Tag Archives: Science

Probability & Statistics for Scientists & Engineers Ninth Edition by Walpole, Myers, Myers, and Ye

Chapter 1: Introduction to Statistics and Data Analysis

Chapter 2: Probability

Chapter 3: Random Variables and Probability Distributions

Chapter 4: Mathematical Expectation

Chapter 5: Some Discrete Probability Distribution

Chapter 6: Some Continuous Probability Distribution

Chapter 7: Functions of Random Variables

Chapter 8: Fundamental Sampling Distributions and Data Descriptions

Chapter 9: One- and Two-Sample Estimation Problems

Chapter 10: One- and Two-Sample Tests of Hypotheses

Chapter 11: Simple Linear Regression and Correlation

Chapter 12: Multiple Linear Regression and Certain Nonlinear Regression Models

Chapter 13: One-Factor Experiments: General

Chapter 14: Factorial Experiments (Two or More Factors)

Chapter 15: 2^k Factorial Experiments and Fractions

Chapter 16: Nonparametric Statistics

Chapter 17: Statistical Quality Control

Chapter 18: Bayesian Statistics

Grantham PHY220 Week 2 Assignment Problem 8

If a car is traveling at 50 m/s and then stops over 300 meters (while sliding), what is the coefficient of kinetic friction between the tires of the car and the road?

SOLUTION:

Draw the free-body diagram of the car

Consider the vertical direction

$\sum F_y=ma_y$

$F_N-mg=0$

$F_N=mg$

Consider the motion in the horizontal direction

Solve for the acceleration of the car.

$v^2=\left(v_0\right)^2+2a_x\Delta x$

$a_x=\frac{v^2-\left(v_0\right)^2}{2\Delta x}=\frac{0-50^2}{2\left(300\right)}=-4.17\:m/s^2$

Solve for the coefficient of kinetic friction

$\sum F_x=ma_x$

$-F_{fr}=ma_x$

$-\mu _kF_N=ma_x$

$\mu _k=\frac{ma_x}{-F_N}=\frac{m\:\left(-4.17\right)}{-m\left(9.80\right)}=\frac{4.17}{9.80}$

$\mu _k=0.43$

Problem 6-1: Odometer reading based on the number of wheel revolutions

Semi-trailer trucks have an odometer on one hub of a trailer wheel. The hub is weighted so that it does not rotate, but it contains gears to count the number of wheel revolutions—it then calculates the distance traveled. If the wheel has a 1.15 m diameter and goes through 200,000 rotations, how many kilometers should the odometer read?

Solution:

The formula for the total distance traveled is

\Delta s=\Delta \theta \times r

Therefore, the total distance traveled is

\begin{align*} \Delta s & =\left(200000\:\text{rotations}\:\times \frac{2\pi \:\text{radian}}{1\:\text{rotation}}\right)\left(\frac{1.15\:\text{m}}{2}\right) \\ \Delta s & =722566.3103\:\text{m} \\ \Delta s & =722.6\:\text{km} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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College Physics by Openstax Chapter 5 Problem 1

A physics major is cooking breakfast when he notices that the frictional force between his steel spatula and his Teflon frying pan is only 0.200 N. Knowing the coefficient of kinetic friction between the two materials, he quickly calculates the normal force. What is it?

Solution:

The formula for friction is

f=\mu _{k\:}N

When we solve for the normal force, $N$ , in terms of the other variables, we have

N=\frac{f}{\mu _k}

The coefficient of kinetic friction is 0.04. Therefore, the normal force is

\begin{align*} N & =\frac{f}{\mu _k} \\ N & =\frac{0.200\:\text{newton}}{0.04} \\ N & =5.00\:\text{newton} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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College Physics by Openstax Chapter 4 Problem 1

A 63.0-kg sprinter starts a race with an acceleration of 4.20 m/s². What is the net external force on him?

Solution:

So, we are given mass, $m = 63.0 \ \text{kg}$ , and acceleration, $a = 4.20 \ \text{m/s}^2$ .

The net force has a formula

\text{F}=\text{m}a

Substituting the given values, we have

\begin{align*} F & = \left( 63.0 \ \text{kg} \right)\left( 4.20 \ \text{m/s}^2 \right) \\ F & = 265 \ \text{kg}\cdot \text{m/s}^2 \\ F & = 265 \ \text{N} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

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College Physics by Openstax Chapter 2 Problem 29

Freight trains can produce only relatively small accelerations and decelerations.

(a) What is the final velocity of a freight train that accelerates at a rate of 0.0500 m/s² for 8.00 minutes, starting with an initial velocity of 4.00 m/s?

(b) If the train can slow down at a rate of 0.550 m/s², how long will it take to come to a stop from this velocity?

(c) How far will it travel in each case?

Solution:

Part A

We are given the the following: $a=0.0500 \ \text{m/s}^2$ ; $t=8.00 \ \text{mins}$ ; and $v_0=4.00 \ \text{m/s}$ .

The final velocity can be solved using the formula $v_f=v_0+at$ . We substitute the given values.

\begin{align*} v_f& = v_0+at \\ v_f & = 4.00\:\text{m/s}+\left(0.0500\:\text{m/s}^2\right)\left(8.00\:\text{min}\times \frac{60\:\sec }{1\:\min }\right) \\ v_f & = 28.0 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

Rearrange the equation we used in part (a) by solving in terms of $t$ , we have

\begin{align*} t & =\frac{{v_f}-v_0}{a} \\ t & = \frac{0\:\text{m/s}-28\:\text{m/s}}{-0.550\:\text{m/s}^2} \\ t & = 50.91\:\sec\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

The change in position for part (a), $\Delta x$ , or distance traveled is computed using the formula $\Delta x=v_0 t+\frac{1}{2} at^2$ .

\begin{align*} \Delta x & =v_0 t+\frac{1}{2} at^2 \\ \Delta x & =\left(4.0\:\text{m/s}\right)\left(480\:\text{s}\right)+\frac{1}{2}\left(0.0500\:\text{m/s}^2\right)\left(480\:\text{s}\right)^2 \\ \Delta x & = 7680\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

For the situation in part (b), the distance traveled is computed using the formula $\Delta x=\frac{v_f^2-v_0^2}{2 a}$ .

\begin{align*} \Delta x & =\frac{\left(0\:\text{m/s}\right)^2-\left(28.0\:\text{m/s}\right)^2}{2\left(-0.550\:\text{m/s}^2\right)} \\ \Delta x & =712.73\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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