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
Answer:
me you same I didn't understand good luck
Answer: c. The Professional Engineers Act and Board Rules
Explanation:
The reference source may be consulted to answer questions regarding the Professional Engineers Act is the The Professional Engineers Act and Board Rules.
The Professional Engineers Act and Board Rules is an Act that was established in order to regulate the qualifications for professional engineered, register them and also make sure that their conducts and behavior are looked into.
Answer:
To run. The machine one at a time
Explanation:
Answer:
6.65 kPa.
- 73.3 kPa.
Explanation:
Without much ado let's jump right into the solution to the problem given. It is given that the pressure = 50 mmHg and the solution to this question is to write out the value of the pressure in kpa and kPa gauge.
P(a) = 0.05m × 133 kN/m³ = 6.65 kPa.
The P(gauge) =( [ 0.05 × 13.6 × 9810] ÷ 1000 ) - 80 = - 73.3 kPa.
