Author Archives: Engineering Math

Physics for Scientists and Engineers 3rd Edition by Randall Knight, Chapter 1 Conceptual Question 6


Determine the signs (positive or negative) of the position, velocity, and acceleration for the particle in Figure Q1.6.

Figure Q1.6

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Physics for Scientists and Engineers 3rd Edition by Randall Knight, Chapter 1 Conceptual Question 5


Does the object represented in Figure Q1.5 have a positive or negative value of \displaystyle a_y? Explain.

Physics for Scientists and Engineers 3rd Edition by Randall Knight, Chapter 1 Conceptual Question 5

Figure Q1.5


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Physics for Scientists and Engineers 3rd Edition by Randall Knight, Chapter 1 Conceptual Question 4


Does the object represented in Figure Q1.4 have a positive or negative value of \displaystyle a_x? Explain.

Figure Q1.4 Physics for Scientists and Engineers 3rd Edition by Randall Knight

Figure Q1.4


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


Is the particle in FIGURE Q1.3 speeding up? Slowing down? Or can you tell? Explain.

Q1.3 Physics for Scientist and Engineers 3rdf Edition by Randall Knight

Figure Q1.3


Solution:

The diagram does not indicate any position in time that should have been represented by numbers on the dots. Without numbers on the dots we cannot tell if the particle in the figure is moving left or right, so we can’t tell if it is speeding up or slowing down. If the particle is moving to the right it is speeding up. If it is moving to the left it is slowing down. 


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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 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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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, 53 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:

  1. Draw A⃗ .  The figure shows vector A that is directed upwards to the right.
  2. Draw B⃗  and place its tail at the tip of A⃗ .The figure shows two vectors. Vector A is on the left and is directed upwards to the right. Vector minus B is on the right and is directed upwards to the left. The tip of vector A and the tail of vector minus B are at the same point.
  3. Draw an arrow from the tail of A⃗ to the tip of B⃗ . This is vector A⃗ B⃗ .     The figure shows two vectors. Vector A is on the left and is directed upwards to the right. Vector minus B is on the right and is directed upwards to the left. The tip of vector A and the tail of vector minus B are at the same point. Vector A minus B starts at the tail of vector A and ends at the tip of vector minus B.
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
3
Part B
4

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:
  1. Draw A⃗ .  The figure shows vector A that is directed upwards to the right.
  2. Place the tail of B⃗  at the tip of A⃗ .  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. The tip of vector A and the tail of vector B are at the same point.
  3. Draw an arrow from the tail of A⃗  to the tip of B⃗ . This is vector A⃗ +B⃗ .  The figure shows three 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. The tip of vector A and the tail of vector B are at the same point. Vector A plus B starts at the tail of vector A and ends at the tip of vector B.
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.
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.
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
1
Part B
2

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

week 2 problem 8

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