3-phase, cage induction motor takes a starting current at normal voltage of 5 times the full-load value and its full-load slip is 4 per cent. What auto-transformer ratio would enable the motor to be started with not more than twice full-load current drawn from the supply? What would be the starting torque under these conditions and how would it compare with that obtained by using a stator resistance starter under the same limitations of line current ?

answer:

i lost in getting someone back.

i learned that people are really two-faced and no matter how angelic someone appears they always have a devil side and that you cant trust anyone anymore no matter how dear they are to you.

if your wondering who it waz it waz mah still miss her tho

~batmans wife dun dun

answer:

try looking it up that is always

explanation:

answer:

1: productivity. productivity is an important reason for developing it systems.

2: information. it systems are key to delivering the information and intelligence you need to improve innovation.

3: cost reduction.

4: collaboration.

5: customer service.

explanation:

which ordered pair is not in the solution

y > -1/2x + 5

y ≤ 3x - 2

a) (5,3) b) (4,3)

c) (3,4) d) (4,4)