Answer:
1st one.
Explanation:
I think that because they the one who is going to get a ticket and also you will not gonna get in a car accident.
In order to develop this problem it is necessary to take into account the concepts related to fatigue and compression effort and Goodman equation, i.e, an equation that can be used to quantify the interaction of mean and alternating stresses on the fatigue life of a materia.
With the given data we can proceed to calculate the compression stress:



Through Goodman's equations the combined effort by fatigue and compression is expressed as:

Where,
Fatigue limit for comined alternating and mean stress
Fatigue Limit
Mean stress (due to static load)
Ultimate tensile stress
Security Factor
We can replace the values and assume a security factor of 1, then

Re-arrenge for 

We know that the stress is representing as,

Then,
Where
=Max Moment
I= Intertia
The inertia for this object is

Then replacing and re-arrenge for 



Thereforethe moment that can be applied to this shaft so that fatigue does not occur is 3.2kNm
Answer with Explanation:
The capillary rise in 2 parallel plates immersed in a liquid is given by the formula

where
is the surface tension of the liquid
is the contact angle of the liquid
is density of liquid
'g' is acceleratioj due to gravity
'd' is seperation between thje plates
Part a) When the liquid is water:
For water and glass we have
Applying the values we get

Part b) When the liquid is mercury:
For mercury and glass we have
Applying the values we get

The negative sign indicates that there is depression in mercury in the tube.
Answer:
a)
1) R16C ; Tn = 17 TMU
2) G4A ; Tn = 7.3 TMU
3) M10B5 ; Tn = 15.1 TMU
4) RL1 ; Tn = 2 TMU
5) R14B ; Tn = 14.4 TMU
6) G1B ; Tn = 3.5 TMU
7) M8C3 ; Tn = 14.7 TMU
8) P1NSE ; Tn = 10.4 TMU
9) RL1 ; Tn = 2 TMU
b) 3.1 secs
Explanation:
a) Determine the normal times in TMUs for these motion elements
1) R16C ; Tn = 17 TMU
2) G4A ; Tn = 7.3 TMU
3) M10B5 ; Tn = 15.1 TMU
4) RL1 ; Tn = 2 TMU
5) R14B ; Tn = 14.4 TMU
6) G1B ; Tn = 3.5 TMU
7) M8C3 ; Tn = 14.7 TMU
8) P1NSE ; Tn = 10.4 TMU
9) RL1 ; Tn = 2 TMU
b ) Determine the total time for this work element in seconds
first we have to determine the total TMU = ∑ TMU = 86.4 TMU
note ; 1 TMU = 0.036 seconds
hence the total time for the work in seconds = 86.4 * 0.036 = 3.1 seconds