## Saturday, August 31, 2013

### Solving Vector Problems, 2

Category: Trigonometry, Physics

"Published in Suisun City, California, USA"

A pilot heads his jet due east. The jet has a speed of 425 mi/hr in still air. The wind is blowing due north with a speed of 40 mi/hr. Find the true speed and direction of the jet.

Solution:

The given word problem above is about getting the true speed and direction of the jet. Since the directions of the jet and the wind are not collinear and have different directions, then we have to get the resultant of the two directions or two vectors as follows Photo by Math Principles in Everyday Life

Since the directions of the jet and the wind are perpendicular to each other, then the two vectors form a right triangle and a rectangle as shown in the figure above. The value of the resultant can be calculated using Pythagorean Theorem as follows

The direction of the resultant is calculated as follows

or

North of East Direction

## Friday, August 30, 2013

### Algebraic Operations - Matrix Operations

Category: Algebra, Arithmetic

"Published in Suisun City, California, USA"

Given the following matrices:

Carry out each indicated operations, or explain why it cannot be performed.

(a) A + B
(b) C - D
(c) C + A
(d) 5A

Solution:

(a) If you add the two matrices with the same dimensions, then each term or entry in a row or column will be added together. Therefore,

(b) If you subtract the two matrices with the same dimensions, then each term or entry in a row or column will be subtracted together. Therefore,

(c) C + A is undefined because we can't add matrices of different dimensions.

(d) If you multiply a matrix with any coefficient, variable, or any single term, then each term or entry will be multiplied by any coefficient, variable, or any single term as follows

## Thursday, August 29, 2013

### Stoichiometry Problem - Material Balance

Category: Chemical Engineering Math, Algebra

"Published in Suisun City, California, USA"

The waste acid from a nitrating process containing 20% HNO3, 55% H2SO4, and 25% H20 by weight is to be concentrated by addition of concentrated sulfuric acid containing 95% H2SO4 and concentrated nitric acid containing 90% HNO3 to get desired mixed acid containing 26% HNO3 and 60% H2SO4. Calculate the quantities of waste and concentrated acids required for 1000 kg of desired mixed acid.

Solution:

The given word problem is about the mixing of different acids which involves the principles of Stoichiometry. The total amount of a substance in the reactants or incoming ingredients must be equal to the total amount of a substance in the final products. In short, the Law of Conservation of Mass must be followed all the time. Since all incoming substances are acids, then there's no chemical reactions involved in the mixture. To illustrate the given problem, it is better to draw the flow diagram as follows Photo by Math Principles in Everyday Life

Basis: 1000 kg of Desired Mixed Acid

Let x = be the amount of waste acid
y = be the amount of concentrated sulfuric acid
z = be the amount of concentrated nitric acid

Overall Material Balance around the Mixer:

Material Balance of Sulfuric Acid:

Material Balance for Nitric Acid:

Substitute the value of y and z to the first equation, we have

Substitute the value of x to the second equation, we have

Substitute the value of x to the third equation, we have

Therefore,

Amount of Waste Acid = 401.6 kg
Amount of Concentrated Sulfuric Acid = 398.65 kg
Amount of Concentrated Nitric Acid = 199.75 kg

## Wednesday, August 28, 2013

### Rate, Distance, Time - Problem, 5

Category: Algebra

"Published in Newark, California, USA"

From Manila to a certain town in Batangas is some 224 miles. How long does it ordinarily take to reach the town, if by traveling 8 miles per hour faster, the trip is reduced by half an hour?

Solution:

The given word problem above is about rate, distance, and time problem with some conditions. Lets analyze the given word problem as follows:

Let x = be the rate in miles per hour
y = be the travel time in hour

We know that

The first working equation will be

If the statement says, "...if by traveling 8 miles per hour faster, the trip is reduced by half an hour?", then the second working equation will be

To eliminate y, substitute the value of y from the first working equation to the second equation, we have

Solve the quadratic equation by completing the square

Take the square root on both sides of the equation

In solving for the rate and time, we have to choose the positive value at the right side of the equation. Hence, the rate is

Therefore, the usual travel time is