1230 Words5 Pages

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 cos40°+(-mg)=0

F_T=mg/(cos40°)

=(0.15)(9.8)/(cos40°)

=1.92 N

So, the tension in the thread is 1.92 N [N 40o W].

b) F_E+(-F_T sin40°)=0

(kq^2)/r^2 =F_T sin40°

q^2=(〖r^2 F〗_T sin40°)/k

q=√((〖r^2 F〗_T sin40°)/k)

=√(((0.40)^2 (1.92) sin40°)/(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

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

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 cos40°+(-mg)=0

F_T=mg/(cos40°)

=(0.15)(9.8)/(cos40°)

=1.92 N

So, the tension in the thread is 1.92 N [N 40o W].

b) F_E+(-F_T sin40°)=0

(kq^2)/r^2 =F_T sin40°

q^2=(〖r^2 F〗_T sin40°)/k

q=√((〖r^2 F〗_T sin40°)/k)

=√(((0.40)^2 (1.92) sin40°)/(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

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