Simplify v(((2)(431*10^-3)(9.8))/((1.3)p(10.5*10^-3)^2))
The problem provided is a mathematical expression that needs to be simplified. It appears to involve physical quantities since it has variables potentially representing velocity (v), gravity (9.8 m/s^2 which is close to Earth's gravitational acceleration), density (p), among other things. The expression inside the radical is a fraction which involves multiplication and division of various numbers, some of which are written in scientific notation (e.g., 10^-3 which denotes a thousandth part of a number). The task is to perform the necessary mathematical operations to simplify this expression to its simplest form, possibly as part of a physics problem dealing with fluid dynamics or related topics given the mention of density (p) and velocity (v).
$v \left(\right. \frac{\left(\right. 2 \left.\right) \left(\right. 431 \cdot \left(10\right)^{- 3} \left.\right) \left(\right. 9.8 \left.\right)}{\left(\right. 1.3 \left.\right) p \left(\left(\right. 10.5 \cdot \left(10\right)^{- 3} \left.\right)\right)^{2}} \left.\right)$
Answer
Solution:
Step 1: Perform the initial multiplication.
Step 1.1: Multiply $2$ by $431$ to get $v \frac{862 \cdot 10^{-3} \cdot 9.8}{1.3 p (10.5 \cdot 10^{-3})^2}$.
Step 1.2: Multiply $862$ by $9.8$ to obtain $v \frac{8447.6 \cdot 10^{-3}}{1.3 p (10.5 \cdot 10^{-3})^2}$.
Step 2: Simplify the numerator.
Step 2.1: Adjust the decimal point in $8447.6$ three places to the left, which gives $v \frac{8.4476 \cdot 10^{0}}{1.3 p (10.5 \cdot 10^{-3})^2}$.
Step 2.2: Convert $8.4476 \cdot 10^{0}$ to its non-scientific form, resulting in $v \frac{8.4476}{1.3 p (10.5 \cdot 10^{-3})^2}$.
Step 3: Calculate the denominator by squaring $10.5 \cdot 10^{-3}$ and multiplying by $1.3$, which simplifies to $v \frac{8.4476}{0.00014332 p}$.
Step 4: Factor out common terms.
Step 4.1: Extract $8.4476$ from the numerator to get $v \frac{8.4476 (1)}{0.00014332 p}$.
Step 4.2: Factor out $0.00014332$ from the denominator, leading to $v \frac{8.4476 (1)}{0.00014332 (p)}$.
Step 5: Separate the fractions to form $v \left( \frac{8.4476}{0.00014332} \cdot \frac{1}{p} \right)$.
Step 6: Divide $8.4476$ by $0.00014332$ to simplify the expression to $v \left( \frac{1}{p} \right)$.
Step 7: Apply the commutative property of multiplication to rewrite as $v \frac{1}{p}$.
Step 8: Combine the variables and fractions.
Step 8.1: Merge $v$ and $\frac{1}{p}$ to form $v \frac{}{p}$.
Step 8.2: Finalize the combination to get $\frac{v \cdot }{p}$.
Step 9: Reorder the terms to present the simplified expression as $\frac{v}{p}$.
Knowledge Notes:
Multiplication of Scientific Notation: When multiplying numbers in scientific notation, you multiply the base numbers and add the exponents of the powers of ten. For example, $a \cdot 10^m \times b \cdot 10^n = (a \times b) \cdot 10^{m+n}$.
Simplifying Numerators and Denominators: To simplify a fraction, you can divide both the numerator and the denominator by the same number (factoring) or perform arithmetic operations to reduce the fraction to its simplest form.
Squaring Numbers in Scientific Notation: When you square a number in scientific notation, you square the base number and multiply the exponent by two. For example, $(c \cdot 10^p)^2 = c^2 \cdot 10^{2p}$.
Commutative Property of Multiplication: This property states that the order in which two numbers are multiplied does not affect the product. For example, $a \times b = b \times a$.
Separating Fractions: When you have a fraction with multiple terms in the numerator or denominator, you can separate them into individual fractions that can be simplified independently, as long as you do not change the overall value of the expression.
Division of Decimals: When dividing decimal numbers, you can move the decimal point in both the numerator and the denominator to make the division easier, as long as you move them by the same number of places.
Final Expression: The goal of simplification is to rewrite the expression in the simplest form possible, which often involves reducing fractions, factoring out common terms, and applying arithmetic properties.
