Two loudspeakers on elevated platforms are at opposite ends of a fie.pdf
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Two loudspeakers on elevated platforms are at opposite ends of a field. Each broadcasts equally in all directions. The sound intensity level at a point halfway between the loudspeakers is 71.6 dB . Q:What is the sound intensity level at a point one-quarter of the way from one speaker to the other along the line joining them? Solution We know that the net intensity level due to multiple source is given as: 71.6 = 10 log ( Inet / I0); 71.6 = 10 log(2I/Io) Now, what we have is that the Inet = 2I where I is the intensity due to one of the speakers. [Since the two loudspeakers are the same, we take Intensity at the centre of the field due be same] Further, we know that Intensity due to a sound source is inversely proportional to the distance from the source. That is the net intensity at a point one quarter away from one of the sources = 4I + 4I/9 = 40 I/9 [Since the distance from one of the speakers will be halved, the intensity becomes 4 times, while for other the distance will become 3/2 times, hence the intensity will become 4/9 times] Hence, the new sound intensity level = 10 log(40 I / 9 Io) = 10log [(20/9)*(2I/Io)] = 10log(20/9) + 71.6 Therefore, the net sound intensity level = 75.0679 dB.
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![Two loudspeakers on elevated platforms are at opposite ends of a field. Each broadcasts equally
in all directions. The sound intensity level at a point halfway between the loudspeakers is 71.6
dB .
Q:What is the sound intensity level at a point one-quarter of the way from one speaker to the
other along the line joining them?
Solution
We know that the net intensity level due to multiple source is given as: 71.6 = 10 log ( Inet / I0);
71.6 = 10 log(2I/Io)
Now, what we have is that the Inet = 2I where I is the intensity due to one of the speakers.
[Since the two loudspeakers are the same, we take Intensity at the centre of the field due be
same]
Further, we know that Intensity due to a sound source is inversely proportional to the distance
from the source.
That is the net intensity at a point one quarter away from one of the sources = 4I + 4I/9 = 40 I/9
[Since the distance from one of the speakers will be halved, the intensity becomes 4 times, while
for other the distance will become 3/2 times, hence the intensity will become 4/9 times]
Hence, the new sound intensity level = 10 log(40 I / 9 Io) = 10log [(20/9)*(2I/Io)] = 10log(20/9)
+ 71.6
Therefore, the net sound intensity level = 75.0679 dB](https://image.slidesharecdn.com/twoloudspeakersonelevatedplatformsareatoppositeendsofafie-230701051350-56148802/85/Two-loudspeakers-on-elevated-platforms-are-at-opposite-ends-of-a-fie-pdf-1-320.jpg)
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Mitotic spindleelongates and gets attached to chromosome.
This phase takes longer than the others because the nuclear envelope fragments and the
microtubules have to attach to the chromosomes.
Yes, time spend in this phase is justified as without spindle fibers attachment to the
chromosomes and nuclear envelope breakdown the cell division is not possible.
Ans 5: The Metaphase of the mitotic phase requires the next longest time for completion about
4% of the cellcycle time is consumed during this phase. In this phase lining up of chromosome at
metaplate occurs, and the check point for two kinetochores of each chromosome are correctly
attached to microtubules from opposite spindle poles or not.
Ans 6.
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