{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Shih final

# Shih final - Version 079 – Final – shih –(56310 1...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Version 079 – Final – shih – (56310) 1 This print-out should have 35 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A rod starts from rest at the top of an inclined plane at t = 0. The rails have negligi- ble resistance. There is a constant, vertically directed magnetic field of magnitude 1 T. The acceleration due to gravitation is 9 . 81 m / s 2 . 3 Ω v . 8 kg 1 T sliding rod 24 m Viewed from above v 30 ◦ 1 T Viewed from the side What will be the terminal speed of the rod? 1. 1.02986 2. 4.84444 3. 3.29554 4. 2.725 5. 1.26157 6. 2.90667 7. 4.63435 8. 0.68125 9. 0.823885 10. 4.74558 Correct answer: 2 . 725 cm / s. Explanation: R v m B sliding rod ℓ Viewed from above Let : B = 1 T , R = 3 Ω , g = 9 . 81 m / s 2 , ℓ = 24 m , and θ = 30 ◦ . v θ B Viewed from the side The induced emf is E = B ℓ v cos θ . Using Ohm’s law, the current in the circuit due to the induced emf is I = E R = B ℓ v cos θ R . Then the force due to the current is F m = I ℓ B = B 2 ℓ 2 v cos θ R . Applying Newton’s 2 nd law to the rod, mg sin θ- F m cos θ = m dv dt g sin θ- F m m cos θ = dv dt . Version 079 – Final – shih – (56310) 2 As the rod reaches the terminal speed, dv dt = 0, and g sin θ = F m m cos θ g sin θ = B 2 ℓ 2 v cos θ mR cos θ . Thus v = mg R sin θ B 2 ℓ 2 cos 2 θ = (0 . 8 kg) (9 . 81 m / s 2 ) (3 Ω) sin 30 ◦ (1 T) 2 (24 m) 2 cos 2 30 ◦ × 100 cm 1 m = 2 . 725 cm / s . keywords: 002 (part 1 of 3) 10.0 points ONLY DO PART 1 OF THIS 3 PART QUESTION!!!!! Consider just four of the energy levels in a certain atom, as shown in the diagram below. n = 1 n = 2 n = 3 n = 4 How many spectral lines will result from all possible transitions among these levels? Which transition corresponds to the highest- frequency light emitted? Which transition corresponds to the lowest-frequency? 1. three; level 2 to level 1 transition; level 4 to level 3 transition. 2. six; level 4 to level 1 transition; level 4 to level 3 transition. correct 3. three; level 4 to level 1 transition; level 4 to level 3 transition. 4. three; level 4 to level 3 transition; level 2 to level 1 transition. Explanation: Six transitions are possible, as shown. n = 1 n = 2 n = 3 n = 4 The highest-frequency transition is from quantum level 4 to level 1. The lowest- frequency transition is from quantum level 4 to level 3. 003 (part 2 of 3) 0.0 points An electron de-excites from the fourth quan- tum level to the third and then directly to the ground state. Two photons are emitted. How does the sum of their frequencies com- pare to the frequency of the single photon that would be emitted by de-excitation from the fourth level directly to the ground state?...
View Full Document

{[ snackBarMessage ]}

### Page1 / 17

Shih final - Version 079 – Final – shih –(56310 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online