Answer:
Explanation:
It is given that,
Frequency for submarine communications, f = 76 Hz
We need to find the wavelength of those extremely low-frequency waves. the relation between the wavelength and the frequency is given by :
or
So, the wavelength of those extremely low-frequency waves is 4000 km. Hence, this is the required solution.
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>