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mars1129 [50]
3 years ago
12

Examine the scenario. Object A has 5 protons and 5 electrons. Object B has 5 protons and 7 electrons. Which option most accurate

ly describes the net charge on the two objects?
Object A has a net charge of -5 because it has 5 electrons. Object B has a net charge of –7 because it has 7 electrons.
Object A has a net charge of 0 because the positive and negative charges are balanced. Object B has a net charge of +2 because there is an imbalance of charged particles.
Object A has a net charge of +5 because it has 5 protons. Object B has a net charge of +5 because it has 5 protons.
Object A has a net charge of 0 because the positive and negative charges are balanced. Object B has a net charge of –2 because there is an imbalance of charged particles.
Physics
1 answer:
Travka [436]3 years ago
4 0
<span>Last choice on the list:
Object A has a net charge of 0 because the positive and negative
charges are balanced.
Object B has a net charge of –2 because there is an imbalance of
charged particles (2 more negative electrons than positive protons).</span>
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A bus is moving and has 500000 joules of kinetic energy. The brakes are applied and the bus stops. How much work is needed to st
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For the work-energy theorem, the work needed to stop the bus is equal to its variation of kinetic energy:
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3 years ago
Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at
andrew-mc [135]

Answer:

1. the electric potential energy of the electron when it is  at the midpoint is - 2.9 x 10^{-17} J

2. the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge is - 5.04 x  10^{-17} J

Explanation:

given information:

q_{1} =  3 nC = 3 x 10^{-9} C

q_{2} =  2 nC = 2 x 10^{-9} C

r = 50 cm = 0.5 m

the electric potential energy of the electron when it is  at the midpoint

potential energy of the charge, F

F = k \frac{q_{e}q}{r}

where

k = constant (8.99 x 10^{9} Nm^{2} /C^{2})

electron charge, q_{e} = - 1.6 x 10^{-19} C

since it is measured at the midpoint,

r = \frac{0.5}{2}

  = 0.25 m

thus,

F = F_{1}+ F_{2}

  = k\frac{q_{e} q_{1} }{r} + k\frac{q_{e} q_{2} }{r}

  = \frac{kq_{e} }{r} (q_{1} +q_{2})

  = (8.99 x 10^{9})( - 1.6 x 10^{-19} )(3 x 10^{-9} +2 x 10^{-9})/0.25

  = - 2.9 x 10^{-17} J

the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge

r_{1} = 10 cm = 0.1 m

r_{2} = 0.5 - 0.1 = 0.4 m

F = k\frac{q_{e} q_{1} }{r} + k\frac{q_{e} q_{2} }{r}

  = kq_{e}(\frac{q_{1} }{r_{1} }+\frac{q_{2} }{r_{2} })

  = (8.99 x 10^{9})( - 1.6 x 10^{-19} )(3 x 10^{-9} /0.1+2 x 10^{-9}/0.4)

  = - 5.04 x  10^{-17} J

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2 years ago
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Answer:

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We also need to know that the radius of curvature is twice the focal length of a curved mirror.

Therefore;

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