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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 car company wants to ensure its newest model can stop in less than 450 ft when traveling at 60 mph. If we assume constant dece
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Answer:

The value of acceleration that accomplishes this is 8.61 ft/s² .

Explanation:

Given;

maximum distance to be traveled by the car when the brake is applied, d = 450 ft

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Read 2 more answers
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Answer:

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f'=f(\frac{v-v_o}{v+v_s})    (1)

where

f: frequency of the source = ?

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vs: speed of the source = 0 m/s (stationary)

You replace the values of all parameters in the equation (1):

To calculate f' you first calculate the frequency of the sound wave, by using the following formula:

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f=\frac{v}{\lambda}=\frac{343m/s}{0.700m}=480Hz

Next, you replace the values of all parameters in the equation (1):

f'=(490Hz)(\frac{343m/s-40.0m/s}{343m/s})=432Hz

hence, the frequency perceived by the car is 432 Hz

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