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Chapter 5 Problems

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cinderella
25 diapositives · 474 Views · Oct 31, 2013

Chapter 5 Problems. Gravitation: 25,26,29,30,37,38 Satellites/Apparent Weight: 39,40,41,43,44,49. Chapter 5 Problem 25. Given r earth surface to spaceship =12800km=1.28x10 7 m r earth center to spaceship =1.28x10 7 m+6.3810 6 m m=1400kg Find F g =? N FRSA

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Chapter 5 Problems
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  1. Chapter 5 Problems • Gravitation: 25,26,29,30,37,38 • Satellites/Apparent Weight: 39,40,41,43,44,49

  2. Chapter 5 Problem 25 • Given • rearth surface to spaceship =12800km=1.28x107m • rearth center to spaceship =1.28x107m+6.38106m • m=1400kg • Find • Fg =? N • FRSA • Fg= G mme = 6.67x10-11 Nm2/kg2(1400kg)(5.98x1024kg) • r2 (1.28x107m+6.38106m)2 • Fg=1520 N

  3. Chapter 5 Problem 29 • Given: • m=2.10kg • g = 12.0 m/s/s • Find • m=? Kg • Weight = Fg=?N • FRSA • m = 2.10 kg • Weight = Fg = mg = 2.10kg( 12.0 m/s/s) = 25.20 N

  4. Chapter 5 Problem 26 • Given Find • mmoon = 7.35x1022kggmoon = ? m/s/s • mearth = 5.98 x1024 kg • rmoon = 1.74x106m mgmoon = G mmm • rearth = 6.38 x106 m r2moon • g = 9.8 m/s/s mgearth = G mmm • r2earth • gmoon = gearthmmoon (r2earth) • mearth (r2moon) • gmoon = (9.8m/s/s)(7.35x1022kg)(6.38 x106 m)2 =1.62m/s/s • (5.98 x1024 kg)(1.74x106m)2

  5. Chapter 5 Problem 30 • Given • rearth surface to 300km =300km=3.00x107m • rearth center to 300km =3.00x105m+6.38106m • m=80kg • Find • Fg =? N • FRSA • Fg= G mme = 6.67x10-11 Nm2/kg2(80kg)(5.98x1024kg)=715 N • r2 (3.00x105m+6.38106m)2 91% • Fg= G mme = 6.67x10-11 Nm2/kg2(80kg)(5.98x1024kg)=784 N • r2 (6.38106m)2

  6. Chapter 5 Problem 37 • Given Fin • mmars =? • mearth = 5.98 x1024 kg • rmoon = 3.20x106m mgmars = G mmm • rearth = 6.38 x106 m r2mars • gearth = 9.8 m/s/s mgearth = G mmm • gmars =.32(9.8m/s/s)r2earth • =3.14m/s/s gmars (r2mars ) mearth = mmoon • gearth (r2earth) • mmars = (314m/s/s)(3.2 x106 m)2 (5.98 x1024 kg)=4.72x1024kg • (6.38x106m)2

  7. Chapter 5 Problem 38 • Given • t=1year (365d)(24hr)(60min)(60sec)=3.15x107s • (1year)(1day)(1hr )(1min ) • r=1.51x1011m • Find • msun • FRSA • Fg=Fc • Gmms= mv2 ms = rv2= • r2 r G msun(1.51x1011m)(2p(1.51x1011m)/3.15x107s)2 6.67x10-11 Nm2 / kg2 msun=2.0x1030kg

  8. Chapter 5 Problem 39 • Given3600km • r=6.38x106m+3.6x106 m =9.98x106m • Find • msun • FRSA • Fg=Fc • Gmms= mv2 Gme = v = 6.67x10-11 (5.98x1024kg) • r2 r r (9.98x106m) • v= 6300 m/s

  9. Chapter 5 Problem 40 • Given • Mmonkey = 17.0kg • FT=220 N • Find • a = ? m/s/s • Fnet = ma • FT – Fg = ma • FT – mg = a = 220N – 17kg(9.8m/s2) = 3.14m/s/s up • m 17kg

  10. Chapter 5 Question 41. • Given • g = ½ g ½ g = v2 • r = 16m r • Find • v=m/s ½ rg = v = ½ (16m) (9.8m/s/s) = 8.85m/s • FRSA • FN=Fc v = 2 p r • FN=mv2 t • r • m ½ g=mv2 t = 2 p r2 p (16m) = 11 seconds per rev • r v 8.85m/s

  11. Chapter 5 Problem 43 • Given100km • r=1.74x106m+1.0x105 m =1.84x106m mmoon=7.35x1022kg • Find • msun • FRSA • Fg=Fc • Gmmm= mv2 Gmm = v = 6.67x10-11 (7.35x1022kg) • r2 r r (1.84x106m) • v= 1632 m/s v=2pr t = 2pr = 2 p (1.84x106m) = 7083s • t v 1632m/s or (2hrs)

  12. Chapter 5 Problem 44 • S Fy = 0 for at rest, constant up, constant down • FN=Fg FN = mg = 58kg(9.8m/s/s) = 568 N • S Fy = ma for accelerating up • FT-Fg = ma FT-mg= ma FT = mg + ma • FT = (58kg*9.8m/s/s) + (58kg*.33*9.8m/s/s)=756N • S Fy = ma for accelerating down • Fg-FT = ma mg-FT = ma mg – ma=FT • (58kg*9.8m/s/s) - (58kg*.33*9.8m/s/s)=381N • S Fy = ma for free fall • Fg = ma mg=ma g=a and FT=0

  13. Bipolar Star System Two stars 3.6x1011m apart rotate about a point 1.8x1011 m from each other in a circular path in 5.0 years. The two stars have the same mass. What is the velocity of the stars? What is the mass of the stars?

  14. Bipolar Star System Two stars 3.6x1011m apart rotate about a point 1.8x1011 m from each other in a circular path in 5.0 years. The two stars have the same mass. What is the velocity of the stars? What is the mass of the stars? 3.6x108m

  15. Bipolar Star System Two stars 3.6x1011m apart rotate about a point 1.8x1011 m from each other in a circular path in 5.0 years. The two stars have the same mass. . What is the velocity of the stars? What is the mass of the stars? 1.8x1011m

  16. Bipolar Star System Two stars 3.6x1011m apart rotate about a point 1.8x1011 m from each other in a circular path in 5.0 years. The two stars have the same mass. What is the velocity of the stars? v = 2 p r t

  17. Bipolar Star System Two stars 3.6x1011m apart rotate about a point 1.8x1011 m from each other in a circular path in 5.0 years. The two stars have the same mass. What is the velocity of the stars? v = 2 p r t v = 2 p ( 1.8x10 11m) = 7170 m/s (5.0yrs(365d)(86400s) ( 1yr )( 1 day )

  18. Bipolar Star System Two stars 3.6x1011m apart rotate about a point 1.8x1011 m from each other in a circular path in 5.0 years. The two stars have the same mass. What is the mass of the stars? Fg Fg

  19. Bipolar Star System Two stars 3.6x1011m apart rotate about a point 1.8x1011 m from each other in a circular path in 5.0 years. The two stars have the same mass. What is the mass of the stars? Fg Fg Fg causes the centripetal acceleration therefore Fg = Fc

  20. Bipolar Star System Two stars 3.6x1011m apart rotate about a point 1.8x1011 m from each other in a circular path in 5.0 years. The two stars have the same mass. What is the mass of the stars? Fg=Fc Fg causes the centripetal acceleration therefore Fg = Fc

  21. Bipolar Star System Two stars 3.6x1011m apart rotate about a point 1.8x1011 m from each other in a circular path in 5.0 years. The two stars have the same mass. What is the mass of the stars? Fg=Fc G mm = m v2 (8x1010m)2 4x1010m Fg causes the centripetal acceleration therefore Fg = Fc Distance between Distance to centers the center of rotation

  22. Bipolar Star System Two stars 3.6x1011m apart rotate about a point 1.8x1011 m from each other in a circular path in 5.0 years. The two stars have the same mass. What is the mass of the stars? Fg=Fc G mm = m v2 (3.6x1011m)2 1.8x1011m Fg causes the centripetal acceleration therefore Fg = Fc Distance between Distance to centers the center of rotation

  23. Bipolar Star System Two stars 3.6x1011m apart rotate about a point 1.8x1011 m from each other in a circular path in 5.0 years. The two stars have the same mass. Fg=Fc G mm = m (7170m/s)2 (3.6x1011m)2 1.8x1011m Fg causes the centripetal acceleration therefore Fg = Fc

  24. Bipolar Star System Two stars 3.6x1011m apart rotate about a point 1.8x1011 m from each other in a circular path in 5.0 years. The two stars have the same mass. What is the mass of the stars? Fg=Fc G m = (7170m/s)2 (3.6x1011m)2 1.8x1011m Fg causes the centripetal acceleration therefore Fg = Fc

  25. Bipolar Star System Two stars 3.6x1011m apart rotate about a point 1.8x1011 m from each other in a circular path in 5.0 years. The two stars have the same mass. Fg=Fc G m = (7170m/s)2 (3.6x1011m)2 1.8x1011m m = (7170m/s)2 (3.6x1011m)2 = 5.5x1029 kg 6.67x10-11Nm2/kg2 (1.8x1011m) Fg causes the centripetal acceleration therefore Fg = Fc