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balandron [24]
3 years ago
15

Two neutral metal spheres on wood stands are touching. A negatively charged rod is held directly above the top of the left spher

e, not quite touching it. While the rod is there, the right sphere is moved so that the spheres no longer touch. Then the rod is withdrawn. which answer is correct?
A) Both the spheres are neutral.
b) Left sphere is negatively charged, another is charged positively.
C) Right sphere is negatively charged, another is charged positively.
D) Both the spheres are charged positively.
E) Both the spheres are charged negatively
Physics
2 answers:
FrozenT [24]3 years ago
7 0

Answer: Option (C) is the correct answer.

Explanation:

As we know that metals are able to conduct electricity so, when a negatively charges rod is kept closer to the left sphere then electrons will enter the sphere.

Since, like charges repel each other. Hence, some of the negative changes from the rod will repel the negative charges of left sphere.

As both left and right spheres are touching each other so, the electrons will move towards the right sphere. As a result, there will be too many electrons (negative charge) present on the right sphere and very less electrons present in the left sphere.

Thus, we can conclude that the statement right sphere is negatively charged, another is charged positively, is true.

Murrr4er [49]3 years ago
4 0

Answer: ITS CCCC

Explanation:

ITS CORRECT

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Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire fe
Aleksandr [31]

Complete question:

Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire feels a repulsive force of 2.4 x 10^-4 N created by the interaction of the 12.5 A current and the magnetic field created by the bottom current (I). Find the magnitude and direction of the bottom current, if the distance between the two wires is 40cm.

Answer:

The bottom current is 12.8 A to the right.

Explanation:

Given;

length of the wires, L = 3.0 m

current in the top wire, I₁ = 12.5 A

repulsive force between the two wires, F = 2.4 x 10⁻⁴ N

distance between the two wires, r = 40 cm = 0.4 m

The repulsive force between the two wires is given by;

F = \frac{\mu_oI_1I_2L}{2\pi r}\\\\I_{2} = \frac{2F\pi r}{\mu_oI_1L}

Where;

I₂ is the bottom current

The direction of the bottom current must be in the same direction as the top current since the force between the two wires is repulsive.

I_{2} = \frac{2F\pi r}{\mu_oI_1L}\\\\I_{2} = \frac{2(2.4*10^{-4})(\pi)(0.4)}{(4\pi*10^{-7})(12.5)(3)}\\\\I_{2} = 12.8 \ A

Therefore, the bottom current is 12.8 A to the right.

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Where V = final velocity

u = initial velocity

Therefore, u = Vi

u = √Vf^2 - 2gs

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u = √0 + 55.468

u = √55.468

u = 7.4476 m/s

u = 7.45m/s

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