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For E = 200 gpa and i = 65. 0(106) mm4, the slope of end a of the cantilevered beam is mathematically given as
A=0.0048rads
<h3>What is the slope of end a of the cantilevered beam?</h3>Generally, the equation for the is mathematically given as
Therefore
A=\frac{10+10^2+3^2}{2*240*10^9*65*10^6}+\frac{10+10^3*3}{240*10^9*65*10^{-6}}
A=0.00288+0.00192=0.0048rads
A=0.0048rads
In conclusion, the slope is
A=0.0048rads
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Answer:
6.0 m/s
Explanation:
According to the law of conservation of energy, the total mechanical energy (potential, PE, + kinetic, KE) of the athlete must be conserved.
Therefore, we can write:
or
where:
m is the mass of the athlete
u is the initial speed of the athlete (at the bottom)
0 is the initial potential energy of the athlete (at the bottom)
v = 0.80 m/s is the final speed of the athlete (at the top)
is the acceleration due to gravity
h = 1.80 m is the final height of the athlete (at the top)
Solving the equation for u, we find the initial speed at which the athlete must jump: