Graphical Method

1.) Using the given, make a scale that would be rational to the quantities.
2.) Use the scale to fit and draw the given direction and distances on the sheet.
3.) Be sure to note the correct distances and angles to the respective direction.
4.) After plotting all given, measure the line and angle from the starting point to the end point. Be sure to use scale and correct direction in determining the total distance and the final direction and angle.

Vector Component method:

In this example we will be adding the two vectors shown below using the component method. The vectors we will be adding are displacement vectors, but the method is the same with any other type of vectors, such as velocity, acceleration, or force vectors. Let's take this all one step at a time. First, let's visualize the x-component and the y-component of d1. Here is that diagram showing the x-component in red and the y-component in green:The two components along with the original vector form a right triangle. Therefore, we can use right triangle trigonometry to find the lengths of the two components. That is, we can use the 'SOH-CAH-TOA' type of definitions for the sine, cosine, and tangent trigonometry functions.Let's find the x-component of d1. Notice that the x-component is adjacent to the angle of 34 degrees, so, we will use the cosine function since it relates an acute angle, the adjacent side to that angle, and the hypotenuse of a right triangle:Now, using trigonometry like this will not tell us the sign, (+ or -), of this component, (or any other). So, we must check the diagram for positive or negative directions. This x-component is aimed to the right, so, it is positive:(Again, remember that these calculations presented here have decimals that have been truncated. Presenting calculations to many more decimals does not help clarify methods, and, also, it violates several rules of significant digits. In other words, these calculations are approximate. The calculator below keeps many more decimal places, so, its outputs will differ slightly.)Now, let's find the y-component of d1. Notice that the y-component is opposite to the angle of 34 degrees, so, we will use the sine function since it relates an acute angle, the opposite side to that angle, and the hypotenuse of a right triangle:Again, check the diagram for positive or negative directions. The y-component aims up, so, it is positive:Here is the diagram now showing the values for the x-component and y-component of d1:Next, let's see the x-component and the y-component of d2. Here is that diagram showing the x-component in red and the y-component in green:We will now find the x-component of d2. Here the x-component is opposite the angle of 64 degrees, so we will use the sine function to find it:Check the diagram for positive or negative directions. This x-component points to the left, so, it is negative:And for our last component we will find the y-component of d2. The x-component is adjacent to the angle of 64 degrees in this diagram, so we will use the cosine function to find it:Check the diagram for positive or negative directions. This y-component is aimed up, so, it is positive:Here is the diagram showing our newly calculated values for the components of d2:Now, we must add up like components to get the components for the total displacement.To get the total x-displacement, add up all of the separate x-components:To get the total y-displacement, add up all of the separate y-components:So, when these two vectors, d1 and d2, are added, the total, or sum, has an x-component of 9.2 meters and a y-component of 30.1 meters.To get the actual 2-D total displacement, add the total x-displacement and the total y-displacement. Here is a diagram with the total x-component shown in red and the total y-component shown in green and the 2_D total shown in blue:Use the Pythagorean theorem to get the magnitude (size) of the total 2-D displacement:Use the arctangent function to get the angle:Check the diagram for NSEW notation:Therefore, our final result for the total 2-D displacement can be stated as:Here's a diagram that shows this result:Again, the component method of addition can be summarized this way:Using trigonometry, find the x-component and the y-component for each vector.

Physics!!!

Expectations: I expect that i will learn more about Physics with the help of sir Mendoza. I want it to be more exciting and enjoyable learning Physics.

What I have learned in Physics this week: I have learned sohcahtoa(sin,cos,tan) although i find it difficult at the beginning with the help of my classmates i find it more easier than before with the help of my classmates i find it more easier than before.

The Mirror Equation

Ray diagrams can be used to determine the image location, size, orientation and type of image formed of
objects when placed at a given location in front of a concave mirror. Ray diagrams provide useful information about object-image relationships, yet fail to provide the information in a quantitative form. While a ray diagram may help one determine the approximate location and size of the image, it will not provide numerical information about image distance and object size. To obtain this type of numerical information, it is necessary to use the Mirror equation and the Magnification equation. The mirror equation expresses the quantitative relationship between the object distance (do), the image distance (di), and the focal length (f). The equation is stated as follows:
The Magnification equation relates the ratio of the image distance and object distance to the ratio of the image height (hi) and object height (ho). The magnification equation is stated as follows:
These two equations can be combined to yield information about the image distance and image height if the object distance, object height, and focal length are known.

CONVEX

The Mirror Equation - Convex Mirrors
Ray diagrams can be used to determine the image location, size, orientation and type of image formed of objects when placed at a given location in front of a mirror. Ray diagrams provide useful information about object-image relationships, yet fail to provide the information in a quantitative form. While a ray diagram may help one determine the approximate location and size of the image, it will not provide numerical information about image distance and object size. To obtain this type of numerical information, it is necessary to use the Mirror equation and the Magnification equation. The mirror equation expresses the quantitative relationship between the object distance (do), the image distance (di), and the focal length (f). The equation is stated as follows:
The Magnification equation relates the ratio of the image distance and object distance to the ratio of the image height (hi) and object height (ho). The magnification equation is stated as follows:
These two equations can be combined to yield information about the image distance and image height if the object distance, object height, and focal length are known.
As a demonstration of the effectiveness of the Mirror equation and Magnification equation, consider the following sample problem and its solution.

Steps in Getting Resultant Vector Using COmponent Method

1. given the vector, find its x and y x component by using SOHCAHTOA..when it has direction we have to determine whether its positive(up and right) of negative (left or down)2.after that we have to get the resultant x ny adding up the x components and so on with the resultant Y.3.then we can now compute for the resultant vector by the use of Pythagorean theorem or simply getting the sQuare root of the sum of the sQuared vectors4. after getting the resultant vector we should get the angle of the resultant vector to know whether it is positive or in the negative side.By using the tan formula you can easily get the measure of the angle.

Malapit na long test!!!
Sr. Menoza said if we fail in the test he will be calling our parents for meeting!!
Aww!!!!!!!!!!!

BUKAS NA!!!