Diameter of a wire on a computer chip is 0.000003m and thickness of a piece of paper is 0.0016 cm. Find the sum of difference of 1000 computer chips and thickness of 20 pieces of paper.
Answers
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Answer:
Total required sum of difference = 3.32 × 10^(-3) m
Step-by-step explanation:
Diameter of a wire on a computer chip = 0.000003 m
= 3 × 10^(-6) m
No. of computer chips = 1000
So, Diameter of 1000 chips = 3 × 10^(-6) × 1000 m
= 3 × 10^(-3) m
Thickness of a piece of paper = 0.0016 cm
1 meter = 100 cm
So, thickness of a piece of paper is = 0.0016 × 10^(-2) m
= 1.6 × 10^(-5) m
No. of pieces of paper = 20
⇒ Total thickness = 1.6 × 10^(-5) × 20
= 3.2 × 10^(-4) m
= 0.32 × 10^(-3) m
Total required sum of difference = 3 × 10^(-3) + 0.32 × 10^(-3) m
= 3.32 × 10^(-3) m