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stich3 [128]
2 years ago
7

Two point charges, A and B, are separated by a distance of 19.0 cm . The magnitude of the charge on A is twice that of the charg

e on B. If each charge exerts a force of magnitude 45.0 N on the other, find the magnitudes of the charges.
Physics
1 answer:
ankoles [38]2 years ago
5 0

Answer:

QA = 19μC

QB = 9.5 μC

Explanation:

  • The force that each charge exerts on the other must obey Coulomb's Law, as follows:

       F_{AB} = \frac{k*Q_{A} * Q_{B}}{r_{AB}^{2}}  (1)

  • We know that the value of the magnitude of FAB  is 45.0 N, the distance between QA and  QB is 0.19 m, and that QA = 2*QB.
  • Replacing in (1), we can solve for QB, as follows:

      Q_{B} = \sqrt{\frac{F_{AB}*r_{AB} ^{2}}{2*k} } = \sqrt{\frac{45.0N*(0.19m) ^{2}}{2*9e9N*m2/C2} } = 9.5e-6 C  (2)

  • Since QA = 2*QB
  • ⇒ QA = 2* 9.5μC = 19.0 μC
  • ⇒ QB = 9.5μC
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3 years ago
A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location
amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

q is the distance of the image from the lens

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p is the distance of the object from the lens

In this problem, we have

f=-16.0 cm (the focal length is negative for a diverging lens)

p=10.0 cm is the distance of the object from the lens

Solvign the equation for q, we find

\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}

q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm

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<h3>Answer:</h3>

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<h3>Explanation:</h3>

<u>We are given;</u>

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We are required to calculate the work done

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Force applied is equivalent to the weight of the dog.

Work done = 24 N × 4.9 m

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more deceleration.

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