Question

The total weight of a spool of number 16 wire is 11.01 pounds. The weight of the spool alone is 1.89 pounds. Find the weight of the wire only. The cold resistance of an armature measures 0.096 ohm. After running for 5 hours, the resistance is 1.25 ohms. What is the difference in the 5. A pin insulator measures 1.312 inches in diameter at the small end and 1.9375 inches at the large end. What is the difference in diameters 6. The field current of a motor is 1.12 amperes at no load. However, when the full load is applied, the current increases to 1.87 amperes. What is UNI12 SUBTRACTION OF DECIMAL FRACTIONS 3. resistance? between the small end and the large end? the difference in amperes BOU es from no load to full load? 7. A bearing bushing has an outside diameter of 3.9375 inches and a wall thickness of 0.28125 inch. Find the inside diameter of the bushing. ISO90.0 8.) A special resistance wire has a diameter of 0.037 inch. The next smaller size is 0.0345 inch in diameter. Find the difference in the diameters. 21 ol NOTE: Use this diagram for problems 9–13. TO CO.O 580200 A

Answer 1

NB...While calculating the difference between two values,we neglect the sign of final value.

3)...
Total weight of pool having 16 wires =11.09 pounds
Weight of the pool =1.89
Therefore the weight of the wire alone =11.09-1.89
=9.12 pounds

4)...
Cold resistance of the armature measured =0.096 ohm
Resistance of the armature after 5 hours =1.25
Difference in the resistance   = initial resistance - final resistance
=0.096-1.25
= 1.154ohm

5)...
Diameter of pin insulator at small end =1.312 inches
Diameter of of pin insulator at larger end =1.9375
Difference in diameter at smaller and larger end =1.312-1.9375
= 0.6255 inches

6)...
Field current of motor at no load =1.12A
Field current of motor at full load =1.87A
Difference from no load to full load in ampere =1.12-1.87
=0.75A

7)...
Outside diameter of a bearing bushing =3.9375
Wall thickness of bearing bushing   = 0.28125
Inside diameter of bushing = outside diameter- wall thickness
= 3.9375- 0.28125
= 3.65625 inches

8)..
Diameter of a resistance wire   = 0.037 inches
Diameter of next smaller resistance wire = 0.0345
Difference in diameter = 0.037 - 0.0345
= 0.0025 inches

THANKYOU..SIR/MADAM..Hope this helped you.