$\sqrt[3]{\frac{\left(9 \times 10^4\right)^2-10^8+1.2 \times 10^{11}}{10^{-1}-9.8 \times 10^{-2}}}$
Answer (2)
Simplify the numerator: ( 9 × 1 0 4 ) 2 − 1 0 8 + 1.2 × 1 0 11 = 1.28 × 1 0 11 .
Simplify the denominator: 1 0 − 1 − 9.8 × 1 0 − 2 = 0.002 .
Divide the numerator by the denominator: 0.002 1.28 × 1 0 11 = 64 × 1 0 12 .
Take the cube root: 3 64 × 1 0 12 = 40000 .
The final answer is 40000 .
Explanation
Initial Analysis We are asked to evaluate the expression 3 1 0 − 1 − 9.8 × 1 0 − 2 ( 9 × 1 0 4 ) 2 − 1 0 8 + 1.2 × 1 0 11 . Let's break this down step by step. First, we'll simplify the numerator and the denominator separately.
Simplifying the Numerator Let's simplify the numerator:
( 9 × 1 0 4 ) 2 = 81 × 1 0 8 = 8.1 × 1 0 9 1 0 8 = 0.1 × 1 0 9 1.2 × 1 0 11 = 120 × 1 0 9 So, the numerator becomes: 8.1 × 1 0 9 − 0.1 × 1 0 9 + 120 × 1 0 9 = ( 8.1 − 0.1 + 120 ) × 1 0 9 = 128 × 1 0 9 = 1.28 × 1 0 11
Simplifying the Denominator Now, let's simplify the denominator:
1 0 − 1 = 0.1 9.8 × 1 0 − 2 = 0.098 So, the denominator becomes: 0.1 − 0.098 = 0.002 = 2 × 1 0 − 3
Dividing Numerator by Denominator Now, we divide the simplified numerator by the simplified denominator: 2 × 1 0 − 3 1.28 × 1 0 11 = 2 1.28 × 1 0 − 3 1 0 11 = 0.64 × 1 0 11 + 3 = 0.64 × 1 0 14 = 64 × 1 0 12
Taking the Cube Root Finally, we take the cube root of the result: 3 64 × 1 0 12 = 3 64 × 3 1 0 12 = 4 × 1 0 12/3 = 4 × 1 0 4 = 40000
Final Answer Therefore, the value of the expression is 40000.
Examples
This type of calculation can be useful in physics when dealing with very large or very small numbers, such as calculating the scale of astronomical distances or the size of subatomic particles. For example, if you were calculating the volume of a cube with sides on the order of 1 0 4 meters and then comparing it to a much larger volume on the order of 1 0 11 cubic meters, you might encounter similar calculations. Understanding how to manipulate scientific notation and cube roots is essential in these scenarios.
The expression simplifies to 40000 by first calculating the numerator and denominator separately, then dividing them and finally taking the cube root of the result.
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