Explanation:
first changing kilo ohm to ohm
860000 = 860 kΩ
and change 34 micro ampare to ampare
34 μA=3.4×10^-5
recalling the equation V=I*R
V= 3.4×10^-5×860000
v=29.24
Answer:
The Estimated uncertainty in a nominal displacement of 2 cm at the design stage is plus or minus 0.0124cm
Explanation:
uncertainty in a nominal displacement
= (u^2 + v^2)^(1/2)
assume from specifications that k = 5v/5cm
= 1v/cm
u^2 = (0.0025*2)^(2) + (0.005*10*2)^2 + (0.0025*2)^2
= 0.01225v
v = 2v * 0.001
= 0.002v
uncertainty in a nominal displacement
= (u^2 + v^2)^(1/2)
= ((0.01225)^2 + (0.002)^2)^(1/2)
= 0.0124 cm
Therefore, The Estimated uncertainty in a nominal displacement of 2 cm at the design stage is plus or minus 0.0124cm
The load is placed at distance 0.4 L from the end of area.
<h3>What is meant by torque?</h3>
The force that can cause an object to rotate along an axis is measured as torque. Similar to how force accelerates an item in linear kinematics, torque accelerates an object in an angular direction. A vector quantity is torque.
Let the beam is of length L
Now the stress on both the end is the same now we can say that torque on the beam due to two forces must be zero
also, we know that stress at both ends are same
Now from two equations we have
solving the above equation we have
so the load is placed at distance 0.4 L from the end of area.
The complete question is:
47. the beam is supported by two rods ab and cd that have cross-sectional areas of and , respectively. determine the position d of the 6-kn load so that the average normal stress in each rod is the same.
To learn more about torque refer to:
brainly.com/question/20691242
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Answer:
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