I need to figure out A but i’m not sure how

Organisms that live on or burrow into the ocean floor are

creativ13 [48]

Benthos

Option b is the answer

8 0

3 years ago

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the magnitude of the dri

nordsb [41]

Complete Question

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm2.) mm/s

Answer:

The drift velocity is $v = 0.0002808 \ m/s$

Explanation:

From the question we are told that

The current on the copper is $I = 20 \ A$

The cross-sectional area is $A = 5.261 \ mm^2 = 5.261 *10^{-6} \ m^2$

The number of copper atom in the wire is mathematically evaluated

$n = \frac{\rho * N_a}Z}$

Where $\rho$ is the density of copper with a value $\rho = 8.93 \ g/m^3$

$N_a$ is the Avogadro's number with a value $N_a = 6.02 *10^{23}\ atom/mol$

Z is the molar mass of copper with a value $Z = 63.55 \ g/mol$

So

$n = \frac{8.93 * 6.02 *10^{23}}{63.55}$

$n = 8.46 * 10^{28} \ atoms /m^3$

Given the 1 atom is equivalent to 1 free electron then the number of free electron is

$N = 8.46 * 10^{28} \ electrons$

The current through the wire is mathematically represented as

$I = N * e * v * A$

substituting values

$20 = 8.46 *10^{28} * (1.60*10^{-19}) * v * 5.261 *10^{-6}$

=> $v = 0.0002808 \ m/s$

8 0

3 years ago

When an electric current flows through a long conductor, each free electron moves

ki77a [65]

A. through a relatively short distance.

The speed is actually called the drift speed of the electron.

7 0

3 years ago

The nose of an ultralight plane is pointed south, and its airspeed indicator shows 28 m/s . the plane is in a 18 m/s wind blowin

leva [86]

<span>Here I think you have to find the velocity in x and y components where x is east and y is north
So as air speed indicator shows the negative speed in y component and adding it in
air speed while multiplying with the direction component we will get the velocity as velocity is a vector quantity so direction is also required
v=-28 m/s y + 18 m/s (- x/sqrt(2) - y/sqrt(2))
solving
v= -12.7 m/s x-40.7 m/s y
if magnitude of velocity or speed is required then
speed= sqrt(12.7^2 + 40.7^2)
speed= 42.63 m/s
if angle is asked
angle = arctan (40.7/12.7)
angle = 72.67 degrees south of west</span>

6 0

3 years ago

an object is acted on by a drag force with a magnitude that is proportional to the speed. the object accelerates downward at 3.0

Nutka1998 [239]

The terminal speed of the object falling down is 66.67 m/s.

The terminal speed acquired by the body when,

Weight of the body = Drag force of the body

It is given,

Drag force is directly proportional to the speed,

So,

F = CV

Where F is drag force,

V is the speed,

C is the constant,

So, it can be written as C = F/V.

The weight of the body = mg

The weight of the body = 10m

M is the mass and g is the acceleration due to gravity,

The drag force when the speed is 20m/s.

Drag force = ma

a is the acceleration during the drag force which is given to be 3m/s²,

Drag force = 3m

Now we can write,

F₁/V₁ = F₂/V₂

F₁ is the drag force at 20m/s speed.

F₂ is the weight of the body and V₂ is the terminal speed,

Now, it can be written,

3m/20 = 10m/V₂

V₂ = 66.67 m/s.

So, the terminal speed is 66.67m/s.

To know more about terminal speed, visit,

brainly.com/question/14605362

#SPJ4

7 0

1 year ago