1. A Faraday ice pail similar to the one you will use in the experiment is used in this question. The principle of its operation
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<h3>Question:5</h3>
The given data can be written in following way in coordinates axes.
(0,0), (2,2), (4,4), (6,4), (8,4), (10,6),(12,4), (14, 2).
a) Average velocity for first 4 seconds
Average velocity = Total Displacement/ Time taken
= (4-0)/(4 - 0)
=4/4
= 1/1
<h3> =1 m/s</h3>b) Average velocity for 4 to 8 seconds
Average velocity = Total Displacement/ Time taken
= (4 - 4)/(8-4)
= 0/4
<h3> = 0</h3>c) Average velocity for last 6 seconds
Last 6 seconds = from 8 to 14 seconds
Average velocity = Total Displacement/Time taken
= (2 - 4)/(14 - 8)
= -2/6
<h3> = -1/3 m/s</h3><h3>Question 6:</h3>The given data can be written in following way in coordinates axes.
(0,0), (10,20), (20,20), (30,20), (40,0)
a) State the kind of motion from Os to 10s and from 30s to 40s
It is obvious from the graph that the velocity between Os to 10s has been increased from 0 m/s to 20m/s. Hence there is a uniform Acceleration in the body.
b) What is the velocity of the body after 10s and 40s ?
It is clear from the graph and table as well that,
<h3>Velocity after 10s is 20m/s</h3>and
<h3>Velocity after 40s is 0 m/s</h3>c) Calculate the distance covered by the body between 10s to 30s.
Distance covered by the body between 10s to 30s will be given by area of rectangle ABCD
Area of rectangle ABCD= (30 - 10) × (20 - O)
=20×20
<h3> = 400 m</h3>
The distance covered by car is equal to (assuming it is moving by uniform motion) the product between the car's speed and the time of the car ride, 4 h:
where
is the car's speed
is the duration of the car ride
Similarly, the distance covered by train is equal to the product between the train's speed and the duration of the train ride, 7 h:
The total distance covered is S=255 km, which is the sum of the distances covered by car and train:
which becomes
(1)
we also know that the train speed is 5 km/h greater than the car's speed:
(2)
If we put (2) into (1), we find
and if we solve it, we find
So, the car speed is 20 km/h and the train speed is 25 km/h.
where
Similarly, the distance covered by train is equal to the product between the train's speed and the duration of the train ride, 7 h:
The total distance covered is S=255 km, which is the sum of the distances covered by car and train:
which becomes
we also know that the train speed is 5 km/h greater than the car's speed:
If we put (2) into (1), we find
and if we solve it, we find
So, the car speed is 20 km/h and the train speed is 25 km/h.