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alina1380 [7]
3 years ago
15

Significant Figures For 7.06 × 10^5 ÷ 5.3 × 10^-2

Chemistry
1 answer:
Kay [80]3 years ago
4 0

Answer:

\boxed{\text{two}}

Explanation:

In multiplication  and division problems, your answer can have no more significant figures than the number with the fewest significant figures.

\dfrac{7.06 \times 10^{5}}{5.3 \times 10^{-2}}= 1.332 075 472 \times 10^{7} (by my calculator)  

There are three significant figures in 7.06 and two in 2.3.

You must round to \boxed{\textbf{two}} significant figures and report the answer as 1.3 × 1.0⁷.

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F = (Kq1q2)/(r^2) where K = (9 × (10^9) Nm(C^-2))

But q1 is the charge on a proton = (1.6 × (10^-19)) C

q2 is charge on an electron = -(1.6 × (10^-19)) C

r = (5.50 × (10^-10))mm = (5.50 × (10^-13))m

Computing all that, F = 0.0007616529 N = (7.62 × 10^-4) N

But the force of attraction is converted to that required for motion when they're released.

F = ma.

For proton, m = (1.67 × 10^-27) kg

a = F/m = 0.000762/(1.67 × 10^-27) = (4.56 × 10^23) m/s2

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a = F/m = 0.000762/(9.11 × 10^-31) = (8.36 × 10^26) m/s2

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7 0
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