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PHYS 1107 Lecture Notes - Lecture 2: Vir Chakra, Gq, University Of Florida

21 views57 pages
ochreeel360
5 Jul 2018
School
UHD
Department
Physics
Course
PHYS 1107
Professor
Ata Zaani

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LECTURES ī­ īŽ˜ ī® ī— THERMAL ENERGYī• TEMPERATURE AND ENERGY TRANSFERS
Eī…¶ī„žī†Œī…ī‡‡ Tī†Œī„‚ī…¶ī†ī„Øī…½ī†Œī…µī„‚ī†šī…ī…½ī…¶ī—
ī» Iī…¶ī†ī…ī„šī„ž ī„‚ ī†ī‡‡ī†ī†šī„žī…µ
ī» Eī…¶ī„žī†Œī…ī‡‡ ī†šī†Œī„‚ī…¶ī†ī„Øī…½ī†Œī…µī„‚ī†šī…ī…½ī…¶ī† ī…½ī„ī„ī†µī†Œ ī„ī„žī†šī‡ī„žī„žī…¶ ī†šī‡‡ī†‰ī„žī† ī…½ī„Ø ī„žī…¶ī„žī†Œī…ī‡‡
ī² K ī…¬ī…ī…¶ī„žī†šī…ī„
ī² U ī†‰ī…½ī†šī„žī…¶ī†šī…ī„‚ī…Æ
ī² Eī†šī…š ī†šī…šī„žī†Œī…µī„‚ī…Æ
ī² Eī„ī…šī„žī…µ ī„ī…šī„žī…µī…ī„ī„‚ī…Æ
ī» Iī…¶ ī„‚ī…¶ ī…ī†ī…½ī…Æī„‚ī†šī„žī„š ī†ī‡‡ī†ī†šī„žī…µ ī†šī…½ī†šī„‚ī…Æ ī„žī…¶ī„žī†Œī…ī‡‡ ī…ī† ī„ī…½ī…¶ī†ī„žī†Œī‡€ī„žī„š
Eī…¶ī„žī†Œī…ī‡‡ Tī†Œī„‚ī…¶ī†ī„Øī„žī†Œī†ī—
ī» Fī…½ī†Œ ī„‚ī…Æī…Æ ī†ī‡‡ī†ī†šī„žī…µī†ī• ī†šī…šī„ž ī„ī…šī„‚ī…¶ī…ī„ž ī…ī…¶ ī…ī…¶ī†šī„žī†Œī…¶ī„‚ī…Æ
ī„žī…¶ī„žī†Œī…ī‡‡ ī…ī† ī†šī…šī„ž ī†ī†µī…µ ī…½ī„Ø ī†šī…šī„ž ī„žī…¶ī„žī†Œī…ī‡‡
ī†šī†Œī„‚ī…¶ī†ī„Øī„žī†Œī† ī„žī‡†ī†‰ī„žī†Œī…ī„žī…¶ī„ī„žī„šī—
ī» īˆ“Eī…ī…¶ī†š ī‘ W ī½ Q
ī» Wī— ī†šī†Œī„‚ī…¶ī†ī„Øī„žī†Œ ī„ī‡‡
ī» Qī— ī†šī†Œī„‚ī…¶ī†ī„Øī„žī†Œ ī„ī‡‡
Eī…¶ī„žī†Œī…ī‡‡ ī“ Iī…¶ī†šī„žī†Œī…¶ī„‚ī…Æ
Sī†µī…µ ī…½ī„Ø Kī›ī† ī„‚ī…¶ī„š Uī›ī†
ī» K ī…½ī„Ø ī†‰ī„‚ī†Œī†šī…ī„ī…Æī„žī†
ī² Tī†Œī„‚ī…¶ī†ī…Æī„‚ī†šī…ī…½ī…¶ī„‚ī…Æī• ī†Œī…½ī†šī„‚ī†šī…ī…½ī…¶ī„‚ī…Æ ī„‚ī…¶ī„š ī‡€ī…ī„ī†Œī„‚ī†šī…ī…½ī…¶ī„‚ī…Æ
ī² Iī„šī„žī„‚ī…Æ ī…ī„‚ī†ī• ī…¶ī…½ ī†Œī…½ī†šī„‚ī†šī…ī…½ī…¶ī„‚ī…Æ ī…½ī†Œ ī‡€ī…ī„ī†Œī„‚ī†šī…ī…½ī…¶ī„‚ī…Æ ī„ī…½ī…¶ī†šī†Œī…ī„ī†µī†šī…ī…½ī…¶
ī» U ī„šī†µī„ž ī†šī…½ ī…ī…¶ī†šī„žī†Œī†‰ī„‚ī†Œī†šī…ī„ī…Æī„ž ī„Øī…½ī†Œī„ī„žī†
ī² Iī„šī„žī„‚ī…Æ ī…ī„‚ī†ī— U ī‘ ī¬
ī² Sī…ī…ī…¶ī…ī„Øī…ī„ī„‚ī…¶ī†š ī„Øī…½ī†Œ ī…Æī…ī†‹ī†µī…ī„šī† ī„‚ī…¶ī„š ī†ī…½ī…Æī…ī„šī† ī‡ī…šī„žī†Œī„ž ī„Øī…½ī†Œī„ī„žī† ī„‚ī†Œī„ž ī†ī†šī†Œī…½ī…¶ī…ī„žī†Œ
Eī…¶ī„žī†Œī…ī‡‡ ī“ Tī…šī„žī†Œī…µī„‚ī…Æ
Aī…¶ ī„‚ī†šī…šī…Æī„žī†šī„žī›ī†ī—
ī» Kī…ī…¶ī„žī†šī…ī„ ī„žī…¶ī„žī†Œī…ī‡‡ī‘ īŖ ī‡† ī²ī¬ ī…¬ī… ī‡† ī¾ī“ī˜īÆ ī…µī¬ī†īæ
ī» Tī…šī„žī†Œī…µī„‚ī…Æ ī„žī…¶ī„žī†Œī…ī‡‡ ī¾ī…¬ī…ī…¶ī„žī†šī…ī„ ī„‚ī…¶ī„š ī†‰ī…½ī†šī„žī…¶ī†šī…ī„‚ī…Æ ī„žī…¶ī„žī†Œī…ī…ī„žī† ī…½ī„Ø ī†‰ī„‚ī†Œī†šī…ī„ī…Æī„žī†ī•
ī„‚ī†‰ī†‰ī†Œī…½ī‡†ī…ī…µī„‚ī†šī„žī…Æī‡‡ ī„žī†‹ī†µī„‚ī…Æīæ ī‘ ī® ī‡† īŽ€īŖ ī‡† ī²ī¬ ī…¬ī… ī‡† ī¾ī²ī±ī¬ ī…µī¬ī†īæī®īŽ
ī» Rī„žī†ī†šī…ī…¶ī… ī†Œī„‚ī†šī„ž ī…½ī„Ø ī„žī…¶ī„žī†Œī…ī‡‡ ī†šī†Œī„‚ī…¶ī†ī„Øī…½ī†Œī…µī„‚ī†šī…ī…½ī…¶ī— ī“ī¬ W ī†šī…½ ī­ī¬ī¬ W
ī» Aī„ī†šī…ī‡€ī„ž ī†Œī„‚ī†šī„ž ī¾Cī‡‡ī„ī…Æī…ī…¶ī… ī°ī“ī¬Wīæ
Sī…½ī†µī†Œī„ī„ž ī…½ī„Ø ī„žī…¶ī„žī†Œī…ī‡‡ ī¾ī…ī…¶ī„ī†Œī„žī„‚ī†ī„ž ī…ī…¶ Eī…ī…¶ī†šīæī—
ī» ī…½ī‡†ī…ī„šī„‚ī†šī…ī…½ī…¶ ī…½ī„Ø ī„Øī…½ī…½ī„š
ī» ī„žī…ī˜ Cī² Hī­ī® Oī² ī½ ī² Oī® ī… ī²ī‘… ī² COī® ī½ ī² Hī®O ī½ ī„žī…¶ī„žī†Œī…ī‡‡
ī» īˆ“Eī…ī…¶ī†š ī¾ī…ī…¶ī†šī„žī†Œī…¶ī„‚ī…Æ ī„žī…¶ī„žī†Œī…ī‡‡ ī„ī…šī„‚ī…¶ī…ī„žīæ ī…µī„žī„‚ī†ī†µī†Œī„žī„š ī‡€ī…ī„‚
ī» Oī®ī…ī…¶ī— īŽ• ī®ī¬ī˜ī® ī…¬Jī¬ī„šī…µīÆ ī…½ī„Ø Oī® ī…½ī†Œ COī® ī…½ī†µī†š
ī» īˆ“Eī…ī…¶ī†šī˜ī•ī†šī…½ī†šī„‚ī…Æ ī‘ īˆ“Eī…ī…¶ī†šī˜ī•ī‡ī…½ī†Œī…¬ī…ī…¶ī… ī½ īˆ“Eī…ī…¶ī†šī˜ī•ī„ī„‚ī†ī„‚ī…Æ
Eī…¶ī„žī†Œī…ī‡‡ ī†šī†Œī„‚ī…¶ī†ī„Øī…½ī†Œī…µī„‚ī†šī…ī…½ī…¶ ī“ ī„žī„Øī„Øī…ī„ī…ī„žī…¶ī„ī‡‡
Eī‡†ī„‚ī…µī†‰ī…Æī„žī—
ī» A ī²ī± ī…¬ī… ī†ī†‰ī†Œī…ī…¶ī†šī„žī†Œ ī‡ī…½ī†Œī…¬ī† ī„‚ī†š ī„‚ ī†Œī„‚ī†šī„ž ī…½ī„Ø ī³ī¬ī¬ W ī„šī†µī†Œī…ī…¶ī… ī„‚ ī„ī…ī„ī‡‡ī„ī…Æī„ž ī„šī„‚ī†ī…š ī…½ī„Ø ī­ī®ī˜ī¬ ī†ī„žī„ī…½ī…¶ī„šī†
ī» Bī„‚ī†ī„‚ī…Æ ī…µī„žī†šī„‚ī„ī…½ī…Æī…ī„ ī†Œī„‚ī†šī„ž ī…½ī„Ø ī³ī¬ Wī–
ī» Wī…½ī†Œī…¬ī…ī…¶ī… ī„žī„Øī„Øī…ī„ī…ī„žī…¶ī„ī‡‡ ī…½ī„Ø ī®ī¬ī¹ī˜
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Tī„žī…µī†‰ī„žī†Œī„‚ī†šī†µī†Œī„ž ī„‚ī…¶ī„š Hī„žī„‚ī†šī…ī…¶ī…ī—
Tī…šī„žī†Œī…µī„‚ī…Æ Eī†‹ī†µī…ī…Æī…ī„ī†Œī…ī†µī…µ
ī» ī…¶ī…½ ī…½ī„ī†ī„žī†Œī‡€ī„‚ī„ī…Æī„ž ī„ī…šī„‚ī…¶ī…ī„žī† ī…ī…¶ ī†šī…šī„ž ī…µī„žī„‚ī†ī†µī†Œī„‚ī„ī…Æī„ž
ī†‰ī†Œī…½ī†‰ī„žī†Œī†šī…ī„žī† ī…½ī„Ø ī†šī…šī„ž ī…ī…¶ī†šī„žī†Œī„‚ī„ī†šī…ī…¶ī… ī„ī…½ī„šī…ī„žī† ī¾ī„žī…ī˜ ī‡€ī…½ī…Æī†µī…µī„žī•
ī†ī†šī„‚ī†šī„žī• ī‡€ī…ī†ī„ī…½ī†ī…ī†šī‡‡ ī™īæ
ī» Oī…¶ī„ī„ž ī‡ī„ž ī†šī…šī…½ī†µī…ī…šī†š ī†šī…šī„‚ī†š īžī…šī„žī„‚ī†šīŽŽīŸ ī‡ī„‚ī† ī„‚ ī„ī…½ī…¶ī†ī„žī†Œī‡€ī„žī„š
ī„Øī…Æī†µī…ī„š ī¾ī„ī„‚ī…Æī…Æī„žī„š ī„ī„‚ī…Æī…½ī†Œī…ī„īæ ī†šī†Œī„‚ī…¶ī†ī„Øī„žī†Œī†Œī„žī„š ī„ī„žī†šī‡ī„žī„žī…¶ ī…½ī„ī…©ī„žī„ī†šī†
ī„ī†µī†š ī…¶ī…½ī†š ī„ī†Œī„žī„‚ī†šī„žī„š ī…½ī†Œ ī„šī„žī†ī†šī†Œī…½ī‡‡ī„žī„šī˜
ī» Fī†Œī…ī„ī†šī…ī…½ī…¶ī„‚ī…Æ ī„žī„Øī„Øī„žī„ī†šī† īžī†Œī„žī…Æī„žī„‚ī†ī„žī„šīŸ ī„ī„‚ī…Æī…½ī†Œī…ī„ ī†ī†šī…½ī†Œī„žī„š
ī†ī…½ī…µī„žī‡ī…šī„žī†Œī„ž ī…ī…¶ ī„‚ī…¶ ī…½ī„ī…©ī„žī„ī†šī˜
Jī„‚ī…µī„žī† Pī†Œī„žī†ī„ī…½ī†šī†š Jī…½ī†µī…Æī„ž Eī‡†ī†‰ī„žī†Œī…ī…µī„žī…¶ī†šī—
ī» Tī…šī„žī†Œī…µī„‚ī…Æ ī„žī„Øī„Øī„žī„ī†šī† ī„‚ī…¶ī„š ī…µī„žī„ī…šī„‚ī…¶ī…ī„ī„‚ī…Æ ī„žī„Øī„Øī„žī„ī†šī† ī„‚ī†Œī„ž ī„ī…½ī†šī…š
ī„žī‡†ī„‚ī…µī†‰ī…Æī„žī† ī…½ī„Ø ī„žī…¶ī„žī†Œī…ī‡‡ ī†šī†Œī„‚ī…¶ī†ī„Øī„žī†Œī†
Eī…¶ī„žī†Œī…ī‡‡ Tī†Œī„‚ī…¶ī†ī„Øī„žī†Œī†ī—
Wī„ž ī„ī„‚ī…¶ ī…ī…¶ī„ī†Œī„žī„‚ī†ī„ž ī†šī„žī…µī†‰ī„žī†Œī„‚ī†šī†µī†Œī„ž ī„ī‡‡ī— ī…šī„žī„‚ī†šī…ī…¶ī…ī• ī‡ī…½ī†Œī…¬ī…ī…¶ī…
ī» Bī…½ī†šī…š ī†‰ī†Œī…½ī„ī„žī†ī†ī„žī† ī†šī†Œī„‚ī…¶ī†ī„Øī„žī†Œ ī„žī…¶ī„žī†Œī…ī‡‡ī• ī„‚ī…¶ī„š ī„ī…šī„‚ī…¶ī…ī„ž ī†šī…šī„ž
ī…ī…¶ī†šī„žī†Œī…¶ī„‚ī…Æ ī„žī…¶ī„žī†Œī…ī‡‡ ī…½ī„Ø ī„‚ ī†ī‡‡ī†ī†šī„žī…µ
īˆ“Eī…ī…¶ī†š ī‘ Q ī½ W
Q ī‘ ī„žī…¶ī„žī†Œī…ī‡‡ ī…ī…¶ī†‰ī†µī†š ī„ī‡‡ ī…šī„žī„‚ī†šī…ī…¶ī…
W ī‘ ī„žī…¶ī„žī†Œī…ī‡‡ ī…ī…¶ī†‰ī†µī†š ī„ī‡‡ ī‡ī…½ī†Œī…¬ī…ī…¶ī…
T ī…ī…¶ī„ī†Œī„žī„‚ī†ī„ž ī² ī‡ī…ī†šī…š ī…¶ī…½ ī…šī„žī„‚ī†šī…ī…¶ī…
ī» Tī„žī…µī†‰ī„žī†Œī„‚ī†šī†µī†Œī„ž ī…ī…¶ī„ī†Œī„žī„‚ī†ī„ž ī„ī‡‡ ī‡ī…½ī†Œī…¬ī…ī…¶ī…īŠ
īˆ“Eī…ī…¶ī†š ī‘ Q ī½ W
Q īŽ• ī¬
ī» Wī„žī…Æī…Æ ī…ī…¶ī†ī†µī…Æī„‚ī†šī„žī„š
ī» Tī…½ī…½ ī„Øī„‚ī†ī†š ī„Øī…½ī†Œ ī†ī…ī…ī…¶ī…ī„Øī…ī„ī„‚ī…¶ī†š ī…šī„žī„‚ī†šī…ī…¶ī… ī†šī…½ ī…šī„‚ī†‰ī†‰ī„žī…¶
īˆ“Eī…ī…¶ī†š ī‘ W
LECTURE īÆīŽ˜ī°ī— THERMAL ENERGYī• TEMPERATURE AND ENERGY TRANSFERS
ī» Hī„žī„‚ī†šī…ī…¶ī…ī— ī„‚ī…¶ ī„žī…¶ī„žī†Œī…ī‡‡ ī†šī†Œī„‚ī…¶ī†ī„Øī„žī†Œ ī„šī†µī„ž ī†šī…½ ī„‚ ī†šī„žī…µī†‰ī„žī†Œī„‚ī†šī†µī†Œī„ž ī„šī…ī„Øī„Øī„žī†Œī„žī…¶ī„ī„ž
ī» Hī„žī„‚ī†š ī†šī†Œī„‚ī…¶ī†ī„Øī„žī†Œī— ī„‚ī…µī…½ī†µī…¶ī†š ī…½ī„Ø ī„žī…¶ī„žī†Œī…ī‡‡ ī†šī†Œī„‚ī…¶ī†ī„Øī„žī†Œī†Œī„žī„š
ī» Pī†Œī…½ī„ī„žī†ī†ī„žī† ī¾ī…µī…½ī†Œī„ž ī„‚ī„ī…½ī†µī†š ī†šī…šī„žī†ī„ž ī…Æī„‚ī†šī„žī†Œīæī—
ī² Cī…½ī…¶ī„šī†µī„ī†šī…ī…½ī…¶ ī¶ ī„žī‡†ī„ī…šī„‚ī…¶ī…ī„ž ī…½ī„Ø ī…¬ī…ī…¶ī„žī†šī…ī„ ī„žī…¶ī„žī†Œī…ī‡‡ ī„ī„žī†šī‡ī„žī„žī…¶ ī†‰ī„‚ī†Œī†šī…ī„ī…Æī„žī†
ī² Cī…½ī…¶ī‡€ī„žī„ī†šī…ī…½ī…¶ ī¶ ī…µī…½ī‡€ī„žī…µī„žī…¶ī†š ī…½ī„Ø ī„‚ ī…šī„žī„‚ī†šī„žī„š ī†ī†µī„ī†ī†šī„‚ī…¶ī„ī„ž
ī² Rī„‚ī„šī…ī„‚ī†šī…ī…½ī…¶ ī¶ ī„žī…Æī„žī„ī†šī†Œī…½ī…µī„‚ī…ī…¶ī„žī†šī…ī„ ī‡ī„‚ī‡€ī„žī†
Tī„žī…µī†‰ī„žī†Œī„‚ī†šī†µī†Œī„ž ī“ ī†šī…šī„žī†Œī…µī…½ī…µī„žī†šī„žī†Œī†
ī» Pī†Œī…½ī‡€ī…ī„šī„ž ī„‚ ī†‹ī†µī„‚ī…¶ī†šī…ī†šī„‚ī†šī…ī‡€ī„ž ī…µī„žī„‚ī†ī†µī†Œī„ž ī…½ī„Ø īžī…šī…½ī†šī…¶ī„žī†ī†īŸī–
ī» ī…šī„žī…Æī†‰ ī†‰ī†Œī„žī„šī…ī„ī†š ī„šī…ī†Œī„žī„ī†šī…ī…½ī…¶ ī…½ī„Ø ī„žī…¶ī„žī†Œī…ī‡‡ ī†šī†Œī„‚ī…¶ī†ī„Øī„žī†Œī–
ī» Fī…½ī†Œ ī„‚ ī†ī„ī„‚ī…Æī„ž ī…¶ī„žī„žī„š ī„‚ ī†šī…šī„žī†Œī…µī…½ī…µī„žī†šī†Œī…ī„ ī†‰ī†Œī…½ī†‰ī„žī†Œī†šī‡‡ī• ī„žī…ī˜
ī² Lī„žī…¶ī…ī†šī…š ī„žī…ī˜ ī…µī„žī†Œī„ī†µī†Œī‡‡ ī„ī…½ī…Æī†µī…µī…¶
ī² Rī„žī†ī…ī†ī†šī„‚ī…¶ī„ī„ž ī„žī…ī˜ ī…½ī„Ø ī†‰ī…Æī„‚ī†šī…ī…¶ī†µī…µ
ī² Vī…½ī…Æī†šī„‚ī…ī„ž ī„žī…ī˜ ī„‚ī„ī†Œī…½ī†ī† ī„‚ ī†šī…šī„žī†Œī…µī…½ī„ī…½ī†µī†‰ī…Æī„ž ī…©ī†µī…¶ī„ī†šī…ī…½ī…¶
Tī„žī…µī†‰ī„žī†Œī„‚ī†šī†µī†Œī„ž Sī„ī„‚ī…Æī„žī†ī—
Aī„ī†ī…½ī…Æī†µī†šī„ž ī¾Kī„žī…Æī‡€ī…ī…¶īæī—
ī» Tī„žī…µī†‰ī„žī†Œī„‚ī†šī†µī†Œī„ž ī…½ī„Ø ī„‚ ī†ī…ī…¶ī…ī…Æī„ž ī†Œī„žī†‰ī†Œī…½ī„šī†µī„ī…ī„ī…Æī„ž ī†ī‡‡ī†ī†šī„žī…µ ī…ī†
ī„šī„žī„Øī…ī…¶ī„žī„šī•
Wī„‚ī†šī„žī†Œī›ī† ī†šī†Œī…ī†‰ī…Æī„ž ī†‰ī…½ī…ī…¶ī†šī— ī†šī…šī„žī†Œī…µī„‚ī…Æ ī„žī†‹ī†µī…ī…Æī…ī„ī†Œī…ī†µī…µ ī„ī„žī†šī‡ī„žī„žī…¶ ī…ī„ī„žī•
ī‡ī„‚ī†šī„žī†Œ ī„‚ī…¶ī„š ī‡€ī„‚ī†‰ī…½ī†µī†Œ
ī» Rī„‚ī†šī…ī…½ī† ī†Œī„žī…Æī„‚ī†šī…ī‡€ī„ž ī†šī…½ ī†šī…šī…ī† T ī„‚ī†Œī„ž ī†µī†ī„žī„š ī„Øī…½ī†Œ ī„‚ī…Æī…Æ ī…½ī†šī…šī„žī†Œ ī†šī„žī…µī†‰ī„žī†Œī„‚ī†šī†µī†Œī„žī†
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Iī„šī„žī„‚ī…Æ Gī„‚ī†ī—
ī» Aī…Æī…Æ ī†šī…šī„ž ī†šī…šī„žī†Œī…µī„‚ī…Æ ī„žī…¶ī„žī†Œī…ī‡‡ ī…ī† ī…ī…¶ ī†šī…šī„ž K ī…½ī„Ø ī†šī…šī„ž ī†‰ī„‚ī†Œī†šī…ī„ī…Æī„žī†
ī» Pī…½ī…ī…¶ī†š ī†‰ī„‚ī†Œī†šī…ī„ī…Æī„žī†
ī» Iī…¶ī†šī„žī†Œī„‚ī„ī†š ī…½ī…¶ī…Æī‡‡ ī„šī†µī†Œī…ī…¶ī… ī„ī…½ī…Æī…Æī…ī†ī…ī…½ī…¶ī†
ī» Cī…½ī…Æī…Æī…ī†ī…ī…½ī…¶ī† ī„‚ī†Œī„ž ī„žī…Æī„‚ī†ī†šī…ī„
Sī„‚ī…µī†‰ī…Æī„ž ī„ī„‚ī…Æī„ī†µī…Æī„‚ī†šī…ī…½ī…¶ ī…½ī„Ø ī‡€ī†Œī…µī†
ī» Cī…½ī…µī†‰ī„‚ī†Œī„ž ī†šī…šī„ž Kī„‚ī‡€ī… ī„‚ī…¶ī„š ī‡€ī†Œī…µī† ī…½ī„Ø Hī„ž ī„‚ī…¶ī„š Oī® ī…ī…¶ ī„‚ī…ī†Œ ī„‚ī†š ī®ī± īŽ£Cī˜
Iī„šī„žī„‚ī…Æ ī…ī„‚ī† ī“ ī†ī†‰ī„žī„žī„š ī„‚ī…¶ī„š ī†šī„žī…µī†‰ī„žī†Œī„‚ī†šī†µī†Œī„žī—
ī» Aī†š ī„‚ ī†‰ī„‚ī†Œī†šī…ī„ī†µī…Æī„‚ī†Œ ī†šī„žī…µī†‰ī„žī†Œī„‚ī†šī†µī†Œī„žī• Tī• ī…ī†Œī…½ī†µī†‰ī† ī…½ī„Ø ī„‚ī†šī…½ī…µī† ī…šī„‚ī‡€ī„ž ī†šī…šī„ž
ī†ī„‚ī…µī„ž Kī„‚ī‡€ī… ī„ī†µī†š ī„šī…ī„Øī„Øī„žī†Œī„žī…¶ī†š ī‡€ī†Œī…µī†
ī» Tī„žī…µī†‰ī„žī†Œī„‚ī†šī†µī†Œī„ž ī†‰ī†Œī„žī„šī…ī„ī†šī† ī†šī…šī„ž ī„šī…ī†Œī„žī„ī†šī…ī…½ī…¶ ī…½ī„Ø ī„žī…¶ī„žī†Œī…ī‡‡ ī†šī†Œī„‚ī…¶ī†ī„Øī„žī†Œ ī„ī‡‡
ī…šī„žī„‚ī†šī…ī…¶ī…
Eī‡†ī„‚ī…µī†‰ī…Æī„žī—
ī» Tī‡ī…½ ī„ī…½ī…¶ī†šī„‚ī…ī…¶ī„žī†Œī† ī…½ī„Ø ī†šī…šī„ž ī†ī„‚ī…µī„ž ī…ī„‚ī† ī¾ī…ī„šī„žī„‚ī…Æīæ ī…šī„‚ī‡€ī„ž ī†šī…šī„žī†ī„ž ī…µī„‚ī†ī†ī„žī† ī„‚ī…¶ī„š ī†šī„žī…µī†‰ī„žī†Œī„‚ī†šī†µī†Œī„žī†ī—
ī² Wī…šī…ī„ī…š ī…ī„‚ī† ī…šī„‚ī† ī„‚ī†šī…½ī…µī† ī‡ī…ī†šī…š ī†šī…šī„ž ī…ī†Œī„žī„‚ī†šī„žī†Œ ī„‚ī‡€ī„žī†Œī„‚ī…ī„ž ī…¬ī…ī…¶ī„žī†šī…ī„ ī„žī…¶ī„žī†Œī…ī‡‡ ī
ī² Wī…šī…ī„ī…š ī„ī…½ī…¶ī†šī„‚ī…ī…¶ī„žī†Œ ī…½ī„Ø ī…ī„‚ī† ī…šī„‚ī† ī†šī…šī„ž ī…ī†Œī„žī„‚ī†šī„žī†Œ ī†šī…šī„žī†Œī…µī„‚ī…Æ ī„žī…¶ī„žī†Œī…ī‡‡ ī
ī» Bī…½ī‡† Qī— Gī†Œī„žī„‚ī†šī„žī†Œ ī„‚ī‡€ī„žī†Œī„‚ī…ī„ž KE
ī» Bī…½ī‡† Pī— Gī†Œī„žī„‚ī†šī„žī†Œ Tī…šī„žī†Œī…µī„‚ī…Æ Eī…¶ī„žī†Œī…ī‡‡
Rī„žī†šī†µī†Œī…¶ ī†šī…½ ī†šī…šī„žī†Œī…µī„‚ī…Æ ī„žī†‹ī†µī…ī…Æī…ī„ī†Œī…ī†µī…µī—
ī» Aī…¶ ī„žī…¶ī„žī†Œī…ī‡‡ ī…ī…¶ī†‰ī†µī†š ī„ī‡‡ ī…šī„žī„‚ī†šī…ī…¶ī… ī„ī…šī„‚ī…¶ī…ī„žī† ī†šī…šī„ž ī†šī„žī…µī†‰ī„žī†Œī„‚ī†šī†µī†Œī„ž ī…½ī„Ø ī„šī…ī„Øī„Øī„žī†Œī„žī…¶ī†š
ī…µī„‚ī†šī„žī†Œī…ī„‚ī…Æī† ī„ī‡‡ ī„šī…ī„Øī„Øī„žī†Œī„žī…¶ī†š ī„‚ī…µī…½ī†µī…¶ī†šī†
ī» Q ī‘ M ī„ īˆ“Tī• ī‡ī…šī„žī†Œī„ž ī„ ī…ī† ī†šī…šī„ž ī†ī†‰ī„žī„ī…ī„Øī…ī„ ī…šī„žī„‚ī†šīŽŽ ī„ī„‚ī†‰ī„‚ī„ī…ī†šī‡‡ī• M ī…ī† ī…µī„‚ī†ī†
ī» ī…½ī†Œ ī…¶ C īˆ“Tī• ī‡ī…šī„žī†Œī„ž C ī…ī† ī†šī…šī„ž ī…µī…½ī…Æī„‚ī†Œ ī…šī„žī„‚ī†šīŽŽ ī„ī„‚ī†‰ī„‚ī„ī…ī†šī‡‡ī• ī…¶ ī…ī† ī…¶ī†µī…µī„ī„žī†Œ ī…½ī„Ø ī…µī…½ī…Æī„žī†
ī² Iī†šī›ī† ī†Œī„žī„‚ī…Æī…Æī‡‡ ī†šī…šī„žī†Œī…µī„‚ī…Æ ī„žī…¶ī„žī†Œī…ī‡‡ ī„ī…šī„‚ī…¶ī…ī„ž ī†‰ī„žī†Œ ī„šī„žī…ī†Œī„žī„ž ī²
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Physics: Principles with Applications
7th Edition, 2013
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Related Questions

Figure 1 illustrates a Keplerian orbit, with Cartesiancoordinates (x,y) and polar coordinates (r,ƏĀ†).

Parametric equations for a general Keplerian orbit are r(ƏĀˆ) = a(1 Ć¢ĀˆĀ’ e cosƏĀˆ)

t(ƏĀˆ) = *(T)/2ƏĀ€+ ( ƏĀˆĆ¢ĀˆĀ’ e sinƏĀˆ)

tan(ƏĀ†/2) = *(1+e)/(1Ć¢ĀˆĀ’e)+1/2 tan(ƏĀˆ/2)

Here ƏĀˆ is an independent variable. The solution is periodic in ƏĀˆwith period Ć¢ĀˆĀ†ĆĀˆ = 2ƏĀ€. Also, a, e, T are constants that determine theparameters of the orbit.

Figure 2 shows the relation between ƏĀˆ and the spatialcoordinates. In the figure, the coordinates (ƎĀ¾,ƎĀ·) are defined by ƎĀ¾= x + ae and ƎĀ· = y a /b,

withb=a*1Ć¢ĀˆĀ’e2 ]1/2.

(A) What kind of curve is the orbit?

(C) Determine x(ƏĀˆ) and y(ƏĀˆ).

(D) Determine ƎĀ¾(ƏĀˆ) and ƎĀ·(ƏĀˆ).

(E) Express r as a function ƏĀ† .

(F) The angular momentum is L = m r2dƏĀ†/dt. Determine L in termsof {a,e,T} Hence verify that L is a constant of the motion.

(G) The energy isE = [1/2] m ( dr/dt )2+ [1/2] m r2(dƏĀ†/dt)2Ć¢ĀˆĀ’ GMm/r.Determine E in terms of {a,e,T}.

(H) E must be a constant of the motion. Hence determine T and Efrom the equation that you obtained in (F).

(I) Write T in terms of the spatial orbit parameters and theforce constant GM.

(J) Now express L and E in terms of a, e, GM.

(K) In Figure 1, determine the coordinates of the points P, A,and R

(L) In Figure 1, determine the time t at the point R.

True or False

(I) Is each one of the following statements true or false for aphoton?

(a) It is a quantum particle, behaving in some experiments like aclassical particle and in some experiments like a classicalwave.TrueYour answer is correct.

(b) Its rest energy is zero. FalseYour answer is incorrect.

(c) Its rest energy is nonzero. TrueYour answer is incorrect.

(d) It carries energy in its motion. 4

(e) It carries momentum in its motion. 5

(f) Its motion is described by a wave function that has awavelength and satisfies a wave equation. 6

(g) For one-dimensional motion, the wave equation is asfollows:
(partial^2 E)/(partial x^2) = mu_0 epsilon_0 (partial^2 E)/(partialt^2)
where the wave function E is the electric field magnitude, with asimilar equation for B = E/c. 7

(h) For one-dimensional motion, the wave equation is asfollows:
-hbar^2/(2m)(d^2 Psi)/(dx^2) + U Psi = E Psi
where E is the constant energy. 8

(i) The intensity of the wave is proportional to the square of itswave function. 9

(j) The intensity of the wave is measured from the rate at whichthe quantum particles bombard a detector. 10



(II) Is each one of the statements true or false for anelectron?

(a) It is a quantum particle, behaving in some experiments like aclassical particle and in some experiments like a classical wave.11

(b) Its rest energy is zero. 12

(c) Its rest energy is nonzero. 13

(d) It carries energy in its motion. 14

(e) It carries momentum in its motion. 15

(f) Its motion is described by a wave function that has awavelength and satisfies a wave equation. 16

(g) For one-dimensional motion, the wave equation is asfollows:
(partial^2 E)/(partial x^2) = mu_0 epsilon_0 (partial^2 E)/(partialt^2)
where the wave function E is the electric field magnitude, with asimilar equation for B = E/c. 17

(h) For one-dimensional motion, the wave equation is asfollows:
-hbar^2/(2m)(d^2 Psi)/(dx^2) + U Psi = E Psi
where E is the constant energy. 18

(i) The intensity of the wave is proportional to the square of itswave function. 19

(j) The intensity of the wave is measured from the rate at whichthe quantum particles bombard a detector.

A boy in a wheelchair (total mass 50.5 kg) has a speed of 1.40 m/s at the crest of a slope 2.40 m high and 12.4 m long. At the bottom of the slope, his speed is 6.50 m/s. Assume air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N. Find the work he did in pushing forward on his wheels during the downhill ride.

Ā 

Ā 
Ā 
Ā 

2 Answers

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  • electron1
    LvĀ 7
    4 years ago

    There are two forces that cause the wheelchair to accelerate. These two forces are the component of the wheelchairā€™s weight that is parallel to the hill and the force the boy exerts. The net force on the wheelchair is equal to the sum of these two forces minus the friction force. Letā€™s use the following equation to determine the acceleration.

    Ā 

    vf2 = vi^2 + 2 * a * d

    6.5^2 = 1.4^2 + 2 * a * 12.4

    24.8 * a = 40.29

    a = 40.29 Ć· 24.8

    This is approximately 1.62 m/s^2.

    Ā 

    Net force = 50.5 * 40.29 Ć· 24.8 = 2034.645 Ć· 24.8

    This is approximately 82 N.

    Ā 

    Force parallel = 50.5 * 9.8 * 2.40 Ć· 12.4 = 1187.76 Ć· 12.4

    Ā 

    This is approximately 95.8 N. This force and the force that the boy exerts are the forces that cause the wheelchair to accelerate.

    Ā 

    Net force = F + (1187.76 Ć· 12.4) ā€“ 41

    Set the two net forces equal to each other.

    Ā 

    F + (1187.76 Ć· 12.4) ā€“ 41 = 2034.645 Ć· 24.8

    F = 41 + (2034.645 Ć· 24.8) ā€“ (1187.76 Ć· 12.4)

    F is approximately 27.3 N

    Ā 

    Work = [41 + (2034.645 Ć· 24.8) ā€“ (1187.76 Ć· 12.4)] * 12.4 = 337.9625 J

    Ā 
  • JupiterSky
    LvĀ 6
    4 years ago

    His change in kinetic energy is due to gravitational- and his own work,

    while the friction does negative work.

    Ā 

    āˆ†K = āˆ†U +āˆ†W + āˆ†Wf

    Ā½m(VfĀ²āˆ’ViĀ²) = mgh + āˆ†W + Ff*d

    āˆ†W = Ā½m(VfĀ²āˆ’ViĀ²) āˆ’ mgh āˆ’ Ff*d

    āˆ†W = Ā½*50.5(6.5Ā²āˆ’1.4Ā²) āˆ’ (50.5*9.8*2.4) āˆ’ (āˆ’41*12.4)

    āˆ†W = 508 J

    Ā 
Still have questions? Get your answers by asking now.
Ā 
Ā 
Next Question
---
asked inĀ Science & MathematicsPhysicsĀ Ā·Ā 8 years ago

What is the spring constant? please help me!!!!?

A block of mass 487 g is pushed against the

spring (located on the left-hand side of the

track) and compresses the spring a distance

4.7 cm from its equilibrium position (as shown

in the ļ¬gure below). The block starts from

rest, is accelerated by the compressed spring,

and slides across a frictionless horizontal track

(as shown in the ļ¬gure below). It leaves the

track horizontally, ļ¬‚ies through the air, and

subsequently strikes the ground.

acceleration = 9.81

Ā 

A) What is the spring constant?

B)What is the speed v of the block when it leaves

the track?

C) What is the total speed of the block when it

hits the ground?

...Show more
Update:

Height Above ground = 1.9 m

Lands = 4.59m away

Ā 
What is the spring constant? please help me!!!!?

3 Answers

Relevance
  • JĆ”noÅ”Ć­k
    LvĀ 7
    8 years ago
    Favorite Answer

    I'm sure there is some crucial information in "the figure below" !!!

    Ā 

    Edit:

    OK... now we got something to work with ... ;-)

    Ā 

    Time needed to fall a vertical distance of 1.9 m is:

    t = āˆš( 2h / g )

    t = āˆš[ 2Ɨ(1.9 m) / (9.81 m/sĀ²) ]

    t = 0.6224 s

    Ā 

    In that time, the block must travel a horizontal distance of 4.59 m, so the horizontal velocity is:

    r = vƗt

    (4.59 m) = vƗ(0.6224 s)

    v = 7.37 m/s < - - - - - - - - - - - - - - - - - answer B

    Ā 

    That is also the speed of the block as it leaves the spring.

    Its kinetic energy at that time is:

    KE = mƗvĀ²/2

    KE = (0.487 kg)Ɨ(7.3749 m/s)Ā²/2

    KE = 13.244 J

    Ā 

    That is also the elastic potential energy stored in the spring at maximum compression. So the spring constant is:

    KE = PEe = kƗdĀ²/2

    (13.244 J) = kƗ(0.047 m)Ā²/2

    k = 12 N/m < - - - - - - - - - - - - - - - - - - answer A

    Ā 

    The gravitational potential energy of the block on the table is:

    PEg = mƗgƗh

    PEg = (0.487 kg)Ɨ(9.81 m/sĀ²)Ɨ(1.9)

    PEg = 9.077 J

    Ā 

    The total energy of the block on the table is:

    TE = PEg + PEe

    TE = (9.077 J) + (13.244 J)

    TE = 22.32 J

    Ā 

    When hitting the ground, all that energy is converted to kinetic energy, so

    TE = KE = mƗvĀ²/2

    (22.32 J) = (0.487 kg)ƗvĀ²/2

    v = 9.57 m/s < - - - - - - - - - - - - - - - - - - answer C

    ...Show more
    Ā 
  • Technobuff
    LvĀ 7
    8 years ago

    There must be a height above ground and a distance from the end of the track on the "as shown in the figure below".??

    Ā 
  • STALKER
    LvĀ 4
    8 years ago

    use f=kl

    Ā 

    where k is the spring constant and l is extension from the natural length.

    Ā 

    k=f/l

    Ā 

    since you have mass in grams. use w=mg

    Ā 

    w=0.487 x 9.8 = 4.77N

    Ā 

    Now use the equation,

    Ā 

    k=4.77/4.7

    Ā 

    k=1.012

    Ā 

    Note: my answer could be wrong since I do not know the springs original length from the diagram shown. But, the method of working out spring constant is same.

    ...Show more
    Ā 
Still have questions? Get your answers by asking now.
Next Question
ć²ć„ć‚‰ćŽ
LvĀ 5
asked inĀ Science & MathematicsPhysicsĀ Ā·Ā 1 decade ago

Physics Momentum, Rolling?

Pat builds a track for his model car out of wood. The track is 10.00 cm wide and 3.00 m long, and is solid. The runway is cut such that it forms a parabola by the equation y=(x - 3)^2/9. Locate the horizontal position of the center of gravity of this track.

Ā 

A washing machine load in the "spin" cycle, (much like a centrifuge used to separate fluids), can be modeled as a cylinder which has zero density toward the center, and starting from some inner radius where the clothes (or fluids) begin, increases in density as you move radially outward. The washing machine has density of clothes (in SI units) of p=14r+5, a height of 0.51 m, and only has clothes present from inner radius 0.2m out to 0.7m from the center.

-What is the moment of Inertia of the wash load?

-What is the power of motor that drives the clothes from rest up to 4 rev/sec, in 14 seconds if the basin has moment of Inertia of 4 kg*.m2

...Show more
Ā 
Physics Momentum, Rolling?

1 Answer

Relevance
  • Ā 
    Anonymous
    1 decade ago
    Favorite Answer
    Ā 

    I can't visualize the first problem.

    Ā 

    The second: Moment of inertia is I = m r^2. In a shell of thickness dr there is a mass of p * 2 * pi * r * h * dr. The moment of inertia of the shell is 2 pi h p r^3 dr. Now integrate this from r = 0.2m to r = 0.7m.

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    Kinetic energy is 1/2 m v^2. In rotations the moment of inertia is like the mass and the angular velocity is like the velocity, so rotational energy is 1/2 I omega^2. Omega is in rad/s. The total inertia of the system is the sum of the inertia of the clothes and the basin. Divide the energy by the time needed to get the power.

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asked inĀ Science & MathematicsPhysicsĀ Ā·Ā 1 decade ago

Physics:You and your friends push a 75-kg...Help!?

You and your friends push a 75-kg greased pig up an aluminum slide at the county fair, starting from the low end of the slide. The coefficient of kinetic friction between the pig and the slide is 0.070.

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(a) All of you pushing together (parallel to the incline) manage to accelerate the pig from rest at the constant rate of 4.9 m/s2 over a distance of 1.8 m, at which point you release the pig. The pig continues up the slide, reaching a maximum vertical height above its release point of 50 cm. What is the angle of inclination of the slide? __degree

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(b) At the maximum height the pig turns around and begins to slip down the slide, how fast is it moving when it arrives at the low end of the slide? __m/s

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Physics:You and your friends push a 75-kg...Help!?

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    Anonymous
    1 decade ago
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    ā™£the distance the pig was pushed up is

    s=0.5*a*t^2, where s=1.8m, a=4.9m/s^2, t is time of push-up;

    speed of the pig gained after s=1.8m is v=a*t; thus v^2=2as;

    ā™  kinetic energy of the pig at the moment itā€™s released after push-up is

    E=0.5*m*v^2, which is enough for the pig to move by itself L distance more along the ramp, where L=h/sin(b), h=50cm=0.5m, b is angle in question;

    ā™¦pigā€™s potential energy at vertex (when stopped) is E1=mgh;

    the energy E2=E-E1 is gone on friction, that is E2=f*L is work of friction, where

    f=Ī¼*w1, w1=mg*cos(b) is component of pigā€™s weight normal to the aluminum ramp; or;

    f= Ī¼*mg*cos(b);

    ā™¦Thus E2=E-E1; or; f*L = 0.5*m*v^2 ā€“mgh; or;

    (Ī¼*mg*cos(b)) * (h/sin(b)) =0.5*m* 2as ā€“mgh; or;

    Ī¼*gh*cos(b)/sin(b) =as ā€“gh, hence cot(b) =(as ā€“gh)/(Ī¼*gh), hence

    (a); b=atan(Ī¼*gh/(as-gh))= atan(0.07*9.8*0.5/(4.9*1.8 -9.8*0.5)) =5Ā°;

    Ā 

    ā™„ pigā€™s potential energy above the low end of the ramp is

    W= mg*(s*sin(b) +h); pigā€™s ramp position is p=s +L = s + h/sin(b);

    Pigā€™s final kinetic energy is W1=W-W2, where W2=f*p is work of friction;

    and W1=0.5*m*v^2; thus

    0.5*m*v^2 = mg*(s*sin(b) +h) ā€“ Ī¼*mg*cos(b)*(s + h/sin(b));

    v^2 =2g*(s*sin(b) +h)*(1 ā€“Ī¼*cot(b)) =2.575;

    (b); v =1.605 m/s;

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