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In Text Questions of Page No 102 Chapter 8 Science Class 9 NCERT Solutions Intext Question Chapter 8 Motion Science

In text Questions of page no 102 Chapter 8: Motion Science Class 9 solutions are developed for assisting understudies with working on their score and increase knowledge of the subjects. Distinguish between speed and velocity,What does the odometer of an automobile measure,What does the path of an object look like when it is in uniform motion is solved by our expert teachers. You can get ncert solutions and notes for class 9 chapter 8 absolutely free. NCERT Solutions for class 9 Science Chapter 8: Motion is very essencial for getting good marks in CBSE Board examinations

In text Questions of page no 102

 

Q. No. 1 Distinguish between speed and velocity.
Ans:

Speed

Velocity

Speed has only magnitude and no direction and hence it is a scalar quantity.

Velocity has magnitude as well as direction and hence it is a vector quantity.

Speed of a moving object cannot be zero.

Velocity of a moving object can be zero.

The rate of change of distance is known as speed.

The rate of change of displacement is known as velocity.

 

Q. No 2: Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Ans:
The condition where the total distance covered by an object is the same as its displacement, then its average speed would be equal to its average velocity.

 

Q. No 3: What does the odometer of an automobile measure?
Ans:
The odometer of an automobile measures the total distance covered by an automobile.

 

Q. No 4: What does the path of an object look like when it is in uniform motion?
Ans:
When an object is in uniform motion then its path will be a straight line straight line.

 

Q. No 5: During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 108 m s−1.

Ans: Time taken by the signal = 5 minute = 5 × 60 = 300 s
Speed of the signal                = 3 × 108 m/s
We know that
Speed= Distance/Time
Therefore Distance= Speed × Time
                          = 3 × 108 × 300
                          = 9 × 1010 m
So the distance of the spaceship from the ground station is 9 × 1010 m.

 

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