9th Class Physics Chapter 2 Comprehensive questions
9th Class Physics Chapter 2 Comprehensive questions. We will solve only 9th class physics revised smart syllabus.
2.1:- How a vector can be represented graphically? Explain
We represent a vector by an arrow head on it or below it or with a bold face letter.
$$\boldsymbol A\boldsymbol\;or\;\overrightarrow A\;or\;\underrightarrow A$$
We represent the magnitude of vector by
$$A\;or\;\left|\boldsymbol A\right|\;or\;\left|\overrightarrow A\right|\;or\;\left|\underrightarrow A\right|$$
We represent a vector graphically by drawing a line segment with an arrow head at its one end. The length of line segment represents the magnitude of vector quantity according to a suitable scale and the direction of arrow represents the direction of vector. First we draw a cartesian coordinate system. A vector is drawn starting from origin towards the given direction. The direction is usually given by an angle with the x-axis. The angle with the x-axis is measured from the right side of x-axis in anti-clockwise direction.
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2.2:- Differentiate between rest and motion and speed and velocity.
Rest and Motion:-
Body is at rest, if a it does not change its position with respect to its surroundings.
Body is in motion, if it is continuously changing its position with respect to its surroundings.
The state of rest or motion of a body is always relative. Two passengers in a moving train are in rest with respect to each other but they are in a state of motion with respect to another observer standing at the platform.
Speed and Velocity:-
Distance covered in unit time in known as speed. It is a scalar quantity.
Speed = Distance/Time
v= s/t
Its SI unit is meter per second.
The net displacement of a body in unit time is called velocity. It is a vector quantity.
If a body moves from a point A to a point B along a curved path, the displacement d is the straight line AB.
The direction of velocity is the same as the direction of displacement. It SI unit is also meter per second
Average velocity = Displacement/Time
$${\boldsymbol v}_{av}\;=\;\frac{\boldsymbol d}t\\Meter\;per\;second\;is\\denoted\;by\;ms^{-1}$$
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2.6:- Prove that the area under speed-time graph is equal to the distance covered by an object.
An object is moving with constant speed v. For a time interval t, the distance s covered by the object will vt.
The area under the graph for time interval t is the area of rectangles of sides t and v. This area is equal to v multiplied by t. So, the area under the speed-time graph up to time axis is numerically equal to the distance covered by the object in time t.
If speed increases uniformly from 0 to v in time interval t, area under the speed time graph is a triangle. Distance covered is equal to area under the graph.
Distance = 1/2 (height) (base) = 1/2 vt
if the speed increases uniformly from 0 to v in time interval t, distance covered is 1/2 vt, which is equal to area under the speed-time graph.
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