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anyanavicka [17]
2 years ago
7

Find A and B I need help like asap so pleasee help

Physics
1 answer:
zloy xaker [14]2 years ago
8 0

Answer:

I don't understand?

Explanation:

Could you please elaborate?

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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
3 years ago
A very long string (linear density 0.7 kg/m ) is stretched with a tension of 70 N . One end of the string oscillates up and down
rewona [7]

To develop this problem it is necessary to apply the concepts related to Wavelength, The relationship between speed, voltage and linear density as well as frequency. By definition the speed as a function of the tension and the linear density is given by

V = \sqrt{\frac{T}{\rho}}

Where,

T = Tension

\rho = Linear density

Our data are given by

Tension , T = 70 N

Linear density , \rho = 0.7 kg/m

Amplitude , A = 7 cm = 0.07 m

Period , t = 0.35 s

Replacing our values,

V = \sqrt{\frac{T}{\rho}}

V = \sqrt{\frac{70}{0.7}

V = 10m/s

Speed can also be expressed as

V = \lambda f

Re-arrange to find \lambda

\lambda = \frac{V}{f}

Where,

f = Frequency,

Which is also described in function of the Period as,

f = \frac{1}{T}

f = \frac{1}{0.35}

f = 2.86 Hz

Therefore replacing to find \lambda

\lambda = \frac{10}{2.86}

\lambda = 3.49m

Therefore the wavelength of the waves created in the string is 3.49m

3 0
3 years ago
I really have no idea where to start with this. Could someone please exaplin the answer and how they got it? Theres a lot of the
LuckyWell [14K]
I believe the answer should be the last option. upon interaction, both objects should have the same charge after the electrons are transferred.
6 0
3 years ago
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True or false? All waves need a medium in order to travel
Art [367]
That's false.  Mechanical waves (like sound and ocean waves) do
need a medium to travel in, but electromagnetic waves (like radio
and light) don't.
4 0
3 years ago
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Picture attached please help! 2 different questions!
Maksim231197 [3]

Wave A would have higher amplitude

Hope this helps :D

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3 years ago
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