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Chapter 0: Problem 2

\(\left(2.5 \times 10^{-7} \times 3.7 \times 10^{-6}\right)+4.2 \times 10^2\) A. \(1.3 \times 10^{-11}\) B. \(5.1 \times 10^{-10}\) C. \(4.2 \times 10^2\) D. \(1.3 \times 10^{13}\)

Short Answer

Expert verified
C. \(4.2 \times 10^2\)

Step by step solution

01

- Multiply the scientific notations

First, multiply the numbers in scientific notation: \((2.5 \times 10^{-7}) \times (3.7 \times 10^{-6})\). Using the properties of exponents, this becomes: \(2.5 \times 3.7 \times 10^{-7-6} = 9.25 \times 10^{-13}\)
02

- Add the product to the constant

Next, add the product obtained from Step 1 to the constant: \(9.25 \times 10^{-13} + 4.2 \times 10^2\). Because the powers of ten are so different, \(10^{-13}\) is negligible compared to \(10^2\). Thus, \(9.25 \times 10^{-13} + 4.2 \times 10^2 \approx 4.2 \times 10^2\).
03

- Choose the correct answer

Finally, compare the result with the provided options: Option C \((4.2 \times 10^2)\) matches our result.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

scientific notation
Scientific notation is a convenient way to express very large or very small numbers. This makes it easier to handle calculations. To express a number in scientific notation, you write it as a product of a number between 1 and 10 and a power of ten. For example, the number 2500 can be written as \(2.5 \times 10^3\).

Scientific notation is especially useful in scientific fields like physics and chemistry, where you often encounter very large or small values.
  • For large numbers, the exponent is positive: \(3.4 \times 10^8\).
  • For small numbers, the exponent is negative: \(5.7 \times 10^{-5}\).
In the exercise, we deal with \(2.5 \times 10^{-7}\) and \(3.7 \times 10^{-6}\), both of which are quite small numbers and thus have negative exponents.
properties of exponents
Understanding the properties of exponents is key to solving problems involving scientific notation. Some core properties are:
  • When multiplying numbers with the same base, you add the exponents: \(a^m \cdot a^n = a^{m+n}\).
  • When dividing numbers with the same base, you subtract the exponents: \(a^m / a^n = a^{m-n}\).
  • When raising a power to another power, you multiply the exponents: \((a^m)^n = a^{m \cdot n}\).
In Step 1 of the exercise, we used the property of adding exponents when multiplying: \((2.5 \times 10^{-7}) \times (3.7 \times 10^{-6})\) becomes \(2.5 \times 3.7 \times 10^{-7-6} = 9.25 \times 10^{-13}\). Knowing these rules simplifies handling these expressions.
approximation in mathematics
Approximations are often necessary in mathematics, especially when comparing numbers that vastly differ in magnitude. In the exercise, after obtaining \(9.25 \times 10^{-13}\), we added it to \(4.2 \times 10^2\). Given that \(10^{-13}\) is extremely small compared to \(10^2\), it becomes negligible. This is why:
  • \(9.25 \times 10^{-13} + 4.2 \times 10^2 \approx 4.2 \times 10^2\)
This approximation technique helps to simplify calculations in many mathematical and scientific contexts. Always consider the orders of magnitude when approximating to decide if a term can be ignored.

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Most popular questions from this chapter

\(\left[\left(1.1 \times 10^{-4}\right)+\left(8.9 \times 10^{-5}\right)\right]^{1 / 2}\) A. \(1.1 \times 10^{-2}\) B. \(1.4 \times 10^{-2}\) C. \(1.8 \times 10^{-2}\) D. \(2.0 \times 10^{-2}\)

\(\frac{5.4 \times 7.1 \times 3.2}{4.6^2}\) A. \(2.2\) B. \(3.8\) C. \(5.8\) D. \(7.9\)

The force of gravity on an any object due to earth is given by the equation \(F=G\left(m_\rho M / r^2\right)\) where \(G\) is the gravitational constant, \(M\) is the mass of the earth, \(m_0\) is the mass of the object and \(r\) is the distance between the center of mass of the earth and the center of mass of the object. If a rocket weighs \(3.6 \times 10^6 \mathrm{~N}\) at the surface of the earth what is the force on the rocket due to gravity when the rocket has reached an altitude of \(1.2 \times 10^4 \mathrm{~km}\) ? \(\left(G=6.67 \times 10^{-11} \mathrm{Nm}^2 / \mathrm{kg}^2\right.\), radius of the earth \(=6370 \mathrm{~km}\), mass of the earth \(=5.98 \times 10^{24} \mathrm{~kg}\) ) A. \(1.2 \times 10^5 \mathrm{~N}\) B. \(4.3 \times 10^5 \mathrm{~N}\) C. \(4.8 \times 10^6 \mathrm{~N}\) D. \(9.6 \times 10^6 \mathrm{~N}\)

The coefficient of surface tension is given by the equation \(\gamma=(F-m g) /(2 L)\), where \(F\) is the net force necessary to pull a submerged wire of weight \(\mathrm{mg}\) and length \(L\) through the surface of the fluid in question. The force required to remove a submerged wire from water was measured and recorded. If an equal force is required to remove a separate submerged wire with the same mass but twice the length from fluid \(x\), what is the coefficient of surface tension for fluid \(x .\left(\gamma_{\text {water }}=0.073 \mathrm{mN} / \mathrm{m}\right)\) A. \(0.018 \mathrm{mN} / \mathrm{m}\) B. \(0.037 \mathrm{mN} / \mathrm{m}\) C. \(0.073 \mathrm{mN} / \mathrm{m}\) D. \(0.146 \mathrm{mN} / \mathrm{m}\)

The kinetic energy \(E\) of an object is given by \(E=1 / 2 m v^2\) where \(m\) is the object's mass and \(v\) is the velocity of the object. If the velocity of an object decreases by a factor of 2 what will happen its kinetic energy? A. Kinetic energy will increase by a factor of 2 . B. Kinetic energy will increase by a factor of 4 . C. Kinetic energy will decrease by a factor of 2 . D. Kinetic energy will decrease by a factor of 4 .
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