The gravitational attraction between a large object and a small object is...
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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!
Hi there!
A.
Since the can was launched from ground level, we know that its trajectory forms a symmetrical, parabolic shape. In other words, the time taken for the can to reach the top is the same as the time it takes to fall down.
Thus, the time to its highest point:
Now, we can determine the velocity at which the can was launched at using the following equation:
In this instance, we are going to look at the VERTICAL component of the velocity, since at the top of the trajectory, the vertical velocity = 0 m/s.
Therefore:
***vsinθ is the vertical component of the velocity.
Solve for 'v':
Now, recall that:
Plug in the expression for velocity:
B.
We can use the same process as above, where T' = 2T and Th = T.
C.
The work done in part B is 4 times greater than the work done in part A.
Here is the complete question
The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine the horizontal shear stress at point H, which is located L = 20 mm below the centriod
The missing image which is the remaining part of this question is attached in the image below.
Answer:
The horizontal shear stress at point H is
Explanation:
Given that :
The internal shear force V = 80 kN = 80 × 10³ N
The moment of inertia = 64,900,000
The length = 20 mm below the centriod
The horizontal shear stress can be calculated by using the equation:
where;
Q = moment of area above or below the point H
b = thickness of the beam = 10 mm
From the centroid ;
Q =
Q =
Q = ( ( 70 × 10) × (55) + ( 210 × 15) (90 + 15/2) ) mm³
Q = ( ( 700) × (55) + ( 3150 ) ( 97.5) ) mm³
Q = ( 38500 + 307125 ) mm³
Q = 345625 mm³
The horizontal shear stress at point H is
