/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 The doozy is defined as the unit... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 18

The doozy is defined as the unit of force required to accelerate a unit of mass, called the cuz, with the gravitational acceleration on the surface of the moon, which is one-sixth of the normal gravitational acceleration on earth. (a) What is the conversion factor that would be used to convert a force from the natural unit to the derived unit in this system? (Give both its numerical value and its units.) (b) What is the weight in doozies of a 3 -cuz object on the moon? What does the same object weigh in Lizard Lick, North Carolina?

Short Answer

Expert verified
The conversion factor is \(1.62 N/doozy\). A 3-cuz object would weigh 3 doozies or \(4.86 N\) on the moon, and 18.17 doozies or \(29.43 N\) in Lizard Lick, North Carolina.

Step by step solution

01

Understand the definition of doozy

Our new unit of force, the doozy, is said to be the force required to accelerate one 'cuz' with the gravity of the moon. This means 1 doozy = 1 cuz * moon gravitational acceleration. The latter is commonly given as \(1.62 m/s^2\). So 1 doozy equals the force to accelerate 1 cuz by \(1.62 m/s^2\)
02

Find the conversion factor to newtons

We can think of a force of 1 doozy as a force of 1 cuz * \(1.62 m/s^2\). To convert this to newtons, we can think of the 'cuz' as a unit of mass and set it equal to one kilogram. This gives us the conversion factor 1 doozy = \(1.62 N\).
03

Calculate the 3-cuz object's weight on the moon

Weight is defined as mass times gravitational acceleration. Therefore, the weight of the object on the moon in doozies is 3 cuz * 1 = 3 doozies. To convert this to newtons, we use our conversion factor 1 doozy = \(1.62 N\), which gives us a weight on the moon of \(3 * 1.62 N = 4.86 N\).
04

Calculate the object's weight in Lizard Lick, North Carolina

On earth, the gravitational acceleration is \(9.81 m/s^2\). Therefore, the weight of the object in Lizard Lick in newtons is 3 kg * \(9.81 m/s^2\) = \(29.43 N\). To convert this to doozies, we divide by our conversion factor to find that the weight in North Carolina equals \(29.43 N / 1.62(doosies/N) = 18.17 doozies\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Gravitational Acceleration
Gravitational acceleration is a fundamental concept in physics. It describes the rate at which an object accelerates due to gravity alone. This acceleration depends on the celestial body you are on. For example, Earth's gravitational acceleration is a constant value of approximately \(9.81 \, m/s^2\). This means that any object in freefall near the Earth's surface will accelerate downwards at this rate.

However, when you are on the moon, gravitational acceleration reduces significantly. It is only about \(1.62 \, m/s^2\). This is roughly one-sixth the strength of Earth's gravity. This difference drastically affects how objects move and how much they "weigh" in terms of gravitational force. In physics problems, knowing the gravitational acceleration on different celestial bodies helps in calculating factors like weight and force.
  • Gravitational acceleration determines how much an object will fall or weigh.
  • On the moon, you experience one-sixth of Earth's gravitational pull.
  • This change is essential in calculations for physics exercises.
Force Measurement
Force is measured based on how much it can accelerate an object. The SI unit of force is the newton (N), defined as the force needed to accelerate one kilogram of mass by one meter per second squared. However, in certain contexts, alternative units are used鈥攍ike the fictional 'doozy' in this exercise.

The doozy is the force unit required to accelerate an object with mass in "cuz" at the moon's gravitational pull. Recognizing that force depends on both mass and acceleration is crucial for understanding unit conversion. For example, 1 doozy can be translated into classical units as \(1.62\, N\), since on the moon, the gravitational acceleration is \(1.62 \, m/s^2\).
  • Force involves both mass and acceleration.
  • The conversion between force units helps in solving problems.
  • Knowing alternative units like doozy expands physics problem-solving strategies.
Weight Calculation
Weight is essentially a measure of the force exerted on an object due to gravitational pull. To calculate weight, we multiply the object's mass by the gravitational acceleration of the location we are considering. Hence, the formula for weight is:\[\text{Weight} = \text{mass} \times \text{gravitational acceleration}\]
On the moon, an object's weight is much lesser than on Earth due to the reduced gravity. In our example, a 3-cuz mass weighs 3 doozies on the moon. In conversion terms, 3 doozies equals about \(4.86 \, N\) on the moon because \(1 \, \text{doozy} = 1.62 \, N\). When the same object is considered on Earth, the weight is greater, calculated at \(29.43 \, N\), and accordingly \(18.17\) doozies. Understanding this conversion is important for interplanetary comparisons.
  • Weight changes with gravitational acceleration.
  • On the moon, objects weigh less than on Earth.
  • Conversion between weight units helps understand these differences.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The following table is a summary of data taken on the growth of yeast cells in a bioreactor: $$\begin{array}{|c|c|}\hline \text { Time, } t(\mathrm{h}) & \text { Yeast Concentration, } X(\mathrm{g} / \mathrm{L}) \\\\\hline 0 & 0.010 \\\\\hline 4 & 0.048 \\\\\hline 8 & 0.152 \\\\\hline 12 & 0.733 \\\\\hline 16 & 2.457 \\ \hline\end{array}$$ The data can be fit with the function \(X=X_{0} \exp (\mu t)\) where \(X\) is the concentration of cells at any time \(t, X_{0}\) is the starting concentration of cells, and \(\mu\) is the specific growth rate. (a) Based on the data in the table, what are the units of the specific growth rate? (b) Give two ways to plot the data so as to obtain a straight line. Each of your responses should be of the form "plot ______ Versus ________ on _______ axes." (c) Plot the data in one of the ways suggested in Part (b) and determine \(\mu\) from the plot. (d) How much time is required for the yeast population to double?

State what you would plot to get a straight line if experimental ( \(x, y\) ) data are to be correlated by the following relations, and what the slopes and intercepts would be in terms of the relation parameters. If you could equally well use two different kinds of plots (e.g., rectangular or semilog), state what you would plot in each case. [The solution to Part (a) is given as an example.] (a) \(y^{2}=a e^{-b / x}\). (b) \(y^{2}=m x^{3}-n\) (c) \(1 / \ln (y-3)=(1+a \sqrt{x}) / b\) (d) \((y+1)^{2}=\left[a(x-3)^{3}\right]^{-1}\) (e) \(y=\exp (a \sqrt{x}+b)\) (f) \(x y=10^{\left[a\left(x^{2}+y^{2}\right)+b\right]}\) (g) \(y=[a x+b / x]^{-1}\)

You arrive at your lab at 8 A.M. and add an indeterminate quantity of bacterial cells to a flask. At 11 A.M. you measure the number of cells using a spectrophotometer (the absorbance of light is directly related to the number of cells) and determine from a previous calibration that the flask contains 3850 cells, and at 5 P.M. the cell count has reached 36,530. (a) Fit each of the following formulas to the two given data points (that is, determine the values of the two constants in each formula): linear growth, \(C=C_{0}+k t ;\) exponential growth, \(C=C_{0} e^{k t} ;\) power-law growth, \(C=k t^{b} .\) In these expressions, \(C_{0}\) is the initial cell concentration and \(k\) and \(b\) are constants. (b) Select the most reasonable of the three formulas and justify your selection. (c) Estimate the initial number of cells present at 8 A.M. \((t=0)\). State any assumptions you make. (d) The culture needs to be split into two equal parts once the number of cells reaches 2 million. Estimate the time at which you would have to come back to perform this task. State any assumptions you make. If this is a routine operation that you must perform often, what does your result suggest about the scheduling of the experiment?

A right circular cone of base radius \(R\), height \(H\), and known density \(\rho_{s}\) floats base down in a liquid of unknown density \(\rho_{f}\). A height \(h\) of the cone is above the liquid surface. Derive a formula for \(\rho_{f}\) in terms of \(\rho_{s}, R,\) and \(h / H,\) simplifying it algebraically to the greatest possible extent. [Recall Archimedes' principle, stated in the preceding problem, and note that the volume of a cone equals (base area)(height)/3.]

The following \((x, y)\) data are recorded: $$\begin{array}{|c|c|c|c|}\hline x & 0.5 & 1.4 & 84 \\\\\hline y & 2.20 & 4.30 & 6.15 \\\\\hline \end{array}$$ (a) Plot the data on logarithmic axes. (b) Determine the coefficients of a power law expression \(y=a x^{b}\) using the method of least squares. (Remember what you are really plotting \(-\) there is no way to avoid taking logarithms of the data point coordinates in this case.) (c) Draw your calculated line on the same plot as the data.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.