Analyze The directions of the individual forces exerted by q, and q, on qz are shown in the figure. The force F23 exerted by q, on q3 is (---Select--because q, and q, have opposite signs. In the coordinate systemshown in the figure, the force F23 is to the left (in the negative x-direction).The force F,3 exerted by q, on 93 is--Select--- +) because both charges are positive. The force F.13makes an angle of 45.0° with the x-axis.Use Coulomb's law to find the magnitude of F23 (in N):192|193|F23 = kea2NFind the magnitude of the force F13 (in N):19.|193|F13 = Ke(vz a)?NFind the x- and y-components of the force F13 (in N):F13xF13y(F13)cos(45.0°) =(F13)sin(45.0°) =NNFind the components of the resultant force (in N) acting on q3:F3x = F13x + F23xN + (-4.37 N)F3y = F13y + F23yN + ONExpress the resultant force (in N) acting on q3 in unit-vector form:F, (5) Nî +3. Consider three point charges located at the corners of a right triangle as shown in the figure, where q, = 933 3.50 иС, q2= -2.00 µC, and a = 0.120 m. Find the resultant force exerted on g3.The force exerted by q, on q, is F13: The forceexerted by q2 on 93 is F23. The resultant force F3is the vector sum F13 + F23·a239293a2a91 +

Answered: Image /qna-images/answer/54b484bb-816e-4481-ae52-0d370556d163.jpg

Analyze The directions of the individual forces exerted by q, and q, on 9z are shown in the figure. The force F23 exerted by 9, on 9z is -Select--- because q, and 9, have opposite signs. In the coordinate sys shown in the figure, the force F23 is to the left (in the negative x-direction). The force F,, exerted by q, on q, is -Select- 9 because both charges are positive. The force F,a makes an angle of 45.0° with the x-axis. 93 Use Coulomb's law to find the magnitude of F22 (in N): 192|193| F23 = ke a2 Find the magnitude of the force F13 (in N): F13 = ketvz a)2 Find the x- and y-components of the force F,3 (in N): F13x - (F13)cos(45.0°) = F13y = (F13)sin(45.0°) = Find the components of the resultant force (in N) acting on 93: F3x = F13x + F23x = F3y = F13y + F23y = N + (-4.37 N) N +ON Express the resultant force (in N) acting on 9z in unit-vector form: i+

Analyze The directions of the individual forces exerted by q, and q, on 9z are shown in the figure. The force F23 exerted by 9, on 9z is -Select--- because q, and 9, have opposite signs. In the coordinate sys shown in the figure, the force F23 is to the left (in the negative x-direction). The force F,, exerted by q, on q, is -Select- 9 because both charges are positive. The force F,a makes an angle of 45.0° with the x-axis. 93 Use Coulomb's law to find the magnitude of F22 (in N): 192|193| F23 = ke a2 Find the magnitude of the force F13 (in N): F13 = ketvz a)2 Find the x- and y-components of the force F,3 (in N): F13x - (F13)cos(45.0°) = F13y = (F13)sin(45.0°) = Find the components of the resultant force (in N) acting on 93: F3x = F13x + F23x = F3y = F13y + F23y = N + (-4.37 N) N +ON Express the resultant force (in N) acting on 9z in unit-vector form: i+

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Newton’s Third Law of Motion

The horse pulls the cart and the cart pulls the horse in response as per Newton third law. So the big question is, “Don’t the two forces cancel each other? How does the cart accelerate?”

Question

Analyze The directions of the individual forces exerted by q, and q, on qz are shown in the figure. The force F23 exerted by q, on q3 is (---Select--
because q, and q, have opposite signs. In the coordinate system
shown in the figure, the force F23 is to the left (in the negative x-direction).
The force F,3 exerted by q, on 93 is
--Select--- +) because both charges are positive. The force F.
13
makes an angle of 45.0° with the x-axis.
Use Coulomb's law to find the magnitude of F23 (in N):
192|193|
F23 = ke
a2
N
Find the magnitude of the force F13 (in N):
19.|193|
F13 = Ke(vz a)?
N
Find the x- and y-components of the force F13 (in N):
F13x
F13y
(F13)cos(45.0°) =
(F13)sin(45.0°) =
N
N
Find the components of the resultant force (in N) acting on q3:
F3x = F13x + F23x
N + (-4.37 N)
F3y = F13y + F23y
N + ON
Express the resultant force (in N) acting on q3 in unit-vector form:
F, (
5) N
î +
3.

Consider three point charges located at the corners of a right triangle as shown in the figure, where q, = 93
3 3.50 иС, q2
= -2.00 µC, and a = 0.120 m. Find the resultant force exerted on g3.
The force exerted by q, on q, is F13: The force
exerted by q2 on 93 is F23. The resultant force F3
is the vector sum F13 + F23·
a
23
92
93
a
2a
91 +

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