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qaws [65]
3 years ago
14

A copper wire has a diameter of 2.05 mm and carries a current of 5 A due solely to electrons. Given that the free (capable to mo

ve) electron concentration in copper is 1023 electrons per cm3, find the average velocity of the electrons in the wire. Give the velocity in mm/s and in miles/hour.
Physics
1 answer:
LekaFEV [45]3 years ago
6 0

To solve this problem we will apply the concepts related to the average velocity in an electron, defined as the value of the current on the product between the charge, the number of electrons, the constancy pi and the squared radius. Our values are given as,

D = 2.05 mm

I = 5A

e = -1.6 x 10^{-19} C

Q = 10^{23} e/cm^3 = 10^{29}e/m^3

The radius would b,

R = \frac{D}{2} = \frac{2.05*10^{-3}m}{2}

R = 1.025*10^{-3}m

Now the average velocity

V = \frac{I}{(Q e \pi R^2)}

Here,

I = Current

Q = Electron concentration

e = Charge of electron

R = Radius

V = \frac{5}{(10^{29})(-1.6*10^{-19})(\pi)(1.025*10^{-3})^2}}

V = 9.46*10^{-5}m/s

Converting to the values required we have that,

V = 9.46*10^{-5} m/s (\frac{1000mm}{1m})

V = 9.46*10^{-2}mm/s

And,

V = 9.46*10^{-5}m/s (\frac{3600s}{1hour})(\frac{0.000621371miles}{1m})

V = 2.116*10^{-4}mph

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You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration
jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time t the ball is in air, we can use the following equation:

y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}

Where:

y=0m is the final height of the ball

y_{o}=40 m is the initial height of the ball

V_{o}=10 m/s is the initial velocity of the ball

t is the time the ball is in air

g=9.8 m/s^{2} is the acceleration due to gravity  

\theta=30\°

Then:

0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}

0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}

Multiplying both sides of the equation by -1 and rearranging:

4.9 m/s^{2}t^{2}-5m/s t-40 m=0

At this point we have a quadratic equation of the form at^{2}+bt+c=0, which can be solved with the following formula:

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}

Where:

a=4.9

b=-5

c=-40

Substituting the known values:

t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}

Solving the equation and choosing the positive result we have:

t=3.41 s  This is the time the ball is in air

5 0
4 years ago
A charged particle of mass m = 5.00g and charge q = -70.0μC moves horizontally to the right at a constant velocity of v = 30.0 k
pychu [463]

Answer:

The magnitude of the force, B = 5 Tesla, Up (North) direction

Explanation:

Magnetic force F= Eq where Electric field, E = 750 NC

and charge, q = -70 μC = -7 ×10^{-5}C

F = 750 ×  -7 ×10^{-5}

F = 0.0525

But F = qvB; B = \frac{F}{qv}

where B is the magnetic field

= 0.0525 ÷ ( -7 ×10^{-5} × 30)

B = 5.0 Teslas

The force on a negative charge is in exactly the opposite direction to that on a positive charge.

Hence the direction of the charge is up (North).

8 0
3 years ago
Read 2 more answers
What should a free-body diagram look like for a skydiver who has opened his parachute and is now slowing down as he falls?
mote1985 [20]
The last choice. Two arrows and the arrow up is shorter than the arrow down. Since the guy is falling and he’s opened his chute, he’s slowing down but he’s still falling meaning the force of gravity is stronger than the air resistance.
8 0
4 years ago
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A square is 1.0 m on a side. Point charges of +4.0 μC are placed in two diagonally opposite corners. In the other two corners ar
finlep [7]

Answer:

<em>B) 1.0 × 10^5 V</em>

Explanation:

<u>Electric Potential Due To Point Charges </u>

The electric potential produced from a point charge Q at a distance r from the charge is

\displaystyle V=k\frac{Q}{r}

The total electric potential for a system of point charges is equal to the sum of their individual potentials. This is a scalar sum, so direction is not relevant.

We must compute the total electric potential in the center of the square. We need to know the distance from all the corners to the center. The diagonal of the square is

d=\sqrt2 a

where a is the length of the side.

The distance from any corner to the center is half the diagonal, thus

\displaystyle r=\frac{d}{2}=\frac{a}{\sqrt{2}}

\displaystyle r=\frac{1}{\sqrt{2}}=0.707\ m

The total potential is  

V_t=V_1+V_2+V_3+V_4

Where V1 and V2 are produced by the +4\mu C charges and V3 and V4 are produced by the two opposite charges of \pm 3\mu\ C. Since all the distances are equal, and the charges producing V3 and V4 are opposite, V3 and V4 cancel each other. We only need to compute V1 or V2, since they are equal, but they won't cancel.

\displaystyle V_1=V_2=k\frac{Q}{r}=9\times 10^9 \frac{4\times 10^{-6}}{0.707}

V_1=V_2=50912\ V

The total potential is

V_t=50912\ V+50912\ V=1\times 10^5\ V

\boxed{V_t=1\times 10^5\ V}

6 0
3 years ago
Select the correct answer.
andrew-mc [135]
I think the answer would be C
7 0
4 years ago
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