| Chapter 13: Do listed problems. |
| Simulations |
| Quiz 13 |
| THE MASTERINGPHYSICS.COM ASSIGNMENT, BASED ON THESE PROBLEMS, ARE POSTED ! GO HERE http://www.masteringphysics.com/site. |
| Helpers!! Click this link then click the links "Practice Questions" and "Practice Problems" in the left hand column. |
2, 8, 13, 14, 17, 18, 25, 28, 33, 34, 40 |
| On these problems, for submission to masteringphysics.com, follow the rules for significant figures. Review your handouts and lab work. Try not to round off to the correct significant figures until the very last step. |
| Click here for 3,000 Solved Problems in Physics (Schaum's Solved Problems Series) |
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| NOTE: In some cases the masteringphysics.com problem has been randomized with different numbers than the book problem. The hints below are done according to book problem. If odd -numbered, you can check your answer in back of book before submitting to masteringphysics.com. |
| highlight #1: Note on #28: FxNET = - GmM/(0.10)2 - [Gm'M/(0.202 + 0.102)]cosØ. FyNET = [Gm'M/(0.202 + 0.102)]sinØ, |
| highlight #2: Note on #25: m + M = 150 kg |
| 2. (a) F = GmM/r2 . Where m and M are the 0.100 kg and 10 kg ball, respectively. r = 0.10 m (b) Compute F' = mg and compare with part (a). It should be very much larger. You can divide answer to part (a) by answer to part (b) . In other words, compute: ( GmM/r2 ) / (mg ) . |
| 8. g = GM/R2 where M and R are the
mass and radius of planet or moon in question. (a) M and R are the mass and radius of the Moon (Earth's moon) (b) M and R are the mass and radius of Jupiter. Jupiter's g is larger than the moon. Not surprising since Jupiter is two and a half times as massive as ALL of the other planets in our Solar System combined. For more information go here : http://en.wikipedia.org/wiki/Jupiter |
| 13. See example 13.2 for the Planet Earth. The formula for other planets is identical; just plug in planet mass and radius. |
| 14. This is a case when the launch speed is larger
than escape speed of example 13.2. Use conservation of
energy: ½mv12 - GmME/RE = ½mv22 + 0 since the potential energy at infinity is zero. Given the first speed, solve for the second. |
| 17. Using equation 13.25, find radius
of orbit with 5-earth-year period T. Please understand the
derivation of equation 13.25. M = sun mass. Then find
speed v = (circumference of circle)/T. See sec.
13.6. Note: one earth year
= (365)(24)(3600) seconds. |
| 18. Look up Earth's orbital radius and
period and use equation 13.25 to find sun mass M. Note: one year
= (365)(24)(3600) seconds. |
| 25. m + M = 150 kg and 8x10-6 N = GmM/r2 , where r = 0.20 m Solve these two equations for m and M. Here is the set up: First off solve for m*M = ( 8x10-6 N)*r2/G = real number. Then substitute m*M = (150 kg - M)*M = real number computed on the above line. Carefully solve this *quadratic* equation for M. Then find m. |
| 28. No good book examples help
explain this problem. This problem is important for Physics 4B. That
is where I step in. (a) This is the easiest because the two forces acting on the mass are perpendicular. Each represents a component of the net force. Plug the components into Pythagorean Theorem to find the magnitude. Let M = 20 kg. The x-component of the force on M is Fx = GmM/(0.10)2 , where m = 5.0 kg. The y-component is Fy = Gm'M/(0.20)2, where m' = 10 kg. The magnitude obeys F2 = Fx2 + Fy2. The vector points in the first quadrant . CW (clockwise) angle with y-axis is Ø = tan-1(| Fx|/| |Fy|) or you could find the angle with the x axis: Ø = tan-1(| Fy|/| |Fx|) (b) Yo, this is more complicated. Let m = 5.0 kg. From M = 20.0 kg, x-component is Fx = - GmM/(0.10)2 . From m ' = 10.0 kg , the x-component of net force is Fx = - [Gmm'/(0.202 + 0.102)]cosØ, where Ø= tan-1(20/10). From m ' = 10.0 kg , the y-component of net force is [Gmm'/(0.202 + 0.102)]sinØ, where Ø= tan-1(20/10). FxNET = - GmM/(0.10)2 - [Gm'M/(0.202 + 0.102)]cosØ. FyNET = [Gm'M/(0.202 + 0.102)]sinØ, The magnitude F obeys equation F2 = FxNET2 + FyNET2. The vector points in the second quadrant . The CCW angle with y-axis is Ø = tan-1(| Fx|/| |Fy|) or you could find the angle with the negative x axis: Ø = tan-1(| Fy|/| |Fx|) |
| 33. (a) ½mv12 - GmME/r1 = ½mv22 - GmME/r2 . The initial speed indexed by 1 is zero, r2 = RE and r1 = RE + 500 km (convert to meters.) Find the second speed. (b) ½mv12 + mgy1 = ½mv22 + mgy2 , where the initial speed indexed by 1 is zero, y2 = 0 and y1 = 500 km (convert to meters.) Find second speed and compare with part (b). |
| 34. ½mv12 - GmME/r1
= ½mv22 - GmME/r2
. The initial speed indexed by 1 is 10,000 km/hr ( convert to m/s), r1 = RE . If the second speed is zero, find r2 . Substract the Earth radius to get the height above the ground. |
| 40. ½mv12 - GmME/(r' + Rmoon) - GmMmoon/Rmoon= 0, where r' = distance between moon and earth centers. Solve for v1 . |
| highlight #1: Note on #28: FxNET = - GmM/(0.10)2 - [Gm'M/(0.202 + 0.102)]cosØ. FyNET = [Gm'M/(0.202 + 0.102)]sinØ, |
| highlight #2: Note on #25: m + M = 150 kg |