Imibuzo Yesibonelo kanye Nengxoxo Ngesinyathelo Esihlobene SikaNewton
I-Pendahuluan
Ukunyakaza okuhlobene kuwumqondo oyinhloko ku-physics ochaza ukuthi ijubane nendawo yento kungashintsha kanjani kuye ngombukeli. USir Isaac Newton, ngemithetho yakhe yokunyakaza kanye namandla adonsela phansi, wabeka isisekelo sokuqonda ukuguquguquka kokunyakaza okuhlobene. Lesi sihloko sizokhuluma ngezibonelo eziningana kanye nezingxoxo zokunyakaza okuhlobene kukaNewton. Sizochaza lezi zinkinga ngezinyathelo zesisombululo ezinemininingwane ukuze kube lula ukuqonda.
Umqondo Oyisisekelo KaNewton Wokunyakaza Okuhlobene
Ku-physics yaseNewtonian, ukunyakaza kwento kulinganiswa njalo uma kuqhathaniswa nohlaka lokubhekisela. Uma sinezinhlaka ezimbili zokubhekisela ezihamba ngokuhambisana ngesivinini esingu-v, khona-ke indawo kanye nesivinini sento kungabonakala ngendlela ehlukile kuhlaka olubili. Eminye imiqondo okufanele uyiqonde yile:
1. Uhlaka Lokubhekisela Olungenasici: Uhlaka lokubhekisela lapho into ihamba ngesivinini esingaguquki uma kungekho mandla asebenzayo entweni.
2. Ijubane Elihlobene: Ijubane lento elilinganiswa uma liqhathaniswa nolunye uhlaka lokubhekisela.
3. Ukufuduka Okuhlobene: Umehluko endaweni phakathi kwezinto ezimbili noma into eyodwa enezinhlaka ezimbili ezihlukene zokubhekisela.
UNewton uchaze kahle kakhulu ukunyakaza okuhlobene ngemithetho yakhe yokunyakaza, futhi singasebenzisa ukuguqulwa kweGalilean ukushintsha phakathi kwezinhlaka ezimbili zokubhekisela ezingasebenzi.
Imibuzo Yengxoxo Eyisibonelo
Umbuzo 1: Ukufuduka Kokunyakaza Okuhlobene
Umbuzo:
Imikhumbi emibili, i-Ship A kanye ne-Ship B, iselwandle olukhulu. I-Ship A iya empumalanga ngesivinini esingama-20 m/s, kanti i-Ship B iya enyakatho ngesivinini esingama-30 m/s. Bala isivinini se-Ship B uma siqhathaniswa ne-Ship A.
Ingxoxo:
Ukuze sixazulule le nkinga, sisebenzisa umqondo wejubane elihlobene. Ijubane elihlobene le-Ship B elihlobene ne-Ship A lingabalwa kusetshenziswa indlela ye-vector.
1. Mela ijubane le-Ship A (\(\vec{v_A}\)) kanye ne-Ship B (\(\vec{v_B}\)) njengezivektha.
\[
\vec{v_A} = 20 \, \text{m/s empumalanga} \kusho \vec{v_A} = 20 \hat{i} \, \text{m/s}
\]
\[
\vec{v_B} = 30 \, \text{m/s north} \kusho \vec{v_B} = 30 \hat{j} \, \text{m/s}
\]
2. Isivinini esilinganiselwe se-Ship B kuya ku-Ship A (\(\vec{v_{BA}}\)) sibalwa ngo:
\[
\vec{v_{BA}} = \vec{v_B} – \vec{v_A}
\]
Faka amanani esikhundleni sika-\(\vec{v_A}\) kanye no-\(\vec{v_B}\):
\[
\vec{v_{BA}} = 30 \hat{j} \, \text{m/s} – 20 \hat{i} \, \text{m/s}
\]
\[
\vec{v_{BA}} = -20 \hat{i} + 30 \hat{j} \, \text{m/s}
\]
3. Ukuze uthole ubukhulu bejubane elihlobene, sebenzisa i-Pythagorean theorem:
\[
|\vec{v_{BA}}| = \sqrt{(-20)^2 + (30)^2}
\]
\[
|\vec{v_{BA}}| = \sqrt{400 + 900}
\]
\[
|\vec{v_{BA}}| = \sqrt{1300} = 10 \sqrt{13} \, \text{m/s}
\]
Ngakho-ke, isivinini se-Ship B uma siqhathaniswa ne-Ship A singu-\(10 \sqrt{13}\) m/s.
Umbuzo 2: Ukunyakaza Okuhlobene Ohlelweni Oluhlanganisiwe
Umbuzo:
Umuntu ohamba ngezinyawo uhamba enyakatho ngamamitha angu-5/s ngaphezu kwesitimela esihamba empumalanga ngamamitha angu-20/s. Thola ijubane lomuntu ohamba ngezinyawo uma kuqhathaniswa nomhlabathi.
Ingxoxo:
Ukuze sithole ijubane lomuntu ohamba ngezinyawo uma kuqhathaniswa nomhlabathi, sisebenzisa umqondo wokwengeza i-vector futhi.
1. Mela isivinini somuntu ohamba ngezinyawo (\(\vec{v_P}\)) kanye nesivinini sesitimela (\(\vec{v_K}\)) njengezimpawu.
\[
\vec{v_P} \text{ related to the train} = 5 \hat{j} \, \text{m/s}
\]
\[
\vec{v_K} \text{ related to the ground} = 20 \hat{i} \, \text{m/s}
\]
2. Ijubane lomuntu ohamba ngezinyawo uma liqhathaniswa nomhlabathi (\(\vec{v_{PT}}\)) yisamba sevektha:
\[
\vec{v_{PT}} = \vec{v_P} + \vec{v_K}
\]
Faka amanani esikhundleni sika-\(\vec{v_P}\) kanye no-\(\vec{v_K}\):
\[
\vec{v_{PT}} = 5 \hat{j} \, \text{m/s} + 20 \hat{i} \, \text{m/s}
\]
3. Ukuthola ubukhulu bejubane elihlobene:
\[
|\vec{v_{PT}}| = \sqrt{(20)^2 + (5)^2}
\]
\[
|\vec{v_{PT}}| = \sqrt{400 + 25}
\]
\[
|\vec{v_{PT}}| = \sqrt{425} = 5 \sqrt{17} \, \text{m/s}
\]
Ngakho-ke, ijubane lomuntu ohamba ngezinyawo uma liqhathaniswa nomhlabathi lingu-\(5 \sqrt{17}\) m/s.
Umbuzo 3: Ukunyakaza Okuhlobene Endizeni Ethambekele
Umbuzo:
Ibhola liphonswa ngesivinini esingu-\( \vec{u} = 10 \hat{i} + 10 \hat{j} \) m/s uma kuqhathaniswa nenqola ehamba ngesivinini esingaguquki esingu-15 m/s. Thola isivinini sebhola uma kuqhathaniswa nomhlabathi.
Ingxoxo:
Sebenzisa isimiso esifanayo ekungezeni i-vector.
1. Mela ijubane lebhola (\(\vec{u}\)) uma liqhathaniswa nenqola kanye nejubane lenqola (\(\vec{v_K}\)) njengamavektha.
\[
\vec{u} = 10 \hat{i} + 10 \hat{j} \, \text{m/s}
\]
\[
\vec{v_K} = 15 \hat{i} \, \text{m/s}
\]
2. Isivinini sebhola uma siqhathaniswa nomhlabathi (\(\vec{v_{BT}}\)) yilesi:
\[
\vec{v_{BT}} = \vec{v_K} + \vec{u}
\]
\[
\vec{v_{BT}} = 15 \hat{i} + (10 \hat{i} + 10 \hat{j})
\]
\[
\vec{v_{BT}} = (15 + 10) \hat{i} + 10 \hat{j}
\]
\[
\vec{v_{BT}} = 25 \hat{i} + 10 \hat{j}
\]
3. Ukuthola ubukhulu bejubane elihlobene:
\[
|\vec{v_{BT}}| = \sqrt{(25)^2 + (10)^2}
\]
\[
|\vec{v_{BT}}| = \sqrt{625 + 100}
\]
\[
|\vec{v_{BT}}| = \sqrt{725} = 5 \sqrt{29} \, \text{m/s}
\]
Ngakho-ke, isivinini sebhola uma siqhathaniswa nomhlabathi singu-\(5 \sqrt{29}\) m/s.
Isiphetho
Umqondo kaNewton wokunyakaza okuhlobene uyisisekelo esiyisisekelo sefiziksi yakudala. Sisebenzisa izimiso eziyisisekelo ezifana nokwengeza i-vector, singanquma ijubane elihlobene kanye nokufuduka kwento eyodwa maqondana nenye noma ozimele abahlukene bezinkomba. Izibonelo ezingenhla zibonisa indlela yokusebenzisa lo mqondo ezimweni ezahlukene, okunikeza ukuqonda okujulile kokunyakaza okuhlobene.
Ngokuqonda nokusebenzisa le mibono, singayiqonda kangcono imithetho kaNewton yokunyakaza nokuthi isebenza kanjani ezweni langempela. Lolu lwazi alugcini nje ngokusiza ekuxazululeni izinkinga zefiziksi kodwa futhi lunikeza ukuqonda okujulile ngendlela indawo yonke esebenza ngayo.
