Isibonelo Semibuzo Yengxoxo Yokuhlobana
Ukuhlobana kungenye yemibono eyisisekelo kakhulu kwifiziksi yanamuhla, eyethulwa ngu-Albert Einstein ekuqaleni kwekhulu lama-20. Lesi sihloko sizoxoxa ngemfundiso yokuhlobana nokuthi isebenza kanjani empilweni yansuku zonke ngezinkinga zesibonelo kanye nezincazelo.
Isingeniso Sokuhlobana
Ithiyori yokuhlobana ihlanganisa izingxenye ezimbili eziyinhloko: iTheory Ekhethekile Yokuhlobana kanye neTheory Ejwayelekile Yokuhlobana. ITheory Ekhethekile Yokuhlobana, eyanyatheliswa ngo-1905, yashintsha ukuqonda kwethu isikhala nesikhathi. Kulo mbono, u-Einstein wathi ijubane lokukhanya liwumkhawulo wejubane omkhulu ongenakudlulwa nokuthi imithetho yefiziksi iyafana kubo bonke ababukeli abahamba ngesivinini esingaguquki.
Okwamanje, i-General Theory of Relativity, eyasungulwa ngo-1915, iphathelene namandla adonsela phansi. Ngaphansi kwalo mbono, amandla adonsela phansi akuwona amandla endabuko, kodwa kunalokho ukugoba kwesikhathi sesikhala okubangelwa ubunzima.
Ukuqonda lo mqondo oyisisekelo kubaluleke kakhulu ngaphambi kokuba singene emibuzweni eyisibonelo kanye nengxoxo yayo.
Imibuzo Eyisibonelo Nengxoxo
Umbuzo 1: Ukwelulwa Kwesikhathi
Umbuzo:
I-astronaut iya enkanyezini ekude ngesivinini esingu-0,8c (lapho u-c kuyisivinini sokukhanya). Uma uhambo luthatha iminyaka eyi-10 yoMhlaba, i-astronaut ibhekana nesikhathi esingakanani ngokwewashi layo (isikhathi esifanele)?
Ingxoxo:
Ukwanda kwesikhathi kuyinto eyenzeka ngenxa yokwehluka kwesivinini esilinganiselwe phakathi kwababukeli ababili. Isikhathi sidlula kancane kakhulu entweni ehambayo uma kuqhathaniswa nombonisi omile.
Ifomula yokwelulwa kwesikhathi yile:
\[ \Delta t' = \frac{\Delta t}{\sqrt{1 – \frac{v^2}{c^2}}}\]
Di mana:
– \(\Delta t'\) yisikhathi esibonwayo sento ehambayo.
– \(\Delta t\) yisikhathi esibonwayo sento engashintshi.
– \(v\) ijubane lento ehambayo.
– \(c\) ijubane lokukhanya.
Faka amanani aziwayo kufomula:
\[v = 0,8c \]
\[ \Delta t = 10 \, \umbhalo{unyaka} \]
\[ \Delta t' = \frac{10}{\sqrt{1 – \frac{(0,8c)^2}{c^2}}}\]
\[ \Delta t' = \frac{10}{\sqrt{1 – 0,64}}\]
\[ \Delta t' = \frac{10}{\sqrt{0,36}}\]
\[ \Delta t' = \frac{10}{0,6}\]
\[ \Delta t' \cishe 16.67 \, \text{year}\]
Ngakho-ke, isikhathi esitholwa yi-astronaut ngokwewashi lakhe cishe siyiminyaka eyi-16,67.
Umbuzo 2: Ukufinyela Ubude
Umbuzo:
Into ingamamitha ayi-100 ubude futhi ilinganiswa lapho iphumule. Uma into ihamba ngesivinini esingu-0,6c, ubude bayo bungakanani ngokusho komqapheli omile?
Ingxoxo:
Ukufinyela kobude kuyisenzakalo lapho ubude bento ehambayo uma kuqhathaniswa nombukeli bufushane kunalapho into iphumule.
Ifomula yokunciphisa ubude yile:
\[ L = L_0 \sqrt{1 – \frac{v^2}{c^2}} \]
Di mana:
– \(L\) ubude bento ehambayo.
– \(L_0\) ubude obufanele (ubude bento uma iphumule).
– \(v\) ijubane lento.
– \(c\) ijubane lokukhanya.
Faka amanani aziwayo kufomula:
\[ L_0 = 100 \, \text{meter} \]
\[v = 0,6c \]
\[ L = 100 \sqrt{1 – \frac{(0,6c)^2}{c^2}}\]
\[ L = 100 \sqrt{1 – 0,36}\]
\[ L = 100 \sqrt{0,64}\]
\[L = 100 \izikhathi 0,8\]
\[ L = 80 \, \text{meter}\]
Ngakho-ke, ubude bento ehambayo ngokusho komqapheli omile bungamamitha angu-80.
Umbuzo 3: Isisindo Esihlobene Nezimo
Umbuzo:
Inhlayiya inesisindo sokuphumula esingu-2 kg. Uma le nhlayiya ihamba ngesivinini esingu-0,9c, iyini inhlayiya ehambisanayo?
Ingxoxo:
Isisindo esivumelanayo yisisindo sento esandayo njengoba into isondela esivivinyweni sokukhanya.
Ifomula yesisindo esivumelanayo yile:
\[ m = \frac{m_0}{\sqrt{1 – \frac{v^2}{c^2}}} \]
Di mana:
– \(m\) yisisindo esivumelana nezimo.
– \(m_0\) isisindo esisele (isisindo esifanele).
– \(v\) ijubane lento.
– \(c\) ijubane lokukhanya.
Faka amanani aziwayo kufomula:
\[ m_0 = 2 \, \umbhalo{kg} \]
\[v = 0,9c \]
\[ m = \frac{2}{\sqrt{1 – \frac{(0,9c)^2}{c^2}}}\]
\[ m = \frac{2}{\sqrt{1 – 0,81}}\]
\[ m = \frac{2}{\sqrt{0,19}}\]
\[ m \cishe \frac{2}{0,436}\]
\[ m \cishe 4,59 \, \umbhalo{kg}\]
Ngakho-ke, isisindo esivumelanayo senhlayiya uma ihamba ngesivinini esingu-0,9c singaba ngu-4,59 kg.
Umbuzo 4: E=mc^2
Umbuzo:
Kukhiqizwa amandla angakanani uma igremu eli-1 lento ibhujiswa ngokuphelele ngokwefomula ka-Einstein \(E=mc^2\)?
Ingxoxo:
Ifomula edumile ka-Einstein \(E=mc^2\) inikeza ubudlelwano obuqondile phakathi kwesisindo (m) namandla (E), kanye \(c\) okuyisivinini sokukhanya.
Kuhlelo lwe-SI (International System of Units):
– Isisindo (m) silinganiswa ngamakhilogremu (kg).
– Isivinini sokukhanya (c) singu-\(3 \times 10^8 \, \text{m/s}\).
Ake sibale amandla akhiqizwa kusuka ku-1 gram yento:
– 1 igremu = 0,001 kg
\[ E = mc^2 \]
\[ E = (0,001) (3 \izikhathi ezingu-10^8)^2 \]
\[ E = (0,001) (9 \izikhathi ezingu-10^{16}) \]
\[ E = 9 \izikhathi 10^{13} \, \umbhalo{ama-joules} \]
Ngakho-ke, amandla akhiqizwayo uma igremu eli-1 lento ibhujiswa ngokuphelele angama-joules angu-(9 \times 10^{13}\).
Isiphetho
Ukuhlobana kuwumqondo oyisisekelo nobalulekile ku-physics, onemiphumela ejulile ezimweni eziningi zomzimba. Ngezibonelo okuxoxwe ngazo ngenhla, sibone ukuthi inkolelo-mbono ekhethekile yokuhlobana ingasetshenziswa kanjani ukuqonda ukwanda kwesikhathi, ukufinyela kobude, ubukhulu bokuhlobana, kanye nobudlelwano phakathi kobukhulu namandla.
Ngokuqonda nokuzijwayeza lezi zinkinga, singabuqonda kangcono ubuhle benkolelo-mbono yokuhlobana kanye nemiphumela yayo ekuqondeni indawo yonke.
