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Xelga [282]
2 years ago
13

A long straight wire has fixed negative charge with a linear charge density of magnitude 4.6 nC/m. The wire is to be enclosed by

a coaxial, thin-walled, nonconducting cylindrical shell of radius 2.2 cm. The shell is to have positive charge on its outside surface with a surface charge density σ that makes the net external electric field is zero. Calculate σ.
Physics
1 answer:
Semmy [17]2 years ago
7 0

Answer: -33.3 * 10^9 C/m^2( nC/m^2)

Explanation: In order to solve this problem we have to use the gaussian law, the we have:

Eoutside =0 so teh Q inside==

the Q inside= 4.6 nC/m*L + σ *2*π*b*L where L is the large of the Gaussian surface and b the radius of the shell.

Then we simplify and get

σ= -4.6/(2*π*b)= -33.3 nC/m^2

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You watch a house being built in your neighborhood. After a few months, it’s ready to be sold. From start to finish this is an e
jenyasd209 [6]
The answer is A of cours
3 0
2 years ago
What is the equivalent resistance between points A and C if R1=1430, R2=1350, R3=1100, R4=1350, and R5=1150.
Marianna [84]

R1 + R4 = 1430 + 1350 = 2780 = R14    series combination of R1 & R4

R2 + R5 = 1350 + 1150 = 2500 = R25

The circuit has been reduced to 3 resistors in parallel

R314 = 2780 * 1100 / (2780 + 1100) = 788  this is the resistance of the parallel combination of R14 and R3

R31425 = 2500 * 788 / (2500 + 788) = 599 which is the equivalent of the circuit  - you can also use the formula for 3 resistors in parallel but this seems simpler

7 0
3 years ago
"Which of the following best describes the circuit shown below?
Anettt [7]
The figure shown above is series combination as the two resistors (bulb) are there which are then connected to the battery
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3 0
3 years ago
Read 2 more answers
Calculate the average drift speed of electrons traveling through a copper wire with a crosssectional area of 80 mm2 when carryin
Vedmedyk [2.9K]

Answer:

The correct answer is 2.8*10^{-5}ms^{-1}

Explanation:

The formula for the electron drift speed is given as follows,

u=I/nAq

where n is the number of of electrons per unit m³, q is the charge on an electron and A is the cross-sectional area of the copper wire and I is the current. We see that we already have A , q and I. The only thing left to calculate is the electron density n that is the number of electrons per unit volume.

Using the information provided in the question we can see that the number of moles of copper atoms in a cm³ of volume of the conductor is 8.93/63.5 molcm^{-3}. Converting this number to m³ using very elementary unit conversion we get 140384molm^{-3}. If we multiply this number by the Avagardo number which is the number of atoms per mol of any gas , we get the number of atoms per m³ which in this case is equal to the number of electron per m³ because one electron per atom of copper contribute to the current. So we get,

n=140384*6.02*10^{23} = 8.45*10^{28}electrons.m^{-3}

if we convert the area from mm³ to m³ we get A=80*10^{-6}m^{2}.So now that we have n, we plug in all the values of A ,I ,q and n into the main equation to obtain,

u=30/(8.45*10^{28}*80*10^{-6}*1.602*10^{-19})\\u=2.8*10^{-5}m.s^{-1}

which is our final answer.

6 0
3 years ago
A hoodlum throws a stone vertically downward with an initial speed of 12.0 m/s from the roof of a building,30.0 m above the grou
bogdanovich [222]
S=ut +1/2at2
30=12t+1/2*10*t2
5t2+12t-30=0
8 0
3 years ago
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