The frequency of an oscillation is equal to the reciprocal of the period:

where
f is the frequency
T is the period
In our problem, the period is T=8.01 s, therefore the corresponding frequency is
Answer:
c
Explanation:
<u>impulse</u><u> </u><u>Is</u><u> </u><u>the </u><u>product</u><u> </u><u>of </u><u>force </u><u>and </u><u>distance</u><u> </u><u>so </u><u>it's </u><u>generally</u><u> </u><u>formula</u><u> </u>
<em>Class I .</em> . .
The fulcrum is between the effort and the load.
The mechanical advantage may be any positive number, more or less than ' 1 '.
<em>Class II .</em> . .
The load is between the fulcrum and the effort.
The mechanical advantage is always greater than ' 1 '.
<em>Class III .</em> . .
The effort is between the fulcrum and the load.
The mechanical advantage is always less than ' 1 '.
Answer:B
Explanation:
Given
Wavelength of light 
Screen distance 
First fringe is at a distance 
No of lines per mm is given by N

where d=slit width
From N-slits Experiment


Position of bright fringe is given by



Put the value of
in eq. 1

Therefore 

for 


