M v m v v v v α α θ φ φ φ we can work with

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m v m v v v v α α θ φ φ φ = = × ° − = = × We can work with relative masses (4 and 192) since the conversion to kg occurs in all terms of the equation and cancels out. Dividing the y -equation by the x -equation gives 1 fn fn sin 2.283 tan 2.152 tan (2.152) 65 cos 1.0609 v v φ φ φ φ = = = = = ° The speed of the recoiling nucleus is 2 2 2 2 5 5 fn fn fn ( ) ( ) (1.0609) (2.203) 10 m/s 2.52 10 m/s x y v v v = + = + × = × Thus the nucleus recoils at 2.52 × 10 5 m/s in a direction 65° below the x -axis.
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38.47. Solve: (a) The Coulomb force between the electron and the proton causes the centripetal acceleration for the electron’s motion in a circular orbit. That is, ( ) 2 2 2 2 0 0 1 1 4 4 r e e mv e F mv r r r πε πε = = = The total energy of the electron is a sum of its kinetic energy and electric potential energy: ( ) ( ) 2 2 2 2 0 0 0 0 1 1 1 1 1 2 4 2 4 4 8 e e e e e E mv r r r r πε πε πε πε = + = + = − (b) The ratio of the potential energy to the kinetic energy is ( ) 2 2 0 0 2 2 1 1 2 0 2 4 4 2 4 U e r e r K mv e r πε πε πε = = = − (c) The part (a) energy is negative because the electron is bound. To remove the electron to infinity, where E = 0, you need to give it energy ionization E E = . Thus ( )( ) ( ) ( ) ( ) 19 2 ionization 0 2 9 2 2 19 11 19 1.60 10 J 13.6 eV 1 eV 8 9.0 10 N m /C 1.60 10 C 1 eV 1 5.29 10 m 2 13.6 eV 1.60 10 J e E r r πε × = × = × × = = × × To calculate the speed, we have ( ) ( ) 2 1 ionization 2 2 . E E U K K K K mv = − = − + = − − + = = Therefore, the speed is ( ) 19 ionization 31 2 13.6 eV 2 1.60 10 J 9.11 10 kg 1 eV E v m × = = × × = 2.19 × 10 6 m/s.
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38.48. Model: Assume that the electric field between the plates is uniform. Visualize: Please refer to Figure 38.9. Solve: (a) The electric force on a positive charge q in a field G E is E . F qE = G G If a droplet is charged with charge q and is suspended between the plates by applying a field E 0 , then the electric force must exactly cancel the weight force: G F net = G F E + G w = 0 N qE 0 mg = 0 N 0 mg q E = (b) A droplet falling at the terminal speed has no net force acting on it. The velocity is negative for a falling drop, or v = − v term . Thus, G F net = G F drag + G w = 0 N bv mg = + bv term mg = 0 N term mg v b = (c) Because F drag = 6 π η rv = bv , b = 6 π η r . The result in part (b) for v term thus simplifies to v term = mg 6 πη r . Noting that 3 4 3 , m V r π ρ ρ = = the expression for the terminal velocity is ( ) 3 4 2 3 term term 2 9 6 9 2 r g r g v v r r g π ρ ρ η πη η ρ = = = (d) We first need to find the mass m of the droplet before we can obtain q from the expression in part (a). Using the expression for r in part (c), the mass of the oil drop is 3/ 2 3 air term oil oil oil oil 4 4 9 3 3 2 v m V r g π π η ρ ρ ρ ρ = = = The terminal velocity is 3 term 3.00 10 m 7.33 s. v = × The mass of the oil drop is ( ) ( ) ( ) ( )( ) 3/ 2 5 3 3 14 3 2 9 1.83 10 kg m/s 3.00 10 m 7.33 s 4 860 kg/m 2.881 10 kg 3 2 860 kg/m 9.8 m/s m π × × = = × ( )( )( ) 14 2 2 18 2.881 10 kg 9.8 m/s 1.0 10 m 2.40 10 C ( / ) 1177 V mg q V d × × = = = × Δ (e)
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