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maria [59]
3 years ago
14

Which of the following would decrease in size during the contraction of a sarcomere? The width of the I-bands The width of the A

-band The width of the myosin filaments The width of the actin filaments

Physics
1 answer:
ANEK [815]3 years ago
8 0

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!

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A stone is thrown horizontally at 60.0 m/sm/s from the top of a very tall cliff. Calculate its horizontal position and vertical
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Answer:

X-Positions:                                         Y-Positions

x(0) = 0                                                   y(0) = 0

x(2) = 120 m                                           y(2) = 19.6 m

x(4) = 240 m                                          y(4) = 78.4 m

x(6) = 360 m                                          y(6) = 176.4 m

x(8) = 480 m                                          y(8) = 313 m

x(10) = 600m                                         y (10) = 490 m

Explanation:

X-Positions

  • First, we choose to take the horizontal direction as our x-axis, and the positive x-axis as positive.
  • After being thrown, in the horizontal direction, no external influence acts on the stone, so it will continue in the same direction at the same initial speed of 60. 0 m/s
  • So, in order to know the horizontal position at any time t, we can apply the definition of average velocity, rearranging terms, as follows:

       x = v_{ox} * t = 60.0 m/s * t(s)

  • It can be seen that after 2 s, the displacement will be 120 m, and each 2 seconds, as the speed is constant, the displacement will increase in the same 120 m each time.

Y-Positions

  • We choose to take the vertical direction as our y-axis, taking the downward direction as our positive axis.
  • As both axes are  perpendicular each other, both movements are independent each other also, so, in the vertical direction, the stone starts from rest.
  • At any moment, it is subject to the acceleration of gravity, g.
  • As the acceleration is constant, we can find the vertical displacement (taking the  height of the cliff as the initial reference level), using the following kinematic equation:

       y = \frac{1}{2} * g* t^{2} = \frac{1}{2} * 9.8 m/s2 * t(s)^{2}

  • Replacing by the values of t, we get the following vertical positions, from the height of the cliff as y = 0:
  • y(2) = 2* 9.8 m/s2 = 19.6 m
  • y(4) = 8* 9.8 m/s2 = 78.4 m
  • y(6) = 18*9.8 m/s2 = 176.4 m
  • y(8) = 32*9.8 m/s2 = 313.6 m
  • y(10)= 50 * 9.8 m/s2 = 490.0 m
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