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Physics for Scientists and Engineers 3rd Edition by Randall Knight, Chapter 1 Conceptual Question 2

How many significant figures does each of the following numbers have?

a) 310

b) 0.00310

c) 1.031

d) 3.10×10⁵

Solution:

Part a

Zeroes before the decimal point merely locate the decimal point and are not significant. Thus, 310 has 2 significant figures.

Part b

Trailing zeros after the decimal point are significant because they indicate increased precision. Therefore, 0.00310 has 3 significant figures.

Part c

Zeroes between nonzero digits are significant. Therefore, 1.031 has 4 significant figures.

Part d

Just like in part b, trailing zeros after the decimal point are significant because they indicate increased precision.Therefore, 3.10×10⁵ has 3 significant figures.

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Physics for Scientists and Engineers: A Strategic Approach Third Edition by Randall D. Knight

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Physics for Scientists and Engineers 3rd Edition by Randall Knight, Chapter 1 Conceptual Question 1

How many significant figures does each of the following numbers have?

a) 53.2

b) 0.53

c) 5.320

d) 0.0532

Solution:

Part a

All nonzero digits are significant, therefore, 53.2 has 3 significant figures.

Part b

The number 0.53 can be written in scientific notation as 5.3×10^-1. Therefore, 0.53 has 2 significant figures.

Part c

Trailing zeros are significant because they indicate increased precision. Therefore, 5.320 has 4 significant figures.

Part d

The leading zeros are not significant but just locate the decimal point. Therefore, 0.0532 has only 3 significant figures.

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Chapter 1 Solution Guides to Physics for Scientists & Engineers 3rd Edition by Randall Knight

Conceptual Questions

Exercises and Problems

Question 29

Question 30

Question 31

Question 32

Question 33

Question 34

Question 35

Question 36

Question 37

Question 38

Question 39

Question 40

Question 41

Question 42

Question 43

Question 44

Question 45

Question 46

Question 47

Question 48

Question 49

Question 50

Question 51

Question 52

Question 53

Question 54

Question 55

Question 56

Question 57

Mechanics of Materials: An Integrated Learning System 4th Edition by Timothy A. Philpot Complete Solution Manual

Chapter 1: Stress

Chapter 2: Strain

Chapter 3: Mechanical Properties of Materials

Chapter 4: Design Concepts

Chapter 5: Axial Deformation

Chapter 6: Torsion

Chapter 7: Equilibrium of Beams

Chapter 8: Bending

Chapter 9: Shear Stress in Beams

Chapter 10: Beam Deflections

Chapter 11: Statically Indeterminate Beams

Chapter 12: Stress Transformations

Chapter 13: Strain Transformations

Chapter 14: Thin-Walled Pressure Vessels

Chapter 15: Combined Loads

Chapter 16: Columns

Chapter 17: Energy Methods

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Mean and Variance | Introduction to Statistics and Data Analysis | Probability & Statistics for Engineers & Scientists | Walpole | Problem 1.10

For the data of Exercise 1.4 on page 13, compute both the mean and variance in “flexibility” for both company A and company B.

Solution:

For the company A:

We know, based on our answer in Exercise 1.4, that the sample mean for samples in Company A is $\displaystyle 7.950$ .

To compute for the sample variance, we shall use the formula

$\displaystyle s^2=\sum _{i=1}^n\frac{\left(x_i-\overline{x}\right)^2}{n-1}$

The formula states that we need to get the sum of $\displaystyle \left(x_i-\overline{x}\right)^2$ , so we can use a table to solve $\displaystyle \left(x_i-\overline{x}\right)^2$ for every sample.

Note: You can refer to the solution of Exercise 1.7 on how to use a table to solve for the variance.

The variance is

$\displaystyle \left(s^2\right)_A=\sum _{i=1}^{10}\:\frac{\left(x_i-7.950\right)^2}{10-1}$

$\displaystyle =1.2078$

The standard deviation is just the square root of the variance. That is

$\displaystyle s_A=\sqrt{1.2078}=1.099$

For Company B:

Using the same method employed for Company A, we can show that the variance and standard deviation for the samples in Company B are

$\displaystyle \left(s^2\right)_B=0.3249$ and $\displaystyle s_B=0.570$ .

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Arizona State University| PHY 121: Univ Physics I: Mechanics|Homework 1-1| Problem 2

Learning Goal:

To practice Tactics Box 1.2 Vector Subtraction.

Vector subtraction has some similarities to the subtraction of two scalar quantities. With numbers, subtraction is the same as the addition of a negative number. For example, 5−3 is the same as 5+(−3). Similarly, A⃗ −B⃗ =A⃗ +(−B⃗ ). We can use the rules for vector addition and the fact that −B⃗ is a vector opposite in direction to B⃗ to form rules for vector subtraction, as explained in this tactics box.

The figure shows two vectors. Vector A is on the left and is directed upwards to the right. Vector B is on the right and is directed downwards to the right.

To subtract B⃗ from A⃗ (Figure 1), perform these steps:

Draw A⃗ .
Draw −B⃗ and place its tail at the tip of A⃗ .
Draw an arrow from the tail of A⃗ to the tip of −B⃗ . This is vector A⃗ −B⃗ .

Part A

Draw vector C⃗ =A⃗ −B⃗ by following the steps in the tactics box above. When drawing −B⃗ , keep in mind that it has the same magnitude as B⃗ but opposite direction.

Part B

Draw vector F⃗ =D⃗ −E⃗ by following the steps in the tactics box above. When drawing −E⃗ , keep in mind that it has the same magnitude as E⃗ but opposite direction.

ANSWERS:

Part A

Part B

Arizona State University| PHY 121: Univ Physics I: Mechanics|Homework 1-1| Problem 1

Learning Goal:
To practice Tactics Box 1.1 Vector Addition.

Vector addition obeys rules that are different from those for the addition of two scalar quantities. When you add two vectors, their directions, as well as their magnitudes, must be taken into account. This Tactics Box explains how to add vectors graphically.

To add B⃗ to A⃗ (Figure 1), perform these steps:

Draw A⃗ .
Place the tail of B⃗ at the tip of A⃗ .
Draw an arrow from the tail of A⃗ to the tip of B⃗ . This is vector A⃗ +B⃗ .

Part A

Create the vector R⃗ =A⃗ +B⃗ by following the steps in the Tactics Box above. When moving vector B⃗ , keep in mind that its direction should remain unchanged.

Part B

Create the vector R⃗ =C⃗ +D⃗ by following the steps in the Tactics Box above. When moving vector D⃗ , keep in mind that its direction should remain unchanged. The location, orientation, and length of your vectors will be graded.

ANSWERS:

Part A

Part B

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