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2 years ago

What is occurring when two waves traveling along the same medium meet and cancel each other out?

Physics

1 answer:

liberstina [14]2 years ago

5 0

Answer:

Destructive interference

Explanation:

Destructive interference occurs when two waves traveling along the same medium meet and cancel each other out. The resultant wave has a smaller amplitude than the individual waves. In this case, the maxima of two waves are 180 degrees out of phase.

Hence, the correct option is (A) "destructive interference"

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I need help with this. I think I got it correct, but I wanna make sure.

kicyunya [14]

Answer:

A because the bigger it is the the more force needs to act apond it

Explanation:

3 0

3 years ago

Read 2 more answers

it is determined that a certain light wave has a wavelength of 3.012x10^-12 m. the light travels at 2.99x10^8 m/s. what is the f

Dahasolnce [82]

Answer:

$9.93\times 10^{19}Hz$

Explanation:

Speed of light is the product of its wavelength and frequency, expressed as

S=fw

Where s represent speed, f is frequency while w is wavelength

Making f the subject of the formula then

f=s/w

Substituting 2.99x10^8 m/s for s and 3.012x10^-12 m for w then

$f=\frac {2.99\times 10^{8}}{3.012\times 10^{-12}}=9.926958831341\times 10^{19}\\f\approx 9.93\times 10^{19}Hz$

Therefore, the frequency equals to $9.93\times 10^{19}Hz$

4 0

2 years ago

Will mark as brainliest if correct!!!!!!!

lutik1710 [3]

C. A could be ruby (speed of light = 170,000 km/s); B could be diamond (speed of light = 120,000 km/s).

Explanation:

Refraction is a phenomenon that occurs when a light rays crosses the boundary between two different mediums.

When this occurs, the light wave changes speed and also direction, according to Snell's law:

$n_1 sin \theta_1 = n_2 sin \theta_2$ (1)

where

$n_1$ is the index of refraction of the first medium

$n_2$ is the index of refraction of the second medium

$\theta_1$ is the angle of incidence (the angle between the incident ray and the normal to the boundary)

$\theta_2$ is the angle of refraction (the angle between the refracted ray and the normal to the boundary)

The index of refraction is the ratio between the speed of light in a vacuum (c) and the speed of light in the medium (v):

$n=\frac{c}{v}$

Using this definition, we can rewrite eq.(1) as

$\frac{\sin \theta_2}{\sin \theta_1} = \frac{v_2}{v_1}$

Where $v_2$ is the speed of light in the 2nd medium and $v_1$ the speed of light in the 1st medium.

Now let's analyze the situation represented in the figure: we see that as the light ray enters the 2nd medium, it bends towards the normal, this means that the angle of refraction is smaller than the angle of incidence:

$\theta_2 < \theta_1$

This means that

$\frac{\sin \theta_2}{\sin \theta_1}$

And therefore,

$\frac{v_2}{v_1}$

So, the speed of light in the second medium is smaller than the speed of light in the first medium: this occurs only in option C), which is therefore the correct choice.

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

6 0

3 years ago

A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex

irina1246 [14]

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=0$

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=3.21\cdot 10^{-3}N$ (+z axis)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=3.21\cdot 10^{-3} N$ (+y axis)

$F_{B_y}=3.21\cdot 10^{-3}N$ (-x axis)

Explanation:

The electric force exerted on a charged particle is given by

$F=qE$

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is:

$F_{B_y}=qvB_y$

where:

$q=+4.9\cdot 10^{-6}C$ is the charge

$v=345 m/s$ is the velocity

$B_y = +1.9 T$ is the magnetic field

Substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And the direction of this force can be found using the right-hand rule:

- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

- Thumb: direction of the force (+z axis)

As in part b), the electric force has not change, since it does not depend on the veocity of the particle:

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

For the field $B_x$ , the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

$F_{B_x}=qvB_x$

And by substituting,

$F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+x axis)

- Thumb: force (+y axis)

For the field $B_y$ , the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

$F_{B_y}=qvB_y$

And by substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+y axis)

- Thumb: force (-y axis)

3 0

3 years ago

A man standing on the Earth can exert the same force with his legs as when he is standing

statuscvo [17]

Answer:

Explanation:

From the analogy of the problem we are made to know that "a man standing on the earth can exert the same force with his legs as when he is standing on the moon".

This force he is exerting is due to his weight. If he can have the same weight on the earth and moon, therefore:

weight = mass x acceleration due gravity

His mass and acceleration due to gravity on both terrestrial bodies are the same.

So, his jump height will be the same on earth and on the moon.

In summary, we have been shown that his mass and the acceleration due to gravity on both planets are the same, therefore, his weight will also be the same. His jump height will also be same.

7 0

2 years ago