Conservation of Mass and EnergyConversion of mass to energy in chemical/nuclear reactionsConversion of mass...
Do items de-spawn?
Proving a statement about real numbers
Why restrict private health insurance?
Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?
Professor forcing me to attend a conference, I can't afford even with 50% funding
Can we track matter through time by looking at different depths in space?
Having the player face themselves after the mid-game
What is the generally accepted pronunciation of “topoi”?
What do *foreign films* mean for an American?
Why is there an extra space when I type "ls" in the Desktop directory?
What is this diamond of every day?
Are there historical instances of the capital of a colonising country being temporarily or permanently shifted to one of its colonies?
Can I use a violin G string for D?
What is better: yes / no radio, or simple checkbox?
I can't die. Who am I?
Plausibility of Mushroom Buildings
Called into a meeting and told we are being made redundant (laid off) and "not to share outside". Can I tell my partner?
Has a sovereign Communist government ever run, and conceded loss, on a fair election?
Minimizing with differential evolution
School performs periodic password audits. Is my password compromised?
Street obstacles in New Zealand
Source permutation
From an axiomatic set theoric approach why can we take uncountable unions?
Specifying a starting column with colortbl package and xcolor
Conservation of Mass and Energy
Conversion of mass to energy in chemical/nuclear reactionsConversion of mass to energy in chemical/nuclear reactionsCan non-free forces change the rest mass?Why does mass change in to energy during a nuclear change?Energy & Mass of a PhotonWhat is the argument for detailed balance in chemistry?Would impact angle matter on relativistic impactor?Hypothetical special relativity with mass conservationDoes the mass of object really increase?Some calculations on the energy consumption of a relativistic rocketIf mass and energy are same what will be the equivalent of a homogeneous ball in terms of energy and information?
$begingroup$
I was thinking about some physics (relativity in particular), when it suddenly occurred to me that all my life I had been balancing chemical equations assuming conservation of mass, but I was disregarding energy!
For example, consider combustion:
$$CH_4 + 2O_2 >>> 2H_2O + CO_2 + {Energy}$$
However, since energy was released, some mass should have been converted to energy right? Why is the equation reflecting a balance in mass?
special-relativity conservation-laws mass-energy physical-chemistry
$endgroup$
add a comment |
$begingroup$
I was thinking about some physics (relativity in particular), when it suddenly occurred to me that all my life I had been balancing chemical equations assuming conservation of mass, but I was disregarding energy!
For example, consider combustion:
$$CH_4 + 2O_2 >>> 2H_2O + CO_2 + {Energy}$$
However, since energy was released, some mass should have been converted to energy right? Why is the equation reflecting a balance in mass?
special-relativity conservation-laws mass-energy physical-chemistry
$endgroup$
$begingroup$
Possible duplicate of physics.stackexchange.com/questions/11449/… ?
$endgroup$
– Shufflepants
14 mins ago
add a comment |
$begingroup$
I was thinking about some physics (relativity in particular), when it suddenly occurred to me that all my life I had been balancing chemical equations assuming conservation of mass, but I was disregarding energy!
For example, consider combustion:
$$CH_4 + 2O_2 >>> 2H_2O + CO_2 + {Energy}$$
However, since energy was released, some mass should have been converted to energy right? Why is the equation reflecting a balance in mass?
special-relativity conservation-laws mass-energy physical-chemistry
$endgroup$
I was thinking about some physics (relativity in particular), when it suddenly occurred to me that all my life I had been balancing chemical equations assuming conservation of mass, but I was disregarding energy!
For example, consider combustion:
$$CH_4 + 2O_2 >>> 2H_2O + CO_2 + {Energy}$$
However, since energy was released, some mass should have been converted to energy right? Why is the equation reflecting a balance in mass?
special-relativity conservation-laws mass-energy physical-chemistry
special-relativity conservation-laws mass-energy physical-chemistry
edited 1 hour ago
Aaron Stevens
12.8k42248
12.8k42248
asked 2 hours ago
Dude156Dude156
1307
1307
$begingroup$
Possible duplicate of physics.stackexchange.com/questions/11449/… ?
$endgroup$
– Shufflepants
14 mins ago
add a comment |
$begingroup$
Possible duplicate of physics.stackexchange.com/questions/11449/… ?
$endgroup$
– Shufflepants
14 mins ago
$begingroup$
Possible duplicate of physics.stackexchange.com/questions/11449/… ?
$endgroup$
– Shufflepants
14 mins ago
$begingroup$
Possible duplicate of physics.stackexchange.com/questions/11449/… ?
$endgroup$
– Shufflepants
14 mins ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Adding to TechDroid's answer, energy is also present in chemical bonds. When some higher energy (less stable) bonds are broken to form lower energy (more stable) ones (i.e. exothermic reactions), that energy difference can be released as energy.
So, almost all of that "+ energy" is due to the energy being released from the bonds themselves, and not the matter.
$endgroup$
add a comment |
$begingroup$
It actually does, but the amount converted is so small it's considered insignificant in the real world context. Based on the Einstein's famous equation ($E=mc^2$), a lot of energy can be extracted from a really small mass, and the reaction of methane and oxygen produces relatively small amount of energy which equates to a lot more smaller merely insignificant mass. The atomic bomb testiments to the amount of energy just some few kilograms of mass can decay into.
In addition the notion of the energy gained to achieve freedom for each atom reacting has to be given up to form a stable bond (that which sounds logical but I'm not entirely certain since I've not explored that domain very much) is also a solid argument to consider.
$endgroup$
$begingroup$
scientificamerican.com/article/…
$endgroup$
– safesphere
1 hour ago
$begingroup$
Well, Einstein has taken it all. But thanks for the link, it added something.
$endgroup$
– TechDroid
34 mins ago
add a comment |
$begingroup$
All the energy released is in the form of potential energy (of the electrons) falling to a lower (in general closer average positions) to the positive nuclei. This is similar to an apple falling off a tree. When this happens photons are released (no mass), molecules/atoms speed up and vibrations within the molecules and atoms increase (kinetic energy). All your chemical equations will have an energy balance but in addition you will need to take into account hidden thermodynamics, such as increased pressure and expansion of gases for example. This stuff is first year university, you will also learn about entropy ( why does salt melt ice?) which is another thermodynamic related energy concept required to balance.
In these reactions NO mass is converted to energy, mass is always conserved. In a nuclear reaction you again get photons, increased atomic/molecular motion but in addition you get high velocity sub-atomic particles like neutrons. Most (like >99% if I recall from wiki) of the mass is again conserved! You just get new types of atoms formed and isotopes (atoms that have absorbed a neutron). A few photons are indeed a result of a complex nuclear reaction where E=mc2 applies. But these are not of the same nature of the photons produced in a chemical reaction.
$endgroup$
add a comment |
$begingroup$
Let's do an analysis and see how much of a difference this makes.
- Methane: −74.87 kJ/mol
- Oxygen: 0
- Water(vapor): −241.818 kJ/mol
- Carbon dioxide: −393.509 kJ/mol
Therefore:
$$CH_4 + 2O_2 >>> 2H_2O + CO_2 + 802.3 text{kJ}$$
The mass of the products and reactants not worrying about the energy would be:
$$12.01 + 4(1.01) + 4(16.00) = 80.04text{g/mol}$$
Now checking the energy released:
$$m/text{mol} = frac{E/text{mol}}{c^2} = frac{802.3text{kJ/mol}}{3.0times 10^8 text{m/s}} = 8.9 times 10^{-9}text{g/mol}$$ or just a bit more than 1 part in $10^{10}$.
So the amount of mass that is missed by not considering the energy is well below the level of precision that is normally used. The amount of mass lost in reactions can generally be ignored until you reach nuclear energies.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f465810%2fconservation-of-mass-and-energy%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Adding to TechDroid's answer, energy is also present in chemical bonds. When some higher energy (less stable) bonds are broken to form lower energy (more stable) ones (i.e. exothermic reactions), that energy difference can be released as energy.
So, almost all of that "+ energy" is due to the energy being released from the bonds themselves, and not the matter.
$endgroup$
add a comment |
$begingroup$
Adding to TechDroid's answer, energy is also present in chemical bonds. When some higher energy (less stable) bonds are broken to form lower energy (more stable) ones (i.e. exothermic reactions), that energy difference can be released as energy.
So, almost all of that "+ energy" is due to the energy being released from the bonds themselves, and not the matter.
$endgroup$
add a comment |
$begingroup$
Adding to TechDroid's answer, energy is also present in chemical bonds. When some higher energy (less stable) bonds are broken to form lower energy (more stable) ones (i.e. exothermic reactions), that energy difference can be released as energy.
So, almost all of that "+ energy" is due to the energy being released from the bonds themselves, and not the matter.
$endgroup$
Adding to TechDroid's answer, energy is also present in chemical bonds. When some higher energy (less stable) bonds are broken to form lower energy (more stable) ones (i.e. exothermic reactions), that energy difference can be released as energy.
So, almost all of that "+ energy" is due to the energy being released from the bonds themselves, and not the matter.
edited 1 hour ago
answered 1 hour ago
F16FalconF16Falcon
3007
3007
add a comment |
add a comment |
$begingroup$
It actually does, but the amount converted is so small it's considered insignificant in the real world context. Based on the Einstein's famous equation ($E=mc^2$), a lot of energy can be extracted from a really small mass, and the reaction of methane and oxygen produces relatively small amount of energy which equates to a lot more smaller merely insignificant mass. The atomic bomb testiments to the amount of energy just some few kilograms of mass can decay into.
In addition the notion of the energy gained to achieve freedom for each atom reacting has to be given up to form a stable bond (that which sounds logical but I'm not entirely certain since I've not explored that domain very much) is also a solid argument to consider.
$endgroup$
$begingroup$
scientificamerican.com/article/…
$endgroup$
– safesphere
1 hour ago
$begingroup$
Well, Einstein has taken it all. But thanks for the link, it added something.
$endgroup$
– TechDroid
34 mins ago
add a comment |
$begingroup$
It actually does, but the amount converted is so small it's considered insignificant in the real world context. Based on the Einstein's famous equation ($E=mc^2$), a lot of energy can be extracted from a really small mass, and the reaction of methane and oxygen produces relatively small amount of energy which equates to a lot more smaller merely insignificant mass. The atomic bomb testiments to the amount of energy just some few kilograms of mass can decay into.
In addition the notion of the energy gained to achieve freedom for each atom reacting has to be given up to form a stable bond (that which sounds logical but I'm not entirely certain since I've not explored that domain very much) is also a solid argument to consider.
$endgroup$
$begingroup$
scientificamerican.com/article/…
$endgroup$
– safesphere
1 hour ago
$begingroup$
Well, Einstein has taken it all. But thanks for the link, it added something.
$endgroup$
– TechDroid
34 mins ago
add a comment |
$begingroup$
It actually does, but the amount converted is so small it's considered insignificant in the real world context. Based on the Einstein's famous equation ($E=mc^2$), a lot of energy can be extracted from a really small mass, and the reaction of methane and oxygen produces relatively small amount of energy which equates to a lot more smaller merely insignificant mass. The atomic bomb testiments to the amount of energy just some few kilograms of mass can decay into.
In addition the notion of the energy gained to achieve freedom for each atom reacting has to be given up to form a stable bond (that which sounds logical but I'm not entirely certain since I've not explored that domain very much) is also a solid argument to consider.
$endgroup$
It actually does, but the amount converted is so small it's considered insignificant in the real world context. Based on the Einstein's famous equation ($E=mc^2$), a lot of energy can be extracted from a really small mass, and the reaction of methane and oxygen produces relatively small amount of energy which equates to a lot more smaller merely insignificant mass. The atomic bomb testiments to the amount of energy just some few kilograms of mass can decay into.
In addition the notion of the energy gained to achieve freedom for each atom reacting has to be given up to form a stable bond (that which sounds logical but I'm not entirely certain since I've not explored that domain very much) is also a solid argument to consider.
edited 36 mins ago
answered 1 hour ago
TechDroidTechDroid
60912
60912
$begingroup$
scientificamerican.com/article/…
$endgroup$
– safesphere
1 hour ago
$begingroup$
Well, Einstein has taken it all. But thanks for the link, it added something.
$endgroup$
– TechDroid
34 mins ago
add a comment |
$begingroup$
scientificamerican.com/article/…
$endgroup$
– safesphere
1 hour ago
$begingroup$
Well, Einstein has taken it all. But thanks for the link, it added something.
$endgroup$
– TechDroid
34 mins ago
$begingroup$
scientificamerican.com/article/…
$endgroup$
– safesphere
1 hour ago
$begingroup$
scientificamerican.com/article/…
$endgroup$
– safesphere
1 hour ago
$begingroup$
Well, Einstein has taken it all. But thanks for the link, it added something.
$endgroup$
– TechDroid
34 mins ago
$begingroup$
Well, Einstein has taken it all. But thanks for the link, it added something.
$endgroup$
– TechDroid
34 mins ago
add a comment |
$begingroup$
All the energy released is in the form of potential energy (of the electrons) falling to a lower (in general closer average positions) to the positive nuclei. This is similar to an apple falling off a tree. When this happens photons are released (no mass), molecules/atoms speed up and vibrations within the molecules and atoms increase (kinetic energy). All your chemical equations will have an energy balance but in addition you will need to take into account hidden thermodynamics, such as increased pressure and expansion of gases for example. This stuff is first year university, you will also learn about entropy ( why does salt melt ice?) which is another thermodynamic related energy concept required to balance.
In these reactions NO mass is converted to energy, mass is always conserved. In a nuclear reaction you again get photons, increased atomic/molecular motion but in addition you get high velocity sub-atomic particles like neutrons. Most (like >99% if I recall from wiki) of the mass is again conserved! You just get new types of atoms formed and isotopes (atoms that have absorbed a neutron). A few photons are indeed a result of a complex nuclear reaction where E=mc2 applies. But these are not of the same nature of the photons produced in a chemical reaction.
$endgroup$
add a comment |
$begingroup$
All the energy released is in the form of potential energy (of the electrons) falling to a lower (in general closer average positions) to the positive nuclei. This is similar to an apple falling off a tree. When this happens photons are released (no mass), molecules/atoms speed up and vibrations within the molecules and atoms increase (kinetic energy). All your chemical equations will have an energy balance but in addition you will need to take into account hidden thermodynamics, such as increased pressure and expansion of gases for example. This stuff is first year university, you will also learn about entropy ( why does salt melt ice?) which is another thermodynamic related energy concept required to balance.
In these reactions NO mass is converted to energy, mass is always conserved. In a nuclear reaction you again get photons, increased atomic/molecular motion but in addition you get high velocity sub-atomic particles like neutrons. Most (like >99% if I recall from wiki) of the mass is again conserved! You just get new types of atoms formed and isotopes (atoms that have absorbed a neutron). A few photons are indeed a result of a complex nuclear reaction where E=mc2 applies. But these are not of the same nature of the photons produced in a chemical reaction.
$endgroup$
add a comment |
$begingroup$
All the energy released is in the form of potential energy (of the electrons) falling to a lower (in general closer average positions) to the positive nuclei. This is similar to an apple falling off a tree. When this happens photons are released (no mass), molecules/atoms speed up and vibrations within the molecules and atoms increase (kinetic energy). All your chemical equations will have an energy balance but in addition you will need to take into account hidden thermodynamics, such as increased pressure and expansion of gases for example. This stuff is first year university, you will also learn about entropy ( why does salt melt ice?) which is another thermodynamic related energy concept required to balance.
In these reactions NO mass is converted to energy, mass is always conserved. In a nuclear reaction you again get photons, increased atomic/molecular motion but in addition you get high velocity sub-atomic particles like neutrons. Most (like >99% if I recall from wiki) of the mass is again conserved! You just get new types of atoms formed and isotopes (atoms that have absorbed a neutron). A few photons are indeed a result of a complex nuclear reaction where E=mc2 applies. But these are not of the same nature of the photons produced in a chemical reaction.
$endgroup$
All the energy released is in the form of potential energy (of the electrons) falling to a lower (in general closer average positions) to the positive nuclei. This is similar to an apple falling off a tree. When this happens photons are released (no mass), molecules/atoms speed up and vibrations within the molecules and atoms increase (kinetic energy). All your chemical equations will have an energy balance but in addition you will need to take into account hidden thermodynamics, such as increased pressure and expansion of gases for example. This stuff is first year university, you will also learn about entropy ( why does salt melt ice?) which is another thermodynamic related energy concept required to balance.
In these reactions NO mass is converted to energy, mass is always conserved. In a nuclear reaction you again get photons, increased atomic/molecular motion but in addition you get high velocity sub-atomic particles like neutrons. Most (like >99% if I recall from wiki) of the mass is again conserved! You just get new types of atoms formed and isotopes (atoms that have absorbed a neutron). A few photons are indeed a result of a complex nuclear reaction where E=mc2 applies. But these are not of the same nature of the photons produced in a chemical reaction.
answered 41 mins ago
PhysicsDavePhysicsDave
94547
94547
add a comment |
add a comment |
$begingroup$
Let's do an analysis and see how much of a difference this makes.
- Methane: −74.87 kJ/mol
- Oxygen: 0
- Water(vapor): −241.818 kJ/mol
- Carbon dioxide: −393.509 kJ/mol
Therefore:
$$CH_4 + 2O_2 >>> 2H_2O + CO_2 + 802.3 text{kJ}$$
The mass of the products and reactants not worrying about the energy would be:
$$12.01 + 4(1.01) + 4(16.00) = 80.04text{g/mol}$$
Now checking the energy released:
$$m/text{mol} = frac{E/text{mol}}{c^2} = frac{802.3text{kJ/mol}}{3.0times 10^8 text{m/s}} = 8.9 times 10^{-9}text{g/mol}$$ or just a bit more than 1 part in $10^{10}$.
So the amount of mass that is missed by not considering the energy is well below the level of precision that is normally used. The amount of mass lost in reactions can generally be ignored until you reach nuclear energies.
$endgroup$
add a comment |
$begingroup$
Let's do an analysis and see how much of a difference this makes.
- Methane: −74.87 kJ/mol
- Oxygen: 0
- Water(vapor): −241.818 kJ/mol
- Carbon dioxide: −393.509 kJ/mol
Therefore:
$$CH_4 + 2O_2 >>> 2H_2O + CO_2 + 802.3 text{kJ}$$
The mass of the products and reactants not worrying about the energy would be:
$$12.01 + 4(1.01) + 4(16.00) = 80.04text{g/mol}$$
Now checking the energy released:
$$m/text{mol} = frac{E/text{mol}}{c^2} = frac{802.3text{kJ/mol}}{3.0times 10^8 text{m/s}} = 8.9 times 10^{-9}text{g/mol}$$ or just a bit more than 1 part in $10^{10}$.
So the amount of mass that is missed by not considering the energy is well below the level of precision that is normally used. The amount of mass lost in reactions can generally be ignored until you reach nuclear energies.
$endgroup$
add a comment |
$begingroup$
Let's do an analysis and see how much of a difference this makes.
- Methane: −74.87 kJ/mol
- Oxygen: 0
- Water(vapor): −241.818 kJ/mol
- Carbon dioxide: −393.509 kJ/mol
Therefore:
$$CH_4 + 2O_2 >>> 2H_2O + CO_2 + 802.3 text{kJ}$$
The mass of the products and reactants not worrying about the energy would be:
$$12.01 + 4(1.01) + 4(16.00) = 80.04text{g/mol}$$
Now checking the energy released:
$$m/text{mol} = frac{E/text{mol}}{c^2} = frac{802.3text{kJ/mol}}{3.0times 10^8 text{m/s}} = 8.9 times 10^{-9}text{g/mol}$$ or just a bit more than 1 part in $10^{10}$.
So the amount of mass that is missed by not considering the energy is well below the level of precision that is normally used. The amount of mass lost in reactions can generally be ignored until you reach nuclear energies.
$endgroup$
Let's do an analysis and see how much of a difference this makes.
- Methane: −74.87 kJ/mol
- Oxygen: 0
- Water(vapor): −241.818 kJ/mol
- Carbon dioxide: −393.509 kJ/mol
Therefore:
$$CH_4 + 2O_2 >>> 2H_2O + CO_2 + 802.3 text{kJ}$$
The mass of the products and reactants not worrying about the energy would be:
$$12.01 + 4(1.01) + 4(16.00) = 80.04text{g/mol}$$
Now checking the energy released:
$$m/text{mol} = frac{E/text{mol}}{c^2} = frac{802.3text{kJ/mol}}{3.0times 10^8 text{m/s}} = 8.9 times 10^{-9}text{g/mol}$$ or just a bit more than 1 part in $10^{10}$.
So the amount of mass that is missed by not considering the energy is well below the level of precision that is normally used. The amount of mass lost in reactions can generally be ignored until you reach nuclear energies.
answered 15 mins ago
BowlOfRedBowlOfRed
17.5k22743
17.5k22743
add a comment |
add a comment |
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f465810%2fconservation-of-mass-and-energy%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Possible duplicate of physics.stackexchange.com/questions/11449/… ?
$endgroup$
– Shufflepants
14 mins ago