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barxatty [35]
3 years ago
6

10.Which of the following is true about atomic and ionic radii? A. In each period, the element with the lowest atomic number has

the smallest ionic radius. B. Ionic radius is greater than atomic radius in an anion. C. The number of protons does not change so there is no difference between radii. D. The element with the lowest atomic number has the largest ionic radius in each group.
Physics
1 answer:
Sav [38]3 years ago
3 0

Answer:

Ionic radius is greater than atomic radius in an anion.

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Which property of substances help to make a thermometer?<br> I NEED HELP.
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What is the magnitude of the force that is exerted on a 20 kg mass to give it an acceleration of 10.0
ANEK [815]

Answer:Mass of the body = 20 kg.

Final Velocity = 5.8 m/s.

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Time = 3 seconds.  

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Acceleration = (v - u)/ t

= (5.8 - 0)/ 3

= 1.6 m /s².

Now, Using the Formula,  

Force = mass × acceleration

= 20 × 1.6

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8 0
2 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

3 0
3 years ago
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