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Brums [2.3K]
2 years ago
13

Two small objects are suspended from threads. When the objects are moved close together, they attract one another. What of the f

ollowing could produce this result?a. On object is positively charged and the other is negatively charged.b. One object is positively charged and the other conductor is uncharged.c. One object is negatively charged and the other conductor is uncharged.d. All of the above could result in an attraction.
Physics
1 answer:
creativ13 [48]2 years ago
4 0

Answer:d

Explanation:

All the given situations are possible because

(a)When particles are oppositely charged then they attract each other

(b)One is Positively charged and other is uncharged: Charged particle will induce charges of opposite nature to attract the other particle

(c)Negatively charged particles will induce the positive charge in the uncharged particle to attract the initially uncharged particle.

                 

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The concept required to solve this problem is associated with potential energy. Recall that potential energy is defined as the product between mass, gravity, and change in height. Mathematically it can be described as

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Here,

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3 years ago
A substance did not change it's chemical nature in reaction which mostly likely describe the reaction
Ad libitum [116K]
If the substance doesn't change chemically, it is a physical reaction.
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When fuel and air are compressed in the compression stroke, ...... a. each molecule of fuel is heated to its flash point b. each
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None of the choices is an appropriate response.

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When the fuel/air mixture is compressed during the compression stroke,
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7 0
3 years ago
If 2.0×10^−4 C of charge passes a point in 2.0×10^−6 s , what is the rate of current flow?1.0×10^−10 A1.0×10^2 A4.0×10^−1 A4.0×1
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This problem is about the rate of the current. It's important to know that refers to the quotient between the electric charge and the time, that's the current rate.

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Where Q = 2.0×10^−4 C and t = 2.0×10^−6 s. Let's use these values to find I.

\begin{gathered} I=\frac{2.0\times10^{-4}C}{2.0\times10^{-6}\sec } \\ I=1.0\times10^{-4-(-6)}A \\ I=1.0\times10^{-4+6}A \\ I=1.0\times10^2A \end{gathered}

<em>As you can observe above, the division of the powers was solved by just subtracting their exponents.</em>

<em />

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3 0
11 months ago
The force generated by a single muscle fiber can be increased by increasing is called
mars1129 [50]

Answer:

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Explanation:

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