Consider the cross section to the right with an overall height h 4.2 inch, bottom flange plate width b1-2.5 inch, top flange pla
te width b2-2.0 inch, and the web and both flange plates have a thickness of t-0.25 inch. The center of coordinate system is located at the centroid of the cross-section area. a) Find the y distance of centroid for the cross section to the bottom edge in inch. b) Find the Sx section modulus in inch3. c) The section is rotated counter-clockwise 20 degrees about its centroid. Find the rotated l moment of inertia in inch4. d) Find the rotated Iky product of inertia in inch4.
1 answer:
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Answer:
angle of incidence for the second mirror is 55.4 degree
Explanation:
Given data;
Angle between two mirror is 90 degree
Incident angle on first mirror is 34.6 degree
For first mirror
Angle of incidence is equal to angle of reflection
34.6 degree = angle of reflection
Now , for the triangle ABC [from figure]
The angle of incidence for
angle of incidence for the second mirror is 55.4 degree
Answer:
10.93m/s with the assumption that the water in the lake is still (the water has a speed of zero)
Explanation:
The velocity of the fish relative to the water when it hits the water surface is equal to the resultant velocity between the fish and the water when it hits it.
The fish drops on the water surface vertically with a vertical velocity v. Nothing was said about the velocity of the water, hence we can safely assume that the velocity if the water in the lake is zero, meaning that it is still. Therefore the relative velocity becomes equal to the velocity v with which the fish strikes the water surface.
We use the first equation of motion for a free-falling body to obtain v as follows;
v = u + gt....................(1)
where g is acceleration due to gravity taken as 9.8m/s/s
It should also be noted that the horizontal and vertical components of the motion are independent of each other, hence we take u = 0 as the fish falls vertically.
To obtain t, we use the second equation of motion as stated;
Given; h = 6.10m.
since u = 0 for the vertical motion; equation (2) can be written as follows;
substituting;
Putting this value of t in equation (1) we obtain the following;
v = 0 + 9.8*1.12
v = 10.93m/s