Question

question 5 (2 points) give the number of significant figures in each number, and the uncertainty 0.00120 mm has 1 significant figures and an uncertainty of 2 mm 2.200 x 10^(-10) μm has 3 significant figures and an uncertainty of 4 a. 3 b. 2 c. 4 d. 5 e. 6 f. 0.001 g. 0.001 x 10^(-10) h. 0.0001 i. 0.00001 j. 0.000001

Explanation:

Step1: Determine significant figures for 0.00120 mm

Leading zeros are not significant, non - zero digits and trailing zeros after a non - zero digit are significant. So 0.00120 has 3 significant figures.

Step2: Determine uncertainty for 0.00120 mm

The uncertainty is in the last significant digit's place. For 0.00120, the last significant digit is in the ten - thousandths place, so the uncertainty is 0.00001 mm.

Step3: Determine significant figures for 2.200×10⁻¹⁰ μm

All digits in the coefficient (2.200) are significant, so there are 4 significant figures.

Step4: Determine uncertainty for 2.200×10⁻¹⁰ μm

The uncertainty is in the last significant digit's place of the coefficient. For 2.200, the last significant digit is in the thousandths place of the coefficient, so the uncertainty is 0.001×10⁻¹⁰ μm.

Answer:

1. A. 3
2. I. 0.00001
3. C. 4
4. G. 0.001×10⁻¹⁰