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drek231 [11]
4 years ago
10

A positive and a negative charge are released from rest in vacuum. They move toward each other. As they do: A positive and a neg

ative charge are released from rest in vacuum. They move toward each other. As they do: A negative potential energy becomes less negative. A positive potential energy becomes a negative potential energy. A positive potential energy becomes more positive. A negative potential energy becomes more negative. A positive potential energy becomes less positive. SubmitRequest Answer
Physics
1 answer:
Pachacha [2.7K]4 years ago
4 0

Answer:

A negative potential energy becomes more negative

Explanation:

Let the charges be - Q₁ and Q₂ . Let the distance between them be d .

Potential energy = k -Q₁x Q₂ / R

= - KQ₁ Q₂ / R

Now if the magnitude of R decreases , the magnitude of potential energy increases . So we see that the negative  potential energy becomes more negative .  

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Norma-Jean [14]

Hi there!

Voltage in a series can be expressed by the following:

V_T = V_1 + V_2

In words, the total voltage is equal to the sum of the individual voltage drops in a SERIES circuit.

We can solve for the total voltage:

V_T = 10 + 10 = \boxed{\text{ D. 20 V}}

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2 years ago
use the graph to answer the question. if the runner continues at the same speed,how far the runner travel in 30 seconds?​
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Answer:

120 Meters

Explanation:

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40 x 3=120

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4 years ago
John is hiking and notices a small stream of water flowing down the side of the mountain. What part of the water cycle is John o
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Answer: Runofff

Explanation: because it said that it is going down the side of the mountain

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3 years ago
A prism is a cut piece of clear glass or plastic in the form of a triangular prism that separates white light into a spectrum th
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8 0
3 years ago
A charge q1 of -5.00X10^-9 C and a charge q2 of -2.00X10^-9 C are separated by a distance of 40.0 cm. Find the equilibrium posit
Alex73 [517]

Answer:

The equilibrium position for the third charge is 69.28 cm

Explanation:

Given;

q₁ = -5.00 x 10⁻⁹ C

q₂ = -2.00 x 10⁻⁹ C

q₃ = 15.00 x 10⁻⁹ C

distance between q₁  and q₂ = 40.0 cm = 0.4 m                                    

(-q₁)--------------------------------------(-q₂)---------------------------------(+q₃)

At equilibrium the repulsive force between q₁ and q₂ must be equal to attractive force between q₂ and q₃

According to Coulomb's law, repulsive or attractive force between charges is calculated as;

F = \frac{Kq_1q_2}{r_1^2} =  \frac{Kq_2q_3}{r_2^2}

where;

F is repulsive or attractive force between charges

K is Coulomb's constant = 8.99 x 10⁹ Nm²/c²

r₁ is the distance between q₁ and q₂

q₁, q₂ and q₃ are the charge

distance between q₂ and q₃, r₂ is calculated as;

\frac{Kq_1q_2}{r_1^2} = \frac{Kq_2q_3}{r_2^2}\\\\\frac{q_1q_2}{r_1^2} = \frac{q_2q_3}{r_2^2}\\\\r_2^2= \frac{r_1^2q_2q_3}{q_1q_2}\\\\r_2^2= \frac{r_1^2q_3}{q_1} = \frac{0.4^2*15*10^{-9}}{5*10^{-9}} = 0.48\\\\r_2 = \sqrt{0.48} = 0.6928 \ m

Therefore, the equilibrium position for the third charge is 69.28 cm

3 0
3 years ago
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