"A is correct answer." The effective length of the tube is responsible for determining the frequency of vibration of the air column in the tube within a wind instrument. "Hope this helps!" "Have a great day!" "Thank you for posting your question!"
Answer:
No, it's not there.
Explanation:
For a machine to be 100% efficient, it has to be with an output which is equal to its input. But machines have an out put less than an input, hence efficiency below 100%.
Answer:
Frequency of oscillation, f = 4 Hz
time period, T = 0.25 s
Angular frequency, 
Given:
Time taken to make one oscillation, T = 0.25 s
Solution:
Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:
f = 
f = 
Frequency of oscillation, f = 4 Hz
The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.
Therefore, time period, T = 0.25 s
Angular frequency of oscillation is given by:
Answer:
Option D
670 Kg.m/s
Explanation:
Initial momentum is given by mv=82*5.6=459.2 Kg.m/s (taking eastward as positive)
Final momentum is also mv but v being westward direction, we take it negative
Final momentum=82*-2.5= -205 Kg.m/s
Change in momentum=Final momentum-Initial momentum=-205-459.2=-664.2 Kg.m/s
Impulse=change in momentum=664.2 Kg.m/s rounded off as 670 Kg.m/s
