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lutik1710 [3]
3 years ago
6

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.

Physics
1 answer:
ICE Princess25 [194]3 years ago
7 0

Answer: a) 1.98inch

b) 2.737 2.44

c) 0.497inch

d) 1.58inch

Explanation:

Solution is attached in the pictures below

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A stone is thrown horizontally from the top of an inclined plane (angle of inclination θ). How would I find the initial speed of
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S = V t     where S is the horizontal distance traveled

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An event occurs in system K' at x' = 2 m, y' = 3.7 m, z' = 3.7 m, and t' = 0. System K' and K have their axes coincident at t =
fiasKO [112]

Answer:

Coordinates of event in system K are (x,y,z,t)=(5.103m , 3.7m , 3.7m , 1.57×10⁻⁸s)

Explanation:

To find the coordinates of event in system K ,we have to use inverse Lorentz transformation

So

x=\frac{x^{|}+vt^{|} }{\sqrt{1-\frac{v^{2} }{c{2} } } } \\x=\frac{2m+0.92c(0) }{\sqrt{1-\frac{(0.92c)^{2} }{c{2} } } }\\x=5.103m\\y=y^{|}\\ y=3.7m\\z=z^{|}\\ z=3.7m

for t

r=\frac{1}{\sqrt{1-v^{2} } } \\r=\frac{1}{\sqrt{1-(0.92)^{2} } } \\r=2.551\\t=r(t^{|}+vx^{|}/c^{2}   )\\t=2.551(0s+(0.92c)(2)/c^{2} )\\t=1.57*10^{-8}s

Coordinates of event in system K are (x,y,z,t)=(5.103m , 3.7m , 3.7m , 1.57×10⁻⁸s)

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