Help with Energy unit Joules EASY

In summary: W [J] = F * d [N * m].In summary, the sun releases 32.4 x 10^25 Joules of energy each second by burning 3.6 x 10^9 kg of matter using the formula E = mc^2, where c is the speed of light at 3 x 10^8 m/s.
  • #1
daNAV1GATOR
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[SOLVED] Help with Energy unit ... Joules... EASY!

Homework Statement



The sun burns up 3.6 x 10[tex]^{9}[/tex] kg of matter each second.
Using the formula E = mc[tex]^{2}[/tex], determine how much energy the
sun releases each second. [Speed of light: 3 x 10[tex]^{8}[/tex] m[tex]/[/tex]s]


Homework Equations



E = mc[tex]^{2}[/tex]


The Attempt at a Solution



E = mc[tex]^{2}[/tex]
. .= (3.6 x 10[tex]^{9}[/tex] kg)(3 x 10[tex]^{8}[/tex] m[tex]/[/tex]s)[tex]^{2}[/tex]
. .= (3.6 x 10[tex]^{9}[/tex] kg)(9 x 10[tex]^{16}[/tex] m[tex]^{2}[/tex][tex]/[/tex]s[tex]^{2}[/tex])
. .= 32.4 x 10[tex]^{25}[/tex] [tex]\frac{kg \cdot m^2}{s^2}[/tex]



Now I'm stumped with what to do with the units... I know it should be Joules per second. But how do I show in my solution that Joules will come out? Joules is N[tex]\cdot[/tex]m!

Do I have to change the kg into N? ... My textbook says that the magnitude of c[tex]^{2}[/tex] is 9 x 10[tex]^{16}[/tex] Joules per kilogram.

Please check my calculations also!
Thank You! :smile:
 
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  • #2
First hint: How is "Joule" defined?
Second hint: If E = mc[tex]^{2}[/tex] is valid, and you have SI units to the right, then you must obtain SI units on the left hand side as well.

The second hint would not be true if you was working with the natural units in sub-atomic phyiscs, where both c and h-bar = 1 i.e E = m, then one must do a more careful unit analysis.
 
  • #3
A Joule is N m and a Newton, N, is kg m/s2. So a Joule is kgm2/s2. That's exactly what you have!
 
  • #4
Hmmm...

well Joule is the unit of work. N*m
Newton x meter
 
  • #5
HallsofIvy said:
A Joule is N m and a Newton, N, is kg m/s2. So a Joule is kgm2/s2. That's exactly what you have!


So that's understood? I'll then just add...

E = mc[tex]^{2}[/tex]
. .= (3.6 x 10[tex]^{9}[/tex] kg)(3 x 10[tex]^{8}[/tex] m[tex]/[/tex]s)[tex]^{2}[/tex]
. .= (3.6 x 10[tex]^{9}[/tex] kg)(9 x 10[tex]^{16}[/tex] m[tex]^{2}[/tex][tex]/[/tex]s[tex]^{2}[/tex])
. .= 32.4 x 10[tex]^{25}[/tex] [tex]\frac{kg \cdot m^2}{s^2}[/tex]
. .= 32.4 x 10[tex]^{25}[/tex] Joules

Simple as that?
Yipee! :biggrin:

Are my calculations correct? exponents...
 
  • #6
Yes, since F [N] = m*a [kg * m/s^2 ]
 

FAQ: Help with Energy unit Joules EASY

1. What is energy and how is it measured?

Energy is the ability to do work or cause change. It can be measured in joules (J), which is the standard unit of energy in the International System of Units (SI).

2. How is energy related to the unit Joules?

The joule is the unit of energy defined as the amount of work done by a force of one newton moving an object one meter in the direction of the force. Therefore, it is used to measure the amount of energy transferred or used in a system.

3. How can I convert between joules and other units of energy?

To convert between joules and other units of energy, you can use conversion factors. For example, 1 joule is equal to 0.239 calories, 0.0002778 kilowatt-hours, and 9.478x10^-4 British thermal units (BTUs).

4. What are some real-life examples of joules and energy?

Joules and energy are present in many aspects of our daily lives. Some examples include the energy used to power electronic devices, the energy stored in food that we consume, and the energy generated by the sun that fuels photosynthesis in plants.

5. How is energy conservation related to the unit joules?

Energy conservation is the practice of reducing energy use to conserve resources and protect the environment. The unit joule is important in energy conservation because it allows us to accurately measure and track energy use, making it easier to identify areas where energy consumption can be reduced.

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