Calculate the energy released when 1.00 kilograms of mass is converted entirely to energy. [(1 J = 1 kg times m squared over s squared ) and the speed of light = 3.00 × 108 m over s ]
Answer

3.00 × 108 J

9.00 × 1016 J

1.00 × 10-7 J

1.11 × 10-17 J

Question

Calculate the energy released when 1.00 kilograms of mass is converted entirely to energy. [(1 J = 1 kg times m squared over s squared ) and the speed of light = 3.00 × 108 m over s ]
Answer
		
3.00 × 108 J
		
9.00 × 1016 J
		
1.00 × 10-7 J
		
1.11 × 10-17 J

abhisar · Answer

LOL i just did it in ur previous question !

abhisar · Answer

C stands for the speed of light. So for loss of say 1Kg we get an energy equivalent to $\sf (3\times 10^8)^2=9\times 10^{16}Joules$

abhisar · Answer

Got it ?

abhisar · Answer

That result is obtained using $\large E=mc^2$

anonymous · Answer

so this is the one you were talking about? hahaha

anonymous · Answer

or are both the answers you gave on both questions correct?

abhisar · Answer

yes ofcrse...!

anonymous · Answer

so to the e=mc2 it was c also? and this one is c?

abhisar · Answer

no for both it's b

anonymous · Answer

dont do that, its late and im sleep deprived lol

abhisar · Answer

lol..m serious for both it's b...what are you doing ?
i solved the whole question and despite that u r not able to choose the correct option ?

anonymous · Answer

ok so this answer is b. what other question are you talking about because i think i have three questions open

abhisar · Answer

What is your other question ?

anonymous · Answer

What is the binding energy for a C-12 nucleus with a mass defect of 0.0993 amu? [(1 amu= 1.66 × 10-27 ,1 J = 1 kg times m squared over s squared ) and the speed of light = 3.00 × 108 m over s ] Answer 1.48 × 10-11 J 1.24 × 10-12 J 4.12 × 10-21 J 8.94 × 1015 J

abhisar · Answer

You can solve this similarly using $\sf E=mc^2$

$\sf E=0.0993\times 1.66 \times 10^{-27}\times (3\times 10^8)^2$

anonymous · Answer

easy way?

abhisar · Answer

That's the easiest way !

abhisar · Answer

Come on u just have to multiply the terms...

kirbykirby · Answer

yep I agree.. lol

abhisar · Answer

$\color{blue}{\text{Originally Posted by}}$ @Abhisar
You can solve this similarly using $\sf E=mc^2$

$\sf E=0.0993\times 1.66 \times 10^{-27}\times (3\times 10^8)^2\ \Rightarrow E=1.48\times 10^{-27}\times 10^{16}\ \Rightarrow E= ?$
$\color{blue}{\text{End of Quote}}$

anonymous · Answer

1.48 X 10 to the negative 11

kirbykirby · Answer

yeah

abhisar · Answer

$\huge \checkmark$

anonymous · Answer

Which of the following are true for E = mc2?
Answer
		
implies that a small change in mass produces a great deal of energy
		
mainly used for chemical reactions
		
implies that a small change is mass produces a small amount of energy

anonymous · Answer

how do i do the comments like you are doing lol

abhisar · Answer

Now, observe urself......change in mass of 1kg produces an energy of $\sf 9\times 10^{16}$. So is it large energy produced or small energy produced ?

anonymous · Answer

large

abhisar · Answer

So the correct option should be ?

anonymous · Answer

for the previous question or the one i just posted here?

abhisar · Answer

for the one you just posted...this is the reason i ask everyone to post one question per thread.

anonymous · Answer

sorry i was trying to keep things simple since im always asking questions

abhisar · Answer

anyways....is every question clear to you ?

anonymous · Answer

it was 1.48 X 10 to the negative 11

anonymous · Answer

yes except the one about what is true for the formula e=mc2

abhisar · Answer

ask it in a new post