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77julia77 [94]
3 years ago
11

Three points A, B, and C of unknown charges are at the corners of an equilateral triangle.

Physics
1 answer:
nasty-shy [4]3 years ago
3 0
Answer: option d. 

Q_A= \frac{1}{2} Q_B=- \frac{1}{3} Q_C

Explanation:


1) The direction of the field lines inform about the sign of the charges.

The field lines <span>extend from the positive charges to the negative charges, so you can conclude that the charge C is positve and both charge A and charge B are negative:
</span><span>
</span><span>
</span><span>Charge C: positive
</span><span>
</span><span>Charge A: negative
</span><span>
</span><span>Charge B: netative
</span>

2) The density of the lines (number of lines in a region) inform about the magnitude of the electric field.

Since the charges are at the same distance, the magnitude of the electric field informs directly about the magnitude of the force and that about the magnitude of the charges.

Since, there are the double of lines between C and B than between C and A, the magnitude of charge B is the double than the magnitud of charge A.

From the five options given (a throug e) the only that is consistent with that charges A and B have the same sign, that charge C has different sign, and that charge B is the double of charge A is:

Q_A= \frac{1}{2} Q_B=- \frac{1}{3} Q_C
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Answer:

C.  changing nuclear energy to radiant energy

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Or think of the atom bomb.  Definitely potential energy until the fuse starts detonation and chain reactions.  The radiant kinetic energy and shock waves were horrendous.

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3 years ago
Two identical charges,2.0m apart,exert forces of magnitude 4.0 N on each other.What is the value of either charge?
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Answer:

\large \boxed{42\, \mu \text{C}}$

Explanation:

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The charges are identical, so we can write the formula as

F=k\dfrac{q^{2}}{r^2}

\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}

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3 years ago
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kodGreya [7K]

The Impulse delivered to the baseball is 89 kgm/s.

To solve the problem above, we use the formula of impulse.

⇒ Formula:

  • I = m(v-u)................. Equation 1

Where:

  • I = Impulse delivered to the baseball
  • m = mass of the baseball
  • v = Final velocity of the baseball
  • u = initial speed of the baseball

From the question,

⇒ Given:

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  • v = -44 m/s

⇒ Substitute these values into equation 1

  • I = 0.8(-44-67)
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Note: The negative tells that the impulse is in the same direction as the final velocity and therefore can be ignored.

Hence, The Impulse delivered to the baseball is 89 kgm/s.

Learn more about impulse here: brainly.com/question/7973509

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Suppose a ball of putty moving horizontally with 1kg.m/s of momentum collides and sticks to an identical ball of puffy moving ve
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Answer:

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Explanation:

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