superman

whaqt is an intentional tort

17 years ago

What is the heat of combustion of ethane, C2H6, in kilojoules per mole of ethane?

12 years ago

Q2_1_2 Answers 1) (TCX+2Lt0)/(G0I) where TCx=-t0L/3 2) t0(3L-x))/(2G0I) 3) x=3L

12 years ago

12

17 years ago

give me your number and i will lmao. you stupid

17 years ago

i just learned this in school today my teacher said the same thing you did that it is not necessary.

17 years ago

is superman the best

17 years ago

jnhgy

17 years ago

preshate it

17 years ago

8.4 - 2 = 6.4

15 years ago

Q2_1_1 TCx=-q0L/3

12 years ago

Q2_1_1 TCx=-t0L/3

12 years ago

Q2_1_2 for0¡Üx<L,d¦Õdx(x)=(TBX+2Lt0)/(G0I) forL<x¡Ü3L,d¦Õdx(x)=(t0(3L-x))/(2G0I) d¦Õdx(x0)=0atx0 in x=3L

12 years ago

Q2_1_2 1) (TCX+2Lt0)/(G0I) where TCx=-t0L/3 2) t0(3L-x))/(2G0I) 3) x=3L

12 years ago

Q2_1_1 TCx=-t0L/3

12 years ago

Q2_1_2 1) (TCX+2Lt0)/(G0I) where TCx=-t0L/3 2) t0(3L-x))/(2G0I) 3) x=3L

12 years ago

Q2_1_1 Answer TCx=-t0L/3

12 years ago

I need to know if this information is correct, because I solved the problems without correspondent images

12 years ago

Let me know if solutions are correct to go with the next problems.

12 years ago

Q2_1_2 Answers 1) (TCX+2Lt0)/(G_0I) where TCx=-t_0L/3 and I=pi*R^4/2 so answer: (2*TCX+4*L*t_0)/(G_0*pi*R^4) 2) t0(3L-x))/(2G0I) where I=i*R^4/2 so answer: ((t_0*(3L-x))/(pi*R^4) 3) x=3L

12 years ago

Where I can see images of the problems ?

12 years ago

OK Let me see Answer Q2_1_1: Sum=0 -Tx_1+INTEGRAL{3L-L}t_0*dx+TXC=0 so Tx_1=2*L*t_0+TXC {0<=x<=L} -Tx_2+INTEFRAL{3L-x}t_0*dx+TXC=0 so Tx_2=t_0*(3L-x)+TXC {L<=x<=3L} d*phi/dx =Tx/GI so in x<=_x<=L d*phi/dx= (TXC+2*L*t_0)/G_0*I in L<=x<=3L d*phi/dx= (t_0*(3...

12 years ago

Is it correct ? If yes I will going with the second

12 years ago

I'm waiting your answers, to follow with next items.

12 years ago

Answers Q2_1_1 TXC=-t_0*L

12 years ago

Answers Q2_1_2 a) T(x) in 0<=x<=L T(X)=t_0*L d*phi/dx=T(x)/(GI) where I=(pi*R^4)/2 so d*phi/dx=(2*t_0*L)/(pi*G_0*R^4) b) T(x) in l<=x<=3L T(x)=t_0*(2*L-x) so d*Phi/dx=(2*t_0+(2*L-x))/(pi*G_0*R^4) c)(d*phi/dx)(x)=0 when 0=(2*t_0+(2*L-x))/(pi*G_0*R^4) so (2...

12 years ago

Q2_1_1 TXC=-t_0*L Q2=1_2 a) 0<=x<L ===>> (2*t_0*L)/(pi*G_0*R4) b)L<=x<=3L =>>(2*t_0*(2*L-))/pi*G_0*R^4) c) x=2*L

12 years ago

yes, I got a green check.

12 years ago

Q2_1_2 a) T(x) in 0<=x<=L T(X)=t_0*L d*phi/dx=T(x)/(GI) where I=(pi*R^4)/2 so d*phi/dx=(2*t_0*L)/(pi*G_0*R^4) b) T(x) in l<=x<=3L T(x)=t_0*(2*L-x) so d*Phi/dx=(2*t_0+(2*L-x))/(pi*G_0*R^4) c)(d*phi/dx)(x)=0 when 0=(2*t_0+(2*L-x))/(pi*G_0*R^4) so (2*L-x)=0...

12 years ago

Q2_1_2 a) T(x) in 0<=x<=L T(X)=t_0*L d*phi/dx=T(x)/(GI) where I=(pi*R^4)/2 so d*phi/dx=(2*t_0*L)/(pi*G_0*R^4) b) T(x) in l<=x<=3L T(x)=t_0*(2*L-x) so d*Phi/dx=(2*t_0+(2*L-x))/(pi*G_0*R^4) c)(d*phi/dx)(x)=0 when 0=(2*t_0+(2*L-x))/(pi*G_0*R^4) so (2*L-x)=0...

12 years ago

Q2-1-1 TXC=-t_0*L

12 years ago

300N

10 years ago

4n

10 years ago

do you mean social studies (hint) **bill of rights**

9 years ago

Excessive bail shall not be required, nor excessive fines imposed, nor cruel and unusual punishments inflicted *****(hint) google search. avoid Wikipedia.

9 years ago

in my opinion i think 2nd one is correct but the correct was to say it would have been something like " the boy walked along the sidewalk safely"

9 years ago

perimeter mean the length of all the sides combined, which in your case is 8. since each side of a square is the same the same length your answer for one side will be the same for the other side. basically 8/4 sides= will give the answer for 1 side which...

9 years ago

according to my calculations the answer is 100% 434 🤓

2 years ago