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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They are formed from erosion and weathering.
The weight of the person is given by:
W = mg
W = weight, m = mass, g = gravitational acceleration
Given values:
m = 40kg, g = 9.81m/s²
Plug in and solve for W:
W = 40(9.81)
W = 390N
Ok. PEMDAS tells us to take care of the square first. When we do that, the denominator becomes
(6.4)^2 x 10^12
= 40.96 x 10^12 .
Now it's just a matter of mashing out the fraction.
The 'mantissa' (the number part) is
6/40.96 = 0.1465
and the order of magnitude is
10^24 / 10^12 = 10^12 .
Put it all together and you've got
1.465 x 10^11 .
