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Chapter 1: Problem 5

Given the quantities \(a=9.7 \mathrm{m}, b=4.2 \mathrm{s}, c=69 \mathrm{m} / \mathrm{s},\) what is the lue of the quantity \(d=a^{3} /\left(c b^{2}\right) ?\)

Short Answer

Expert verified
The value of \( d \) is approximately 0.7504.

Step by step solution

01

Understand the Formula

The given formula is \( d = \frac{a^3}{c \cdot b^2} \). This means we need to cube \( a \) and square \( b \), before dividing \( a^3 \) by the product of \( c \) and \( b^2 \).
02

Cube the Value of a

Calculate \( a^3 \). Given the value of \( a = 9.7 \), cube it as follows: \( a^3 = 9.7 \times 9.7 \times 9.7 = 912.673 \).
03

Square the Value of b

Calculate \( b^2 \). Given the value of \( b = 4.2 \), square it as follows: \( b^2 = 4.2 \times 4.2 = 17.64 \).
04

Calculate the Product of c and b^2

Find \( c \cdot b^2 \) by multiplying the given \( c = 69 \) with the previously calculated \( b^2 = 17.64 \). Thus, \( c \cdot b^2 = 69 \times 17.64 = 1216.16 \).
05

Calculate the Value of d

Divide the result from Step 2 by the result from Step 4: \( d = \frac{912.673}{1216.16} \approx 0.7504 \).

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Key Concepts

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

Mathematical Calculations in Physics
Mathematical Calculations in Physics involve working with various mathematical operations to solve problems related to physical quantities. In this context, we are dealing with functions such as cubing, squaring, and division. Familiarity with these operations allows students to transition between theoretical concepts and real-world applications.

  • **Cubing**: This involves multiplying a number by itself three times. In our exercise, the quantity \(a\) is cubed, resulting in \(a^3\). This illustrates how quantities can grow quickly as we apply higher powers.
  • **Squaring**: This is the process of multiplying the number by itself. When we square the time \(b\) in our problem, it helps demonstrate changes over the time squared scale, common in acceleration and energy formulas.
  • **Division**: After computing these values, the final step is division, which combines the calculations into an interpretable physical quantity, often representing a rate or density in physics.

These basic mathematical techniques are foundational in the analysis and solution of physics problems, reinforcing both arithmetic skills and conceptual understanding.
Step-by-Step Calculations
Breaking down the problem into step-by-step calculations helps demystify the process and provides a clear path from input values to the final solution. In physics, this methodology is invaluable for systematically solving complex equations.

  • **First Step**: Review and understand the formula. Knowing what mathematical transformation is required for each variable ensures that each step is calculated correctly.
  • **Cubing of \(a\)**: Begin by isolating and solving part by part. Calculating \(a^3\) gives a standalone numeric result simplifying further calculations.
  • **Squaring of \(b\)**: Next, calculate \(b^2\). Each solved component provides a necessary check-point, ensuring accuracy before proceeding.
  • **Final Calculation**: By multiplying \(c\) by \(b^2\), you create the divisor for the last division. Division in the final step combines all separate calculations into a single result \(d\), making the task of double-checking against errors manageable.

Following these steps ensures that each component of the calculation is properly addressed, reinforcing the learning of precise mathematical processes.
Physics Formulas
Physics relies heavily on formulas to predict and understand the behavior of physical systems. These mathematical expressions are pathways connecting observed phenomena with theory. In our exercise, the formula \(d = \frac{a^3}{c \cdot b^2}\) serves to amalgamate several measurable quantities into one coherent expression.

  • **Structure of Formulas**: Formulas often indicate mathematical operations like addition, multiplication, and division. Mastery of these operations is essential in understanding and deriving new relations in physics.
  • **Units and Dimensions**: While numerical calculations result in a raw number, in physics, interpreting this result depends greatly on the units involved (meters, seconds, etc.). Ensuring dimensional consistency assures that the formula translates logically to real-world scenarios.
  • **Conceptual Interpretation**: Beyond computation, it is critical to understand what the formula is expressing. For instance, by cubing \(a\) and manipulating \(b\) and \(c\), we refine a quantity that often aligns with movement, force, or energy in a broader physical context.

Grasping these elements of physics formulas enables students not only to solve given problems but also to comprehend and innovate within the broader discipline of physics.

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Most popular questions from this chapter

The volume of liquid flowing per second is called the volume flow rate \(O\) and has the dimensions of \([\mathrm{L}]^{3} /[\mathrm{T}]\) . The flow rate of a liquid through a hypodermic needle during an injection can be esti- mated with the following equation: $$Q=\frac{\pi R^{n}\left(P_{2}-P_{1}\right)}{8 \eta L}$$ The length and radius of the needle are \(L\) and \(R,\) respectively, both of which have the dimension \([\mathrm{L}]\) . The pressures at opposite ends of the nee- dle are \(P_{2}\) and \(P_{1},\) both of which have the dimensions of \([\mathrm{M}] /\left\\{[\mathrm{L}][\mathrm{T}]^{2}\right\\}\) The symbol \(\eta\) represents the viscosity of the liquid and has the dimen- sions of \([\mathrm{M}] /\\{[\mathrm{L}] \mathrm{T}]\) . The symbol \(\pi\) stands for pi and, like the number 8 and the exponent \(n\) , has no dimensions. Using dimensional analysis, determine the value of \(n\) in the expression for \(Q .\)

A force vector \(\overrightarrow{\mathbf{F}}_{1}\) points due east and has a magnitude of 200 newtons. A second force \(\overline{F}_{2}\) is added to \(\overline{F}_{1}\) . The resultant of the two vectors has a magnitude of 400 newtons and points along the east/west line. Find the magnitude and direction of \(\overline{\mathbf{F}}_{2}\) . Note that there are two answers.

Three deer, \(A, B,\) and \(C,\) are grazing in a field. Deer \(B\) is located 62 \(\mathrm{m}\) from deer \(\mathrm{A}\) at an angle of \(51^{\circ}\) north of west. Deer \(\mathrm{C}\) is located \(77^{\circ}\) north of east relative to deer \(\mathrm{A}\) . The distance between deer \(\mathrm{B}\) and \(\mathrm{C}\) is 95 \(\mathrm{m}\) . What is the distance between deer \(\mathrm{A}\) and \(\mathrm{C} ?\) (Hint: Consider the law of cosines given in Appendix E.)

A circus performer begins his act by walking out along a nearly horizontal high wire. He slips and falls to the safety net, 25.0 \(\mathrm{ft}\) below. The magnitude of his displacement from the beginning of the walk to the net is 26.7 \(\mathrm{ft}\) . (a) How far out along the high wire did he walk? (b) Find the angle that his displacement vector makes below the horizontal.

A pilot flies her route in two straight-line segments. The dis- placement vector \(\vec{A}\) for the first segment has a magnitude of 244 \(\mathrm{km}\) and a direction \(30.0^{\circ}\) north of east. The displacement vector \(\overrightarrow{\mathbf{B}}\) for the second segment has a manitude of 175 \(\mathrm{km}\) and a direction due west. The resultant displacement vector is \(\overrightarrow{\mathrm{R}}=\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}\) and makes an angle \(\theta\) with the direction due east. Using the component method, find the magnitude of \(\overrightarrow{\mathbf{R}}\) and the directional angle \(\theta .\)
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