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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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Doss [256]

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There are so many questions which one you don't know

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2 years ago
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What does MA mean for you in real life?
andrezito [222]
To me "Ma" mean in real life is either mama or Massachusttes.
8 0
2 years ago
As a way of determining the inductance of a coil used in a research project, a student first connects the coil to a 5.62 V batte
Reptile [31]

Answer:

Its inductance L = 166 mH

Explanation:

Since a current, I = 0.698 A is obtained when a voltage , V = 5.62 V is applied, the resistance of the coil is gotten from V = IR

R = V/I = 5.62/0.698 = 8.052 Ω

Since we have a current of I' = 0.36 A (rms) when a voltage of V' = 35.1 V (rms) is applied, the impedance Z of the coil is gotten from

V₀' = I₀'Z where V₀ = maximum voltage = √2V' and I₀ = maximum current = √2I'

Z = V'/I' = √2 × 35.1 V/√2 × 0.36 V = 97.5 Ω

WE now find the reactance X of the coil from

Z² = X² + R²

X = √(Z² - R²)

= √(97.5² - 8.05²)

= √(9506.25 - 64.8025)

= √9441.4475

= 97.17 Ω

Now, the reactance X = 2πfL where f = frequency of generator = 93.1 Hz and L = inductance of coil.

L = X/2πf

= 97.17/2π(93.1 Hz)

= 97.17 Ω/584.965 rad/s

= 0.166 H

= 166 mH

Its inductance L = 166 mH

5 0
3 years ago
A proton travels with a speed of 5.02×10⁶ m/s in a direction that makes an angle of 60.0° with the direction of a magnetic ficld
Triss [41]

The magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.

Speed of the proton = 5.02 × 10 ⁶ m /a

Angel of between the velocity and the magnetic force = 60 °

The magnitude of magnetic field B = 0.180 T

The magnitude of the magnetic force on the proton is,

F = q(v \times B)

F = qvB \: sin \:  θ

F = 1.6 \times 10 ^{ - 19}  \times 5.02 \times 10 ^{6}  \times 0.180 \times  \: sin \: 60°

= 1.25 \times 10 ^{ - 13}  \: N

Therefore, the magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.

To know more about magnetic force, refer to the below link:

brainly.com/question/23096032

#SPJ4

4 0
1 year ago
A hill is 132 m long and makes an angle of 12.0 degrees with the horizontal. As a 54 kg jogger runs up the hill, how much work d
TEA [102]

Answer:

14523.55J

Explanation:

The work done by the jogger against gravity is given by the following equation;

W=mgh.................(1)

where m is the mass, g is acceleration due to gravity taken as 9.8m/s^2 and h is the height of the hill.

Since the length of the hill is 132m and it is inclined at 12 degrees to the horizontal, the height is thus given as follows;

h=132sin12^o\\h=27.44m

Substituting this into equation (1) with all other necessary parameters, we obtain the following;

W=54*9.8*27.44\\W=14523.55J

4 0
3 years ago
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