If you multiply current by time, you get electric charge. In this case, because of the "milli" prefix for the current, you would also get "milli" in the answer, that is, millicoulomb.
1000
Get a and b // where a and b are 2 numbers taken from the user Set Ma to a // Ma is the multiple of a Set Mb to b // Mb is the multiple of b While Ma<>Mb //(condition is that Ma not equal Mb) { if (Ma > Mb) then // finding which multiple is smaller Set Mb to Mb + b // we add to the smaller multiple the input value Else Set Ma to Ma + a End if } Print "the LCM of the numbers is" + Ma
My mother - ma mère
Well it depends on what you are using for a pulley the mechanical advantage is equal to the number of ropes lifting the object such as if you have one pulley the MA (mechanical advantage) is equal to 1 if you have two pullies the MA is 2 if you are using a lever such as a seesaw you have to move the fulcrum as close to the object being lifted and have to longest possible input arm. If you modify a seesaw a 60 pound child can lift a 200 pound adult. That is about all i know hope it helps if you are using a different simple machine or need more help email me at : icecbejohn@yahoo.com
The ideal MA is 47.
0.1 A = 100 mA
20
To state the obvious- 1000 mA is 10x the amount of 100 mA 1 A = 1000 mA therefore 100 mA = 0.1 A
% / Sqrt output mA / Non square output mA 0% / 4 / 4 10% / 9.059644256 / 5.6 20% / 11.15541753 / 7.2 25% / 12 / 8 30% / 12.76356092 / 8.8 40% / 14.11928851 / 10.4 50% / 15.3137085 / 12 60% / 16.39354671 / 13.6 70% / 17.38656042 / 15.2 75% / 17.85640646 / 16 80% / 18.31083506 / 16.8 90% / 19.17893277 / 18.4 100% / 20 / 20 if u want to find mA at x% Then let Y= sq. rt(x/100) now u can to find mA of Y directly i.e mA= Y*100*.16+4 thats the mA for sq. of x eg: you want to find mA at 49% at a range of 0 to 300 cubic.meter/hr (or anything) then Y= sq.rt (49/100) =0.7 mA---> 0.7*100*.16+4 = 15.2 mA
Force, you get? Force = ma or mass x acceleration or your ma!
If you are talking about 4 mA = 0% and 20 mA = 100% then the formula would be a linear equation: y=mx + b y=6.25(x) - 25 y= percentage x=mA ouput So for example put 4 mA in place of x and you get 0% for y and if you use 20 mA in place of x you get 100% for y. You can rewrite the equation if only Percent input is known to find mA output it would be X= (Y + 25)/6.25 this way you input the percent in for (Y) and you have the mA output for X
Following Too Closely V.T.L. Sec. 1129Passing Red LightV.T.L. Sec. 1111dPassing Stop SignV.T.L. Sec. 1172Unsafe Lane ChangeV.T.L. Sec. 1128Failure To YieldV.T.L. Sec. 1140, 1141, 1142, 1143Improper PassingV.T.L. Sec. 1124Failure To Obey Traffic Control DeviceV.T.L. Sec. 1110(a)
The 4-20 ma is used more often in analog signaling than 0-20 ma because sufficient power dissipation. The hence is the upper rage and was reduced to 20 ma.
F = ma, so we have 100 = m x 5, so m = 20 kg.
Yes a 4 - 20 ma loop can be tested.
because 4 ma for live zero and at 20 ma no spark will generate and it is safe current this is why we use 4-20 ma and another thing is tx and cotrollers even plc and Dcs desinged to accept 4- 20 ma singal which is standard
One of the newtons laws(i think) says F=ma so 100=5a which then leads to 100/5=20=a. So, assuming no friction, it is 20 m/s/s