Ask question

Ask question

All categories

3 years ago

11

The density of mobile electrons in copper metal is 8.4 × 1028 m-3. Suppose that i= 4.4 × 1018 electrons/s are drifting through a

copper wire. (This is a typical value for a simple circuit.) The diameter of the wire is 2.5 mm. In this case, about how many minutes would it take for a single electron in the electron sea to drift from one end to the other end of a wire 26 cm long?

1 answer:

neonofarm [45]3 years ago

4 0

Answer: 405.3 minutes

Explanation: In order to explain this problem we have to use the following:

Fisrtly we calculate the volume of the wire, this is given by:

Vwire=π*r^2*L where r and L are the radius and L the length of teh wire, respectively.

Vwire=π*1.25*10^-3*0.26=1.27*10^-6 m^3

then the number of the total electrons in tthe wire volume is given by;

n° electrons in the wire=ρ*Vwire=8.4*10^28*1.27*10^-6 m^3=1.07 *10^23

Finally, considering the current in the wire equal to 4.4*10^18 electrons/s

the time consuming to extract all the electrons from the wire is given by:

t= total electrons in the wire/ current=1.067*10^23/4.4*10^18=24,318 s

equivalent to 405.3 minutes

You might be interested in

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!<br><br> The two main types of circuits are:

Gre4nikov [31]

Answer:

series and parallel

Explanation:

4 0

3 years ago

Why are there shadows?

Deffense [45]

Shadows are formed when an opaque object or an object that doesn't allow light to pass through is in the way or infront of etc. a source of light.

6 0

3 years ago

Read 2 more answers

Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma

AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

$m_1$ = Mass of first car = 120 kg

$m_2$ = Mass of second car

$u_1$ = Initial Velocity of first car = 14 m/s

$u_2$ = Initial Velocity of second car = 0 m/s

$v_1$ = Final Velocity of first car = -2 m/s

$v_2$ = Final Velocity of second car

For perfectly elastic collision

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}$

Applying in the next equation

$v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s$

$m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg$

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0

3 years ago

A certain runner averages 4.1 m/s over a 10

Ghella [55]

Answer: 40.650406504065 or 40 minutes and 39 seconds.

Explanation:

1 k = 1000m

race = 10000m

runner time = 10000 / 4.1

runner time = 2439.0243902439024 seconds

runner time = 2439.0243902439024/60 = 40.650406504065 or 40 minutes and 39 seconds.

6 0

3 years ago

Three masses are located in the x- y plane as follows: a mass of 6 kg is at (0 m, 0 m), a mass of 4 kg is at (3 m, 0 m), and a m

enot [183]

Answer:

The center of mass of three mass in the x-y plane is located at (1,0.5).

Explanation:

It is given that, a mass of 6 kg is at (0,0), a mass of 4 kg is at (3,0), and a mass of 2 kg is at (0,3). We need to find the center of mass of the system. Center of mass in x direction is :

$C_x=\dfrac{6\times 0+4\times 3+2\times 0}{6+4+2}\\\\C_x=1$

The center of mass in y direction is :

$C_y=\dfrac{6\times 0+4\times 0+2\times 3}{6+4+2}\\\\C_y=0.5$

So, the center of mass of three mass in the x-y plane is located at (1,0.5).

3 0

3 years ago