PHYSICS IS EASY !

assalamualaikum .
harinie kami ade kelas waktu malam . haiiihh . .bulan pon bagi semangat taw nak belajar mlm2 . haha .

tp x pe la . demi untuk belajar .kami pegi jugak .naseb la haritu mOOd baek .yg dtg lambat x perlu denda .
tq so much dr . =) ape yg kami belajar malam tue . ini diaaa . .tajuk baru chapter 3 .KINEMATICS .
dengar cerita masuk test kan . OMG . semoga kami dapat buat . .AMINNN . . nie la yg kami belajar . .

"Kinematics is the study of how things move. Here, we are interested in the motion of normal objects in our world. A normal object is visible, has edges, and has a location that can be expressed with (x, y, z) coordinates. We will not be discussing the motion of atomic particles or black holes or light.
We will create a vocabulary and a group of mathematical methods that will describe this ordinary motion. Understand that we will be developing a language for describing motion only. We won't be concerned with what is causing or changing the motion, or more correctly, the momentums of the objects. In other words, we are not concerned with the action of forces within this topic. "

ini ialah kinematic equation

dan ini

equation for linear motion .

$v = v_i + a \Delta t \,$ $s = s_i + v_i\Delta t + \tfrac{1}{2} a(\Delta t)^2 \,$

$s = s_i + \tfrac{1}{2} (v + v_i)\Delta t \,$

$v^2 = v_i^2 + 2a(s - s_i) \,$

Equations for Vertical Motion

Case I:

When a body is projected vertically downwards with initial velocity u then the equation describing motions are

v = u + gt ……(1)

h = ut + 1/2 gt² ……(2)

v² = u² + 2gh ……(3)

h_n = distance covered in n^th second = u + 1/2 g(2n –1)

Case II:

When body is falling down freely, then

v = gt ……(1)

h = 1/2 gt² ……(2)

v² = 2gh ……(3)

h_n = 1/2 g(2n –1) ……(4)

Case III:

When body is projected vertically upward with initial velocity u, then

v = u –gt ……(1)

h = ut –1/2gt² ……(2)

v² = u² –2gh ……(3)

h_n = u –1/2g(2n –1) ……(4)

Greatest height attained = u²/2g ……(5)

Time to the greatest height = u/g ……(6)

Time to a given height h = u/g ± √u²–2gh/g.

Notes:

Time from any point on the path to the highest point is same as the time from the highest point to the given point when the body is returning.

Solved Example 1:

A body is thrown vertically upward and rises to a height of 10 meters. Calculate (i) the velocity with which the body was thrown upwards and (ii) the time taken by the body to reach the highest point.

Solution:

Let the body be thrown upwards with a velocity of 4 m/sec. Thus greatest height attained = u²/g meters.

(i) We have, greatest height = 10 meters

=> u²/2g = 10 => u² = 196 => u = 14 m/sec

Hence, the required velocity of projection = 14 m/sec

(ii) Time taken by the body to reach the greatest height = u/g sec = 14/9.8 = 10/7 sec.

Solved Example 2:

The greatest height attained by a particle projected vertically upwards is 19.6 meters. Find how soon after projection the particle will be at a height of 14.7 meter.

Solution:

Let u be the velocity of projection since at the greatest height, the velocity of the particle is zero.

0² = u² –2g × 19.6 => u² = 19.6 × 19.6 => u = 19.6 m/sec

Suppose after time t the particle is at a height of 14.7 meter, then

14.7 = 19.6t – 1/2 × 9.8 × t²

= t² –4t+ 3 = 0 => (t –1) (t –3) = 0 => t = 1, 3

Thus, the particle will be at a height of 14.7 meter after 1 sec of its projection. It will also attain the same height after 3-seconds while coming downward after attaining the maximum height.

dan selepas itu kami di beri tugasan .

membuat sebanyak 22 soalan . for guys jwab soalan genap . n for girls jwab soalan ganjil .

soalan yg mencabar minda . not easy but take it easy . hahahaha . .

that's ALL . smoga rakan2 yang lain dpat menjawab soaln dgn jayanya . GOOD LUCK GUYS !

THANK YOU FOR READING .

assalamualaikum .

ini lah yang kami belajar semasa kelas petang tadi .

1.A force is a pull on a small object resulting from the interaction with another object .

2.Newton's first law

Lex I: Corpus omne perseverare in statu suo quiescendi vel movendi uniformiter in directum, nisi quatenus a viribus impressis cogitur statum illum mutare.

Law I: Every body persists in its state of being at rest or of moving uniformly straight forward, except insofar as it is compelled to change its state by force impressed.

This law states that if the resultant force (the vector sum of all forces acting on an object) is zero, then the velocity of the object is constant. Mathematically stated:

$\sum \mathbf{F} = 0 \Rightarrow \frac{d \mathbf{v} }{dt} = 0.$

Consequently:

An object that is at rest will stay at rest unless an unbalanced force acts upon it.
An object that is in motion will not change its velocity unless an unbalanced force acts upon it.

Newton placed the first law of motion to establish frames of reference for which the other laws are applicable. The first law of motion postulates the existence of at least one frame of reference called a Newtonian or inertial reference frame, relative to which the motion of a particle not subject to forces is a straight line at a constant speed. Newton's first law is often referred to as the law of inertia. Thus, a condition necessary for the uniform motion of a particle relative to an inertial reference frame is that the total net force acting on it is zero.

example .

1. What will happen to an object at rest if no force is applied?

2. What will happen to an object at rest if it is pushed, but there is a large frictional force acting on the object?

3. What will happen to the motion of an object that is pushed, and there is very little frictional force acting on the object?

4. What will happen to the motion of an object if there is a constant net force applied to it?

answers :

-nothing
-depends on the frictional force
-an object that has a force applied to it temporarily (a push) will travel at a constant speed infinitely until an outside force disrupts it such as friction. under most circumstances where friction is present an object will slowly decelerate and eventually stop
-if a constant net force is applied to an object, it will accelerate at a uniform rate until the force is removed. example: if you drop an object it will constantly accelerate downward due to the constant force of gravity pulling down on the object. the net force is always the same, yet the objects speed keeps increasing

THAT'S ALL .

THANK YOU . =)