Test #2-solutions

# Test #2-solutions - Version 086/ABBBC Test#2...

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

Version 086/ABBBC – Test #2 – Antoniewicz – (57420) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A satellite circles planet Roton every 8 . 1 h in an orbit having a radius oF 1 . 9 × 10 7 m. IF the radius oF Roton is 8 . 17 × 10 6 m, what is the magnitude oF the Free-Fall acceleration on the surFace oF Roton? 1. 5.39001 2. 26.9821 3. 47.4022 4. 4.77091 5. 8.9794 6. 50.0457 7. 23.3898 8. 444.318 9. 1.82421 10. 6.34291 Correct answer: 4 . 77091 m / s 2 . Explanation: Basic Concepts: Newton’s law oF gravi- tation F g = G m 1 m 2 r 2 . Kepler’s third law T 2 = p 4 π 2 G M P r 3 . The Free-Fall acceleration a on the surFace oF the planet is the acceleration which a body in Free Fall will Feel due to gravity F g = G M m R 2 = m a , where M is the mass oF planet Roton. This acceleration a is a = G M R 2 , (1) the number which is g on Earth. Here, how- ever, the mass M is unknown, so we try to fnd this From the inFormation given about the satellite. Use Kepler’s third law For the period oF the orbit T 2 = p 4 π 2 G M P r 3 . (2) By multiplying both sides with R 2 and com- paring to equation (1), we can identiFy our a in the right hand side T 2 R 2 = p 4 π 2 a P r 3 . IF we solve For a , we obtain a = p 4 π 2 T 2 R 2 P r 3 = 4 . 77091 m / s 2 which is our answer. Although identiFying a in this way is a “quick” way oF solving the problem, we could just as well have calculated the planet mass M explicitly From equation (2) and inserted into equation (1). 002 10.0 points A bucket Full oF water is rotated in a verti- cal circle oF radius 1 . 118 m (the approximate length oF a person’s arm). What must be the minimum speed oF the pail at the top oF the circle iF no water is to spill out? 1. 3.13675 2. 3.31005 3. 2.49656 4. 3.71989 5. 3.84044 6. 2.52389 7. 2.34683 8. 3.5265 9. 3.34831 10. 3.39626 Correct answer: 3 . 31005 m / s. Explanation: IF we analyze the Forces on the water at the top oF a vertical circle (using the notation F ji For a two body Force on ˆ i From ˆ j ), we see that N pail,water + W earth,water = m a centripetal = m v 2 r .

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

View Full Document
Version 086/ABBBC – Test #2 – Antoniewicz – (57420) 2 To minimize the velocity, we minimize the lhs of the equation. Since we can’t change the weight of the water, we use the lowest Normal force that we can, 0. Then, m g = m v 2 r and so, v = g r = r (9 . 8 m / s 2 ) (1 . 118 m) = 3 . 31005 m / s . 003 10.0 points The horizontal surface on which the objects slide is frictionless. The acceleration of gravity is 9 . 8 m / s 2 . 1 kg 8 kg 6 kg F F r If F = 23 N and F r = 7 N, what is the magnitude of the force exerted on the block with mass 8 kg by the block with mass 6 kg? 1. 12.0769 2. 8.84615 3. 7.94118 4. 12.5714 5. 10.4286 6. 11.5625 7. 7.8 8. 10.0 9. 9.57143 10. 13.4 Correct answer: 13 . 4 N. Explanation: m 1 m 2 m 3 F F r Given : V F = +23 N ˆ ı , V F r = 7 N ˆ ı , m 1 = 1 kg , m 2 = 8 kg , m 3 = 6 kg , and g = 9 . 8 m / s 2 . Note: F is acting on the combined mass of the three blocks, resulting in a common acceleration after accounting for friction.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 10

Test #2-solutions - Version 086/ABBBC Test#2...

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

View Full Document
Ask a homework question - tutors are online