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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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A paint can is sitting on a ladder 20 m high. It has a mass of 7 kg. What is the
loris [4]

Answer:

<h3>The answer is 1400 J</h3>

Explanation:

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From the question we have

PE = 20 × 10 × 7

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<h3>1400 J</h3>

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3 years ago
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3 years ago
Charge q1 = +2.00 μC is at -0.500 m along the x axis. Charge q2 = -2.00 μC is at 0.500 m along the x axis. Charge q3 = 2.00 μC i
Kobotan [32]

The magnitude of <em>electrical</em> force on charge q_{3} due to the others is 0.102 newtons.

<h3>How to calculate the electrical force experimented on a particle</h3>

The vector <em>position</em> of each particle respect to origin are described below:

\vec r_{1} = (-0.500, 0)\,[m]

\vec r_{2} = (+0.500, 0)\,[m]

\vec r_{3} = (0, +0.500)\,[m]

Then, distances of the former two particles particles respect to the latter one are found now:

\vec r_{13} = (+0.500, +0.500)\,[m]

r_{13} = \sqrt{\vec r_{13}\,\bullet\,\vec r_{13}} = \sqrt{(0.500\,m)^{2}+(0.500\,m)^{2}}

r_{13} =\frac{\sqrt{2}}{2}\,m

\vec r_{23} = (-0.500, +0.500)\,[m]

r_{23} = \sqrt{\vec r_{23}\,\bullet \,\vec r_{23}} = \sqrt{(-0.500\,m)^{2}+(0.500\,m)^{2}}

r_{23} =\frac{\sqrt{2}}{2}\,m

The resultant force is found by Coulomb's law and principle of superposition:

\vec R = \vec F_{13}+\vec F_{23} (1)

Please notice that particles with charges of <em>same</em> sign attract each other and particles with charges of <em>opposite</em> sign repeal each other.

\vec R = \frac{k\cdot q_{1}\cdot q_{3}}{r_{13}^{2}}\cdot \vec u_{13}  +\frac{k\cdot q_{2}\cdot q_{3}}{r_{23}^{2}}\cdot \vec u_{23} (2)

Where:

  • k - Electrostatic constant, in newton-square meters per square Coulomb.
  • q_{1}, q_{2}, q_{3} - Electric charges, in Coulombs.
  • r_{13}, r_{23} - Distances between particles, in meters.
  • \vec u_{13}, \vec u_{23} - Unit vectors, no unit.

If we know that k = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}, q_{1} = 2\times 10^{-6}\,C, q_{2} = 2\times 10^{-6}\,C, q_{3} = 2\times 10^{-6}\,C, r_{13} =\frac{\sqrt{2}}{2}\,m, r_{23} =\frac{\sqrt{2}}{2}\,m, \vec u_{13} = \left(-\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}  \right) and \vec u_{23} = \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}  \right), then the vector force on charge q_{3} is:

\vec R = \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}  \right) + \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}  \right)

\vec R = 0.072\cdot \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}  \right) + 0.072\cdot \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}  \right)\,[N]

\vec R = 0.072\cdot \left(0, -\sqrt{2}\right)\,[N]

And the magnitude of the <em>electrical</em> force on charge q_{3} (R), in newtons, due to the others is found by Pythagorean theorem:

R = 0.102\,N

The magnitude of <em>electrical</em> force on charge q_{3} due to the others is 0.102 newtons. \blacksquare

To learn more on Coulomb's law, we kindly invite to check this verified question: brainly.com/question/506926

8 0
2 years ago
While visiting the Albert Michelson exhibit at Clark University, you notice that a chandelier (which looks remarkably like a sim
aksik [14]

Answers:

a) 0.144 Hz

b) 0.904 rad/s

c) 11.818 m

d)9.77 m/s^{2}

Explanation:

The rest of the question is written below:

a) Calculate the frequency of oscillation (in Hertz) of the chandelier

b) Calculate the angular frequency \omega of the chandelier in radians/ second

c) Determine the length L in meters of the chandelier

d) That evening, while hanging out in JJ. Thompson's House O' Blues, you notice that (coincidentally) there is a chandelier identical in every way to the one at the Michelson exhibit except this one swings back and forth 0.01 seconds slower, so the period is T+0.01 s. Determine the acceleration due to gravity in m/s^{2} at the club.

a) The frequency f has an inverse relation with the period T:

f=\frac{1}(T} (1)

Where T=6.9 s

f=\frac{1}(6.9s}=0.144 Hz (2)

b) The angular frequency \omega is given by:

\omega=2\pi f=\frac{2 \pi}{T} (3)

\omega=2\pi (0.144 Hz) (4)

\omega=0.904 rad/s (5)

c) Another expression for the period is:

T=2 \pi \sqrt{\frac{L}{g}} (6)

Where:

L is the length of the pendulum

g=9.8 m/s^{2} is the mean acceleration due gravity

Isolating L:

L=\frac{T^{2} g}{4 \pi^{2}} (7)

L=\frac{(6.9)^{2} (9.8 m/s^{2})}{4 \pi^{2}} (8)

L=11.818 m (9)

d) In this case the period of the pendulum is T_{p}=T+0.01 s. So, we will use equation (7) with this period and find g:

g=\frac{4 \pi^{2}L}{(T+0.01 s)^{2}} (10)

g=\frac{4 \pi^{2}(11.818 m)}{(6.9 s+0.01 s)^{2}} (11)

g=9.77 m/s^{2} (12) This is the acceleration due gravity at the place, which is near the mean value of 9.8 m/s^{2}

5 0
3 years ago
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