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Korvikt [17]
4 years ago
13

PLEASE HELP AND HURRY!!!!!!!!!

Physics
1 answer:
never [62]4 years ago
4 0

Answer:

I believe the answer would be A. point x

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Two identical conducting spheres each having a radius of 0.500 cm are connected by a light, 1.90-m-long conducting wire. A charg
Ivan

Answer: 4.19 N

Explanation: In order to determinate the tension applied on the wire we have to calculate the electric force between the conductor spheres connected by the wire.

As the wire is a conductor the spheres are at same potential so we have:

V1=V2

V1=k*Q1/r1 and V2=k*Q2/r2

where r1=r2, then

Q1=Q2

so the electric force is given by:

F=k*Q^2/d^2  where d is the distance between the spheres.

Finally replacing the values,  we have

F=9*10^9(41*10^-6)^2/(1.9)^2= 4.19 N

8 0
3 years ago
to what circuit element is an ideal inductor equivalent for circuits with constant currents and voltages?
Marat540 [252]

Answer:

Short circuit

Explanation:

In an ideal inductor circuit with constant current and voltage, it implies that the voltage drop in the circuit is zero (0).

Also, In circuit analysis, a short circuit is defined as a connection between two nodes that forces them to be at the same voltage.

In an ideal short circuit, this means there is no resistance and thus no voltage drop across the connection. That is voltage drop is zero (0).

Therefore, the circuit element is short circuit.

8 0
3 years ago
2x6573858493.8404? please
Studentka2010 [4]

Answer:

13,147,716,987.681

if you multiply 2 by 6573858493.8404, you get 13,147,716,987.681

4 0
1 year ago
S A particle with charge q is located at x=-R, and a particle with charge -2 q is located at the origin. Prove that the equipote
miss Akunina [59]

Let the distance form origin of the point on x-axis be x.

then (-k*2*q/x)+(k*q/(2R-x))=0

solving for x,

we get x=4R/3.

so, the curve is a sphere with center at (4R/3,0,0)

radius=2R-4R/3=2R/3.

To learn more about charge and potential ,visit:

https://brainly.in/question/3525362?

#SPJ4

4 0
2 years ago
Two very large parallel sheets are 5.00 cm apart. Sheet A carries a uniform surface charge density of -8.10 μC/m2 , and sheet B,
gogolik [260]

Answer:E=1.19×10^6N/C

Explanation:see attachment

5 0
3 years ago
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