Ask question

Ask question

All categories

3 years ago

11

A copper wire has a circular cross section with a raduis of 1.25mm. If the wire carries a current of 3.70A, find the drift speed

of electron in this wire.

1 answer:

alexdok [17]3 years ago

4 0

Answer: 0.043mm/s

Explanation:

The drift speed of electron in this wire will ba calculated as:

= I/nqA

= I/nqπr²

= 3.7/[(1.1 × 1^29) × (1.6 × 10^19) × (3.14 × 0.00125²)

= 4.3 × 10^-5

= 0.043mm/s

The drift speed of the electron in this wire is 0.043mm/s.

You might be interested in

What are the parts of a wedge and their function?

Svet_ta [14]

Answer:

A wedge is a machine that consists of two inclined planes, giving it a thin end and thick end. A wedge is used to cut or split apart objects. Force is applied to the thick end of the wedge, and the wedge applies force to the object along both of its sloping sides. This force causes the object to split apart

5 0

3 years ago

Read 2 more answers

A boat starts moving across a river at velocity v perpendicular to the river bank.

Rina8888 [55]

D because you are dump to ask that questions that is so East

5 0

3 years ago

When analyzing the results of a scientific experiment, Kevin should base his conclusion on

elena-14-01-66 [18.8K]

The best answer would be b

5 0

2 years ago

6) A penny is dropped from the roof of a house. The penny is traveling at 35 m/s right before it hits

nadezda [96]

Wait what is m/s?
Can you tell me that so i could help

8 0

3 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

Nina [5.8K]

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:

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

$t'=\frac{1116.344}{2.4073}$

$t'\approx463.733\ s$

3 0

4 years ago