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Ede4ka [16]
3 years ago
13

A small, positively charged ball is moved close to a large, positively charged ball. which describes how the small ball likely r

esponds when it is released?it will move toward the large ball because like charges repel.it will move toward the large ball because like charges attract.it will move away from the large ball because like charges repel.it will move away from the large ball because like charges attract.
Physics
2 answers:
NNADVOKAT [17]3 years ago
6 0

Answer;

-it will move away from the large ball because like charges repel.

Explanation;

-Electric force is the force that pushes apart two like charges, or that pulls together two unlike charges. The basic law of electrostatics Like charges of electricity repel each other, whereas unlike charges attract each other.

When small, positively charged ball is moved close to a large, positively charged ball it would be pushed away from the large positively charged ball since they are both positively charged. One has to put in energy to try to move the small ball closer to the large ball. The closer one try to move it to the large ball, the more energy one has to put in, so the more electrical potential energy the small ball would have.

MariettaO [177]3 years ago
4 0

Answer:

C. It will move away from the large ball because like charges repel.

Explanation:

I just did the test and got it right

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(a) The ratio of turns in the primary and secondary coils of her transformer is 1.833

(b) The ratio of input to output current is 0.55

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\frac{V_p}{V_s} = \frac{N_p}{N_s} = \frac{I_s}{I_p}

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(a) The ratio of turns in the primary and secondary coils of her transformer;

\frac{N_p}{N_s} = \frac{V_p}{V_s} \\\\\frac{N_p}{N_s} = \frac{220}{120} = 1.833

(b) The ratio of input to output current;

\frac{I_p}{I_s} = \frac{V_s}{V_p} \\\\\frac{I_p}{I_s} = \frac{120}{220} \\\\\frac{I_p}{I_s} = 0.55

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