A piano string having a mass per unit length equal to 4.80 ✕ 10−3 kg/m is under a tension of 1,300 N. Find the speed with which
1 answer:
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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>Answer:
Explanation:
Answer:
Explanation:
Here two charges are placed at distance "d" apart
now the net value of electric field at some position between two charges will be ZERO
so we will have
electric field due to charge 1 = electric field due to charge 2
Let the position where net field is zero will lie at distance "r" from q1
now we will have
now square root both sides
now we have
so we have