Answer:
31.2 m/s
Explanation:
= Frequency of approach = 480 Hz
= Frequency of going away = 400 Hz
= Speed of sound in air = 343 m/s
= Speed of truck
Frequency of approach is given as
eq-1
Frequency of moving awayy is given as
eq-2
Dividing eq-1 by eq-2
= 31.2 m/s
Hi!
The correct answer would be: the width of I-bands
The sacromere is the smallest contractile unit of striated muscles. These units comprise of filaments (fibrous proteins) that, upon muscle contraction or relaxation, slide past each other. The sacromere consists of thick filaments (myosin) and thin filaments (actin).
<em>Refer to the attached picture to clearly see the structure of a sacromere.</em>
<u>When a sacromere contracts, a series of changes take place which include:</u>
<em>- Shortening of I band, and consequently the H zone</em>
<em>- The A line remains unchanged</em>
<em>- Z lines come closer to each other (and this is due to the shortening of the I bands) </em>
The only changes that take place occur in the zones/areas in the sacromere (as mentioned), not in the filaments (actin and myosin) that make the up the sacromere; hence all other options are wrong.
Hope this helps!
