Recent content by haengbon

I can't seem to understand why both the numerator and denominator needs to be multiplied by y2 :(

I tried it again and got -3 for 4x=1/64 :D is that correct now? ^^ thank you for the help by the way :)

yes xD !

Homework Statement solve for x using the expansion method x + y + z =1 x + y + 2z =2 x + y + 3z = 1 Homework Equations none The Attempt at a Solution 1 1 1 1 1 1 1 2 1 1 1 1 3 1 1 1 1 1 1 1 1 1 2 1 1 1 1 3 1 1 1 1 1 1 1 1...

Homework Statement log39 + log41/64 + log51 Homework Equations none The Attempt at a Solution I tried answering it by... (3x=9) + (4x=1/64) + (5x=1) because of this, I got 5. Is that correct?

Homework Statement 9 - 9y-2 ____________ y-1 - y0 this whole thing is then raised to 2 (I can't seem to format it that way:( ) Homework Equations none The Attempt at a Solution I'm hoping someone would guide me through this :(

um, if that's so, is it then possible to be solved? I need to use this formula :D ? Fe = \frac{ Kq1q2 }{ r2 }

I'm not really sure since I only remembered this problem by memory :( but if the third body WAS neutral and they ARE in equilibrium, how would you answer this? :D

Homework Statement Three point charges, 3.0 \muc, -5.0\muc and 0 \muc. If the distance from point 1 (3.0) and point 2 (-5.0) is 40m, what is the distance from point 2 to point 3? Homework Equations coulombs law [b The Attempt at a Solution I don't know how to start...

Homework Statement A man jumps on a trampoline with an angle of 53\circ with respect to the horizontal. What velocity does he need in order to be caught by the other man who is 5.2 m away from him, and 6.1 m above the ground? Homework Equations so I guess... X= 5.2 m Y= -6.1 m...

so that's the only problem ? :D so If I switched/corrected the two, then my answer would immediately be correct? :D

Homework Statement What is the Empirical and Molecular Formula of caffeine, given these information: 0.376g caffeine would yield to 0.682 CO2 , 0.174g H2O and 0.110g N. The molecular weight if caffeine is 194 g/mole. Homework Equations --- The Attempt at a Solution 0.682 CO2...

Homework Statement A box weighing 20 kg is on an inclined plane with an angle of 30\circ.The coefficient of friction is 0.3. What is the acceleration of the box?Homework Equations Newton's Second law ( I think? ) g= 10 m/s2 The Attempt at a Solution\sumy = 0 NF - Wysin30\circ = 0 NF...

oops! I'm very sorry for that :)) the one in my paper was 0.1 moles of A :D haha xD thank you! um, but the part that I'm really confused about is this one ... 4A + 3O2 -> 2 A2O3 if we take the balancing off.. A + O2 -> A2O3 Where did we get A + O2 ? Is this common knowledge...

8.0 - 5.6 = 2.4 gO 2.4 gO x 1 mole O / 16 gO = 0.15 moles O A:O = 2:3 A:0.15 = 2:3 \frac{A}{0.15} x \frac{2}{3} 3A = (0.15)(2) 3A = 0.30 A = 0.1 moles Oxygen a) AW = g/mole AW = 5.6 gO / 0.1 moles O AW = 56 g/mole b) 2:3 A2O3 = (56)(2) + (16)(3) = 160 g...