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ivanzaharov [21]
3 years ago
13

Compare and contrast sound waves and electromagnetic waves. Any bs answers will be reported. Please hurry!!!

Physics
1 answer:
Trava [24]3 years ago
3 0

Answer:

They are both forms of energy.  One has to do with hearing and the other a little bit of light.

Explanation:

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What is the direction of the magnetic field if an electron moving in the positive x direction experiences a magnetic force in th
Alex787 [66]

Given :

An electron moving in the positive x direction experiences a magnetic force in the positive z direction.

To Find :

The direction of the magnetic field.

Solution :

We know, force is given by :

\vec{F}=q(\vec{v}\times \vec{B)}

Here, q = -e.

\vec{F}=(-e)(\vec{v}\times \vec{B)}\\\\\hat{k}=(-e)(\hat{i}\times \vec{B})

Now, for above condition to satisfy :

\hat{i}\times \vec{B}=-\hat{k}

So, \vec{B}=-\hat{j}

Therefore, direction of magnetic field is negative y direction.

Hence, this is the required solution.

7 0
4 years ago
Based on the TIA/EIA 568-B structured cabling standard, the cabling that runs from the telecommunications closet to each work ar
lana [24]

Answer:

False.

Explanation:

Backbone Cabling: A system of cabling that connects the equipment rooms and telecommunications rooms.

Horizontal Cabling: The system of cabling that connects telecommunications rooms to individual outlets or work areas on the floor.

4 0
4 years ago
Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

4 0
3 years ago
If you're studying the science of sound, you're studying____?
timurjin [86]
The study of sound is called sonics and the study of sound waves are acoustics
3 0
4 years ago
What force is needed to give a 4.5-kg bowling ball an acceleration of 9 m/s2?
Sloan [31]
F=ma  
f=4.5*9                                           

40.5 N
hope this help
8 0
3 years ago
Read 2 more answers
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