## The Relationship between Variables (Part II)

This article is a continuation from Part I, which discusses about direct proportion. Inverse ProportionTwo sets of readings for the quantities x and y are given in the table below.x34612y4321Here, when x is doubled, y is halved; when x is trebled, y is...

## Problem: Kinematics #3

The figure above shows a displacement versus time squared (t2) graph for the motion of an object.Which of the following motion can be represented by this graph?A. A ball that is travelling at terminal velocity.B. A ball that is falling freely from a stationary position.C. A ball that bounces back from the floor.D. A ball that is travelling on a rough surface.SolutionShow solution >> The answer is

## The Relationship between Variables (Part I)

One of the most important mathematical operations in physics is finding the relationship between variables. Through the study of these relationships, we can know how a change in one variable affects another variable, thus enabling us to make prediction...

## Problem: Kinematics #2

The figure above shows two objects, P and Q moving with velocities 30 m s-1 and 20 m s-1 respectively towards each other on a straight line.How long, after that instant, will P and Q meet?A. 100.0 sB. 83.3 sC. 20.0 sD. 13.3 sSolutionShow solution >> The answer is C.Let the distance travelled by P and Q before they meet each other be d1 and d2 respectively. Here, we can form an equation:d1 + d2=

## Problem: Materials #2

The density of an object is ¾ of the density of water. If the object is floating on the water surface, then the ratio between the volume of object above the water surface and the volume of object below the water surface isA. 1:4B. 1:3C. 3:4D. 4:3SolutionShow solution >> The answer is B.This is a problem that requires the application of Archimedes’ principle.We know that the density of the object

## Problem: Materials #1

An object has a weight of 4 N in air, 3 N in water and 2.8 N in a salt solution. If the density of water is 1000 kg m-3, the density of the salt solution isA. 830 kg m-3B. 1070 kg m-3C. 1200 kg m-3D. 1430 kg m-3SolutionShow solution >> The answer is C.This is a problem that requires the application of Archimedes’ principle. The object has a weight of 4 N in air. Because the upthrust on the object