{"id":13127,"date":"2020-02-11T04:41:31","date_gmt":"2020-02-11T09:41:31","guid":{"rendered":"https:\/\/library.bc.edu\/answerwall\/?p=13127"},"modified":"2020-02-12T16:11:23","modified_gmt":"2020-02-12T21:11:23","slug":"if-pmc-and-e-mc2-how-does-p-%e2%88%9ae","status":"publish","type":"post","link":"https:\/\/library.bc.edu\/answerwall\/2020\/02\/11\/if-pmc-and-e-mc2-how-does-p-%e2%88%9ae\/","title":{"rendered":"If P=MC and E = MC^2 how does P = \u221aE ?"},"content":{"rendered":"\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"543\" height=\"460\" src=\"https:\/\/library.bc.edu\/answerwall\/wp-content\/uploads\/2020\/02\/20200211_042013_2.jpg\" alt=\"If P-MC and E = MC^2 how does P = \u221aE ?\" class=\"wp-image-13117\" srcset=\"https:\/\/library.bc.edu\/answerwall\/wp-content\/uploads\/2020\/02\/20200211_042013_2.jpg 543w, https:\/\/library.bc.edu\/answerwall\/wp-content\/uploads\/2020\/02\/20200211_042013_2-300x254.jpg 300w\" sizes=\"auto, (max-width: 543px) 100vw, 543px\" \/><figcaption>If P=MC and E = MC^2 how does P = \u221aE ? <\/figcaption><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"536\" height=\"466\" src=\"https:\/\/library.bc.edu\/answerwall\/wp-content\/uploads\/2020\/02\/20200211_042025_2.jpg\" alt=\"E-= MC^2 -&gt; \u221a both sides therefore \u221aE= MC since P=MC -&gt; =\u221aE Hope this helps!\" class=\"wp-image-13118\" srcset=\"https:\/\/library.bc.edu\/answerwall\/wp-content\/uploads\/2020\/02\/20200211_042025_2.jpg 536w, https:\/\/library.bc.edu\/answerwall\/wp-content\/uploads\/2020\/02\/20200211_042025_2-300x261.jpg 300w\" sizes=\"auto, (max-width: 536px) 100vw, 536px\" \/><figcaption>E-= MC^2 -&gt; \u221a both sides therefore \u221aE= MC since P=MC -&gt; =\u221aE Hope this helps!<\/figcaption><\/figure>\n\n\n\n<p>\n\nUnfortunately, this sounds a bit like a homework question, so I&#8217;m going to decline to answer for now. However, I will note that my friend&#8217;s proof above is not the correct path. The correct form of the equation is E=M*(C^2), but the helper has assumed (or incorrectly stated), that E=(M*C)^2 in their proof. You cannot root both sides cleanly: \u221aE=(\u221aM)*C.\n\n<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"314\" height=\"304\" src=\"https:\/\/library.bc.edu\/answerwall\/wp-content\/uploads\/2020\/02\/AW02122020-05-e1581466202174.jpeg\" alt=\"1. E=MC^2 M=E\/C^2 2. P=MC M=C\/P 3. E\/C^2=C\/P E=C^3\/P P=C^3\/E\" class=\"wp-image-13142\" srcset=\"https:\/\/library.bc.edu\/answerwall\/wp-content\/uploads\/2020\/02\/AW02122020-05-e1581466202174.jpeg 314w, https:\/\/library.bc.edu\/answerwall\/wp-content\/uploads\/2020\/02\/AW02122020-05-e1581466202174-300x290.jpeg 300w\" sizes=\"auto, (max-width: 314px) 100vw, 314px\" \/><figcaption>1. E=MC^2  M=E\/C^2    2. P=MC M=C\/P   3. E\/C^2=C\/P  E=C^3\/P  P=C^3\/E<\/figcaption><\/figure>\n\n\n\n<p>\n\nThese are all correct. I&#8217;m honestly unsure that P <em>can <\/em>equal \u221aE. But any physics faculty member can probably help answer that definitively.\n\n<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Unfortunately, this sounds a bit like a homework question, so I&#8217;m going to decline to answer for now. However, I will note that my friend&#8217;s proof above is not the correct path. The correct form of the equation is E=M*(C^2), but the helper has assumed (or incorrectly stated), that E=(M*C)^2 in their proof. You cannot &hellip; <\/p>\n<p class=\"link-more\"><a href=\"https:\/\/library.bc.edu\/answerwall\/2020\/02\/11\/if-pmc-and-e-mc2-how-does-p-%e2%88%9ae\/\" class=\"more-link\">Continue reading<span class=\"screen-reader-text\"> &#8220;If P=MC and E = MC^2 how does P = \u221aE ?&#8221;<\/span><\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-13127","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/library.bc.edu\/answerwall\/wp-json\/wp\/v2\/posts\/13127","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/library.bc.edu\/answerwall\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/library.bc.edu\/answerwall\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/library.bc.edu\/answerwall\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/library.bc.edu\/answerwall\/wp-json\/wp\/v2\/comments?post=13127"}],"version-history":[{"count":4,"href":"https:\/\/library.bc.edu\/answerwall\/wp-json\/wp\/v2\/posts\/13127\/revisions"}],"predecessor-version":[{"id":13156,"href":"https:\/\/library.bc.edu\/answerwall\/wp-json\/wp\/v2\/posts\/13127\/revisions\/13156"}],"wp:attachment":[{"href":"https:\/\/library.bc.edu\/answerwall\/wp-json\/wp\/v2\/media?parent=13127"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/library.bc.edu\/answerwall\/wp-json\/wp\/v2\/categories?post=13127"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/library.bc.edu\/answerwall\/wp-json\/wp\/v2\/tags?post=13127"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}