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The Fundamental Unit of Life class 9 Notes

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  The Fundamental Unit of Life class 9 Notes Introduction All living organisms like plants, animals, and human beings are made up of tiny microscopic structures called cells . The cell is the basic structural and functional unit of life . This means: It forms the structure of all living bodies. It performs all the functions necessary for life such as respiration, nutrition, and reproduction. Discovery of Cell The cell was first discovered by Robert Hooke in 1665 while observing a thin slice of cork under a microscope. Later, Anton van Leeuwenhoek observed living cells in pond water. Cell Theory The cell theory was proposed by: Matthias Schleiden Theodor Schwann According to the Cell Theory:      1) All living organisms are made up of cells.      2) The cell is the basic unit of life.      3) All cells arise from pre-existing cells. Why Are Cells Important? Every plant, animal, and human body is made up of cells. Cells carry out ...
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An athlete completes one round of circular track of diameter 50 m in 10 second. What will be the distance cover and the displacement at the end of 45 second?

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  An athlete completes one round of circular track of diameter 50 m in 10 second. What will be the distance cover and the displacement at the end of 45 second? Solution: Given: Diameter of circular track = 50 m So radius ( r ) = 25 m Time for 1 round = 10 s Total time = 45 s 1) Distance Covered First, find circumference of the track: Circumference = 2 π r = 2 π ( 25 ) = 50 π  m \text{Circumference} = 2\pi r = 2\pi (25) = 50\pi \text{ m} 50 π ≈ 157  m So, in 10 s → distance = 157 m In 45 s: Distance = 157 10 × 45  = 157 × 4.5 = 157 × 4.5                                  = 706.5  m (approx) Distance covered ≈ 706.5 m 2) Displacement After 45 Seconds In 40 seconds → athlete completes 4 full rounds (comes back to starting point). Remaining time = 5 seconds In 5 seconds → athlete completes half round. Half round means he reaches the diametrically opposite point . Displac...
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A body moves with an initial velocity of 2 m/s and uniform acceleration of 3 m/s2. Calculate the velocity when it has traveled a distance of 77 m.

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  A body moves with an initial velocity of 2 m/s and uniform acceleration of 3 m/s 2 . Calculate the velocity when it has traveled a distance of 77 m. Solution: Given: Initial velocity ( u ) = 2 m/s Acceleration ( a ) = 3 m/s² Displacement ( s ) = 77 m Since time is not given, we use the third equation of motion: v 2 = u 2 + 2 a s Step 1: Substitute values v 2 = ( 2 ) 2 + 2 ( 3 ) ( 77 ) v^2 = (2)^2 + 2(3)(77) v 2 = 4 + 6 × 77 v^2 = 4 + 6 × 77 v 2 = 4 + 462 v^2 = 4 + 462 v 2 = 466 v^2 = 466 Step 2: Take square root v = 466 v = \sqrt{466} ​ v ≈ 21.56  m/s
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A body starts from rest and moves with a uniform acceleration of 2 m/s2. What will be its velocity and displacement at the end of 10 seconds?

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  A body starts from rest and moves with a uniform acceleration of 2 m/s 2 . What will be its velocity and displacement at the end of 10 seconds? Given: Initial velocity ( u ) = 0 m/s (starts from rest) Acceleration ( a ) = 2 m/s² Time ( t ) = 10 s Velocity after 10 seconds Using the formula: v = u + a t v = u + at v = 0 + ( 2 × 10 ) v = 0 + (2 × 10) v = 20  m/s v = 20 \text{ m/s} Displacement after 10 seconds Using the formula: s = u t + 1 2 a t 2 s = ut + \frac{1}{2}at^2 s = 0 + 1 2 × 2 × ( 10 ) 2 s = 0 + \frac{1}{2} × 2 × (10)^2 s = 1 × 100 s = 1 × 100 s = 100  meters s = 100 \text{ meters}
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Inequality Notes class 8

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  Inequality Notes class 8 What is Inequality? Inequality refers to a mathematical statement that shows the relationship between two expressions that are not equal. In inequalities, the two expressions are connected using inequality symbols. Inequality Symbols: > (Greater Than) : Shows that the value on the left is larger than the value on the right. Example: 5 > 3 5 > 3 < (Less Than) : Shows that the value on the left is smaller than the value on the right. Example: 3 < 5 3 < 5 ≥ (Greater Than or Equal To) : Indicates that the value on the left is either greater than or equal to the value on the right. Example: 5 ≥ 3 5 \geq 3 ≤ (Less Than or Equal To) : Indicates that the value on the left is either less than or equal to the value on the right. Example: 3 ≤ 5 3 \leq 5 ≠ (Not Equal To) : Indicates that two values are not equal. Example: 5 ≠ 3 5 \neq 3 Solving Inequalities: Just like solving equations, inequalities can be solved to find th...
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An object dropped from a leaf false with a constant acceleration of 10 m/s2. Find its Speed 2 second after it was dropped.

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  An object dropped from a leaf false with a constant acceleration of 10 m/s 2 . Find its Speed 2 second after it was dropped.                            Solution: Given: Initial speed, u = 0 u = 0 (object is dropped) Acceleration due to gravity, a = 10   m/s 2 a = 10 \, \text{m/s}^2 Time, t = 2 \, \text{s} t  = 2 s Formula used: v = u + a t Calculation: v = 0 + ( 10 × 2 ) = 20   m/s Answer: The speed of the object after 2 seconds is 20 m/s (downwards) .
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A train travels 20 km at a uniform speed of 60 km/h and the next 20 km at a uniform speed of 80 km/hr. calculate its average speed.

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                                A train travels 20 km at a uniform speed of 60 km/h and the next 20 km at a uniform speed of 80 km/hr. calculate its average speed. Given: First distance d 1 = 20   km d_1 = 20 \, \text{km} , speed v 1 = 60   km/h v_1 = 60 \, \text{km/h} Second distance d 2 = 20   km d_2 = 20 \, \text{km} , speed v 2 = 80   km/h v_2 = 80 \, \text{km/h} Step 1: Time taken for each part t 1 = 20 60 = 1 3  h t_1 = \frac{20}{60} = \frac{1}{3} \text{ h} t 2 = 20 80 = 1 4  h t_2 = \frac{20}{80} = \frac{1}{4} \text{ h} Step 2: Total distance and total time Total distance = 20 + 20 = 40  km \text{Total distance} = 20 + 20 = 40 \text{ km} Total time = 1 3 + 1 4 = 4 + 3 12 = 7 12  h \text{Total time} = \frac{1}{3} + \frac{1}{4} = \frac{4 + 3}{12} = \frac{7}{12} \text{ h} Step 3: Average speed Average speed = Total distance Total time = 40 7 / 12 =...
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