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Kobotan [32]
3 years ago
5

Multiply these measurements using significant digit rules: 1.27kg x 9.8m/s

Physics
1 answer:
Leviafan [203]3 years ago
6 0

Answer:

12.45kgm/s

Explanation:

1.27kg x 9.8m/s = 12.446kgm/s

The sum of decimal point is 3 and the unit kg multiplies m divided by s

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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

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