Answer:
The shortest distance d to the edge of the plate is 66.67 mm
Concepts and reason
Moment of a force:
Moment of a force refers to the propensity of the force to cause rotation on the body it acts upon. The magnitude of the moment can be determined from the product of force’s magnitude and the perpendicular distance to the force.
Moment(M) = Force(F)×distance(d)
Moment of inertia ( I )
It is the product of area and the square of the moment arm for a section about a reference. It is also called as second moment of inertia.
First prepare the free body diagram of sectioned plate and apply moment equilibrium condition and also obtain area and moment of inertia of rectangular cross section. Finally, calculate the shortest distance using the formula of compressive stress (σ) in combination of axial and bending stress
Solution and Explanation:
[Find the given attachments]
Answer:
Answer below :)
Explanation:
<u>SIZE OF THEODOLITE:</u> A theodolite is designated by diameter of the graduated circle on the lower plate. The common sizes are 8 cm to 12 cm while<em> 14 cm</em> to <em>25 cm</em> instrument are used for triangulation work.
Complete Question:
A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm if the tool moves at 400 mm per second.
Answer:
26 mins 40 secs
Explanation:
Reduction in depth, Δd = 20 mm
Depth of cut, 
Number of passes necessary for this reduction, 
n = 20/0.5
n = 40 passes
Tool width, w = 5 mm
Width of metal plate, W = 200 mm
For a reduction in the depth per pass, tool will travel W/w = 200/5 = 40 times
Speed of tool, v = 100 mm/s
minimum time required to reduce the depth of the plate by 20 mm:
number of passes * Time/pass
n * Time/pass
40 * 40
1600 = 26 mins 40 secs
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
Answer:
what we have to answer please mention questions
