# Lesson 9 : Chapter 2

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Unit 3 Lesson 9 29. Draw the FBD first: From the FBD, F_net=(Gm_1 m_3)/r^2 -(Gm_2 m_3)/r^2 F_net=(6.67×〖10〗^(-11) )(5.98×〖10〗^24 )(1200)/(3.0×〖10〗^8 )^2 -(6.67×〖10〗^(-11) )(7.35×〖10〗^22 )(1200)/(0.84×〖10〗^8 )^2 F_net=5.32-0.83375 F_net=4.48 N So, the net gravitational force on the rocket from the earth and the moon is 4.48 N. 30. a) By using g=Gm/r^2 r^2=Gm/g r=√(Gm/g) =√((6.67×〖10〗^(-11) )(5.98×〖10〗^24 )/7.2) = 7.44×〖10〗^6 7.44×〖10〗^6 m-6.38×〖10〗^6 m=1.06×〖10〗^6 m So, the meteor is 1.06×〖10〗^6 m high. b) Let m_1 be mass of meteor and m_2 be mass of earth. F_G=(Gm_1 m_2)/r^2 =(6.67×〖10〗^(-11) )(30)(5.98×〖10〗^24 )/(7.44×〖10〗^6 )^2 =216 N So, a 30 kg meteor will experience a force of…show more content…
So, the total electric field at charge 1 is 1.91 N/C [N 45o E]. 34.a) Draw the FBD first, let up and right be positive. From the FBD, the charge 1 is at rest and have a net force of 0. F_T cos⁡40°+(-mg)=0 F_T=mg/(cos⁡40°) =(0.15)(9.8)/(cos⁡40°) =1.92 N So, the tension in the thread is 1.92 N [N 40o W]. b) F_E+(-F_T sin⁡40°)=0 (kq^2)/r^2 =F_T sin⁡40° q^2=(〖r^2 F〗_T sin⁡40°)/k q=√((〖r^2 F〗_T sin⁡40°)/k) =√(((0.40)^2 (1.92) sin⁡40°)/(9.0×〖10〗^9 )) =±4.7×〖10〗^(-6) C The charge on 2 is +4.7×〖10〗^(-6) C. The charge is positive on 2 by Coulomb’s Law because it attracts 1, which holds a negative charge. 35. Let right and down be positive. a) F_net=F_E ma=qε a=qε/m =(1.6×〖10〗^(-19) )(4.0×〖10〗^2 )/(9.11×〖10〗^(-31) ) =7.0×〖10〗^13 So, the acceleration of the electron is7.0×〖10〗^13 m/s^2 [down]. b) ∆d_y=v_1y ∆t+1/2 a_y ∆t^2 0.02=1/2 (7.0×〖10〗^13 )∆t^2 ∆t=√((2)(0.02)/(7.0×〖10〗^13)) =2.4×〖10〗^(-8) s ∆d_x=v_1x ∆t =(4.0×〖10〗^6 )(2.4×〖10〗^(-8) ) =0.096 m So, the horizontal distance travelled by the electron when it hits the plate is 0.096 m. c) v_2^2=v_1^2+2a∆d v_2=√((4.0×〖10〗^6 )^2+2(7.0×〖10〗^13 )(0.020) ) =4.3×〖10〗^6 m/s So, the velocity of the electron as it strikes the plate is 4.3×〖10〗^6 m/s. Lesson 11 36. a) E_Etotal=(kq_1 q_2)/r_12 +(kq_1 q_3)/r_13 +(kq_2 q_3)/r_23