2007-06-14 09:22:35 · 7 answers · asked by ali859us 1 in Politics & Government ➔ Other - Politics & Government
no...why would they do this?according to the NNPT they have every right to a nuclear program as long as it is in peaceful purposes...and since there is nothing to show the contrary they are not doing anything illegal...
2007-06-14 09:25:57 · answer #1 · answered by Tinkerbell05 6 · 1⤊ 0⤋
No reason Iran can not have a program for establishing nuclear power. The uranium require for fuel for a nuclear power plant is completely different than weapons grade plutonium. Should their plants be inspected by outside agencies? Why not but then should we also be inspecting the nuclear power plants of Pakistan, North Korea, etc?
2007-06-14 10:38:55 · answer #2 · answered by ndmagicman 7 · 0⤊ 0⤋
Iran will continue to develop nuclear weapons with the help of Russia. They will never stop until they turn Israel into a parking lot. That is they only thing they are committed to. Sick huh?
2007-06-14 09:36:40 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Iran will not stop its nuclear program unless we (or the Israelis) stop it for them. Iran will continue to push for dominance of the middle east by supplying arms and munitions to the terrorist groups and by pushing them to fight the US and Israel. Iran's goal is to achieve power in the middle east to control all of it and they are preceding in an open manner. How long will we take their s**t and do nothing but talk?
2007-06-15 03:14:52 · answer #4 · answered by Anonymous · 0⤊ 0⤋
As soon as Iran get a deal with the US then some other small country in need will start a program to get something from the US. It's a vicious cycle. Although as we see if you don't even start a program you get invaded.
2007-06-14 09:32:48 · answer #5 · answered by Jose R 6 · 0⤊ 0⤋
Not until the United States of America stops them. And about time we did.
2007-06-14 09:25:56 · answer #6 · answered by SallyJM 5 · 1⤊ 0⤋
Why should they if U.S.A doesn't?
2007-06-14 09:47:46 · answer #7 · answered by Bart Simpson 1 · 0⤊ 0⤋