Answer:
It is a non profit organization that dedicates to licensing professional engineers and surveyors
Explanation:
Answer:
8861.75 m approximately 8862 m
Explanation:
We need to remember Newton's 2nd Law which says that the force experienced by an object is proportional to his acceleration and that the constant of proportionality between those two vectors correspond to the mass of the object.
for the weight of an object (which is a force) we have that the acceleration experienced by that object is equal to the gravitational acceleration, obtaining that 
For simplicity we work with
despiting the effect of the height above sea level. In this problem, we've been asked by the height above sea level that makes the weight of an object 0.30% more lighter.
In accord with the formula
the "normal" or "standard" weight of an object is given by
when
, so we need to find the value of
that makes
meaning that the original weight decrease by a 0.30%, so now we operate...
now we group like terms on the same sides
we cancel equal tems on both sides and obtain that 
Answer:
The dial bore gauge measures the inside of round holes, such as the bearing journals. This one tool can measure 2″ up to 6″ diameter holes. Both tools are needed in order to check the interior and exterior dimensions of the crankshaft, rods and engine block journals, as well as the thickness of the bearings themselves.
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Answer:
Any engineering job would be good YOU should be the one choosing which job.
Explanation:
Engineering is a great outlet for the imagination, and the perfect field for independent thinkers.
Answer:
a) 246.56 Hz
b) 203.313 Hz
c) Add more springs
Explanation:
Spring constant = 12000 N/m
mass = 5g = 5 * 10^-3 kg
damping ratio = 0.4
<u>a) Calculate Natural frequency </u>
Wn = √k/m = 
= 1549.19 rad/s ≈ 246.56 Hz
<u>b) Bandwidth of instrument </u>
W / Wn = 
W / Wn = 0.8246
therefore Bandwidth ( W ) = Wn * 0.8246 = 246.56 * 0.8246 = 203.313 Hz
C ) To increase the bandwidth we have to add more springs