/cm^2$

When the third polarizer is removed the second polarizer becomes the

the final polarizer and the intensity of light emerging from the stack would be

$I_f_2 = \frac{I_p}{2} cos ^2 \theta_1$

$I_f_2$ is the intensity of the light emerging from the stack

Substituting values

$I_f_2 = \frac{234.622}{2} cos^2 (21)$

$I_f_2 = 102.24 W/cm^2$

7 0

3 years ago

How are reversibility, time, and entropy connected

kondor19780726 [428]

Answer:

As one goes "forward" in time, the second law of thermodynamics says, the entropy of an isolated system can increase, but not decrease. ... Thus, entropy measurement is a way of distinguishing the past from the future.

Explanation:

6 0

3 years ago