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OverLord2011 [107]
3 years ago
9

An electron moving in the y direction, at right angles to a magnetic field, experiences a magnetic force in the -x direction. Th

e direction of the magnetic field is in the
Physics
1 answer:
Brut [27]3 years ago
5 0

Answer:

The direction of magnetic field is along + Z axis.

Explanation:

The direction of motion of electron is along y axis.

The magnetic  force is along - X axis.

The force on the charged particle moving in the magnetic field is

\overrightarrow{F} = q (\overrightarrow{v}\times \overrightarrow{B})\\\\- F \widehat{i} = - q (v \widehat{j}\times \overrightarrow{B})\\

So, the direction of the magnetic field is along + Z axis.

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Why does it take the water in a swimming pool some time to heat up during a hot day?
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4 years ago
A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to
White raven [17]

Answer:

1) \Delta s=1000\ ft

2)  \Delta s'=998.11\ ft.s^{-1}

3) t\approx125\ s

t'\approx463.733\ s

Explanation:

Given:

width of river, w=500\ ft

speed of stream with respect to the ground, v_s=8\ ft.s^{-1}

speed of the swimmer with respect to water, v=4\ ft.s^{-1}

<u>Now the resultant of the two velocities perpendicular to each other:</u>

v_r=\sqrt{v^2+v_s^2}

v_r=\sqrt{4^2+8^2}

v_r=8.9442\ ft.s^{-1}

<u>Now the angle of the resultant velocity form the vertical:</u>

\tan\beta=\frac{v_s}{v}

\tan\beta=\frac{8}{4}

\beta=63.43^{\circ}

  • Now the distance swam by the swimmer in this direction be d.

so,

d.\cos\beta=w

d\times \cos\ 63.43=500

d=1118.034\ ft

Now the distance swept downward:

\Delta s=\sqrt{d^2-w^2}

\Delta s=\sqrt{1118.034^2-500^2}

\Delta s=1000\ ft

2)

On swimming 37° upstream:

<u>The velocity component of stream cancelled by the swimmer:</u>

v'=v.\cos37

v'=4\times \cos37

v'=3.1945\ ft.s^{-1}

<u>Now the net effective speed of stream sweeping the swimmer:</u>

v_n=v_s-v'

v_n=8-3.1945

v_n=4.8055\ ft.s^{-1}

<u>The  component of swimmer's velocity heading directly towards the opposite bank:</u>

v'_r=v.\sin37

v'_r=4\sin37

v'_r=2.4073\ ft.s^{-1}

<u>Now the angle of the resultant velocity of the swimmer from the normal to the stream</u>:

\tan\phi=\frac{v_n}{v'_r}

\tan\phi=\frac{4.8055}{2.4073}

\phi=63.39^{\circ}

  • Now let the distance swam in this direction be d'.

d'\times \cos\phi=w

d'=\frac{500}{\cos63.39}

d'=1116.344\ ft

<u>Now the distance swept downstream:</u>

\Delta s'=\sqrt{d'^2-w^2}

\Delta s'=\sqrt{1116.344^2-500^2}

\Delta s'=998.11\ ft.s^{-1}

3)

Time taken in crossing the rive in case 1:

t=\frac{d}{v_r}

t=\frac{1118.034}{8.9442}

t\approx125\ s

Time taken in crossing the rive in case 2:

t'=\frac{d'}{v'_r}

t'=\frac{1116.344}{2.4073}

t'\approx463.733\ s

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Light travels slowest in high density media, faster in lower density media, and fastest in low density media.
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Based on what we know, we can confirm that the wave behavior being observed when light passes through media of different densities is refraction.

<h3>What is refraction?</h3>

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