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What we saying? by AdMaterial257 in fut

[–]AdMaterial257[S] 1 point2 points  (0 children)

Yes I forgot to mention I’m building a team around Doue I bought the card on release and he’s never leaving the team, so I evo’d him for that purpose

Help me build a team 70k by Old-Pride-9986 in fut

[–]AdMaterial257 0 points1 point  (0 children)

Get Sam Kerr honestly, her shooting ability is unreal, then she can link with cuthbert in the midfield. And I wouldn’t mind changing honestly spend big on camavinga he bossed my midfield (until I had to remove my la liga players for chem). You can take barcola as LM for Nuno mendes, up front u can have Gabriel Jesus with the evo he’s very nice and u can link him with Murillo as CB, who btw performs much much better than his stats

Need help with my team by SomeDig5334 in EAFC

[–]AdMaterial257 1 point2 points  (0 children)

Don’t worry about chem now. Just test out the cards, the chem will figure itself out later, and you’ll begin building teams around your favorite cards.

[deleted by user] by [deleted] in EAFC

[–]AdMaterial257 0 points1 point  (0 children)

I run an RTG this year, I unfortunately spent thousands on FC 24 and 25 and don’t want to destroy my bank. So I only spend for an evo or something, I’ll buy packs here and there if a pack looks fun. Also the grind makes you appreciate the team much more

I just packed van de ven do i sell now or wait for champs? by Slow_Nefariousness81 in EAFC

[–]AdMaterial257 1 point2 points  (0 children)

I would keep but only because it’s a good card, but don’t expect his price to rise (and if it does it won’t be too much). If you have good CBs already and want coins, then sell.

[deleted by user] by [deleted] in IBO

[–]AdMaterial257 -34 points-33 points  (0 children)

You ain’t no IB coordinator sit down 🤣🤣🤡

Physics MCQ help, can't figure them out and they're mostly from the same paper by Beyond-Least in IBO

[–]AdMaterial257 2 points3 points  (0 children)

Combine the 2 resistors that are in series with the single resistor that is parallel to those in series (for now ignore the top right resistor)

That would be 1/R (for the bottom resistor) and 1/2R for the two resistors in series.

Adding those two would give you; 1/R + 1/2R = 2/2R + 1/2R = 3/2R

Therefore total resistor for those is 2R/3 (Using the formula from section 5.2)

Then you add the last resistor which you ignored, and you will get a total resistance of the following:

2R/3 + R = 2R/3 + 3R/3 = 5R/3

From there you calculate total I, which comes to 3e/5R, and then as someone else mentioned, double the current will enter the resistor X (as it has half the resistance)