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 Posted: Fri 15:36, 07 Jan 2011 Post subject: ugg boots Electric field in the class of flat Proj |
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Electric field in the class of flat Projectile Motion
Sub-levels to do a similar throwing motion. Direction: L-vt direction: Y E A No. 2 lost thirty q (L). Two charged particles incident offset ratio of one to 1Eq (L) 2 f soil one by one 4 Dan, 22, thirty, 2vE a 2Ek-E = EI a report (2)-E equations have a solution 4E: E . One 4.25E. Summary of changes when the initial conditions, it should be the correct problem solving steps, and then analyzed again from scratch. Rather than assuming the conclusions of the previous copy to the next question to ask. Thus, strictly in accordance with the basic problem-solving steps,[link widoczny dla zalogowanych], to ensure the accuracy of solving problems. Improve efficiency. The reason is that steps are summarized from the application of the law out of need. 2 gravity can not ignore the charged body in the direction of vertical uniform electric field in the class level 2 Projectile Motion Problems on the quality of the thirty = 5 × lO kg discussion of charged particles h h e learning teaching ∞ P Deep Li o Volume 24 VOLUME 266 4, 2006 (second half) Physics Teaching JournalofPhysicsTeachingVoI_24No. 266 (X) 04.2006.19 son,[link widoczny dla zalogowanych], with a rate of 73o 2m / s speed from horizontal parallel plates A,[link widoczny dla zalogowanych], B flying into the central level of the electric field, known to the metal plate length L = 0. im, the distance between the plates d: 2 × 10m, when the U cents = IO00V,[link widoczny dla zalogowanych], the charged particles exactly along a straight line through the field, as shown in Figure 3, if two potential difference between plates adjustable, to make the particles fly from the cubicle a, U number of changes in range (g get 10m / s). Analysis and Solutions When J,[link widoczny dla zalogowanych], Xian a 1000V, by the equilibrium conditions: rng = qU/dmgd5 × 1O a × 1OX2 × 10I2 Ding == I:! Eleven U1OO a 1 × 10 a \L / v = 0.1/2-0.05s when the plate voltage changes, the maximum lateral acceleration of particles: to = z \Second Law are: a state for a Caesar f U:-m-(-g-n-U-a-)-d: 1800Vq downward deflection when the particles are: mg a ==: delete U:-m (g- - a) d the value of a 200VqU cents to 200V <Ux, <1800V3 gravity can not ignore the charged body in uniform electric field in the class level 3 Projectile Motion Problems on distance d = 0.1m, a potential difference U erection of two parallel plates 10V intermediate, with a long thin line Z === 0.01m hanging a mass = 0.1: c: 2g, charge q = 10C band won a positively charged ball, the just thread the ball was pulled to a horizontal position A shown in Figure 3 after the release gently, and asked: (1) placed the ball to the lowest point B the speed and line pull how much? (2) If the ball put to the B point thread suddenly broken through after the ball just to point B is below the C point, the / 3 (7 How far apart? (g = lOm / s) of the ball with a positive charge, just point at A When released, the electrostatic force in the horizontal and vertical right down the gravity, taut thread immediately, since the ball in the electric force, gravity, thread tension speed circular motion under the action of doing. In a small ball B after graduation, the horizontal electric field only by a constant force, as \acting was mgl a = 1 Ⅲ u2B (1) in the B point, according to circular movement of the transient characteristics. by a rope in tension and gravity force as the centripetal force, that is T-mg: (2) E a U (3) and simultaneous (1 ) ~ (3) c-type, have u = √ 2 一一% / 2X10X0.01 - 2X10aX10.7X0.01m / s === 0.32m / S Ding g + 3g one of a pregnant 2Eq = (3X0.2 10 A X10-2 × sowing × lON: 4 × 10 a N (2) in the B point out, a small ball in the horizontal direction of the electric field by a constant force, so that the ball have the right level of acceleration.: fish :: d === 5m / s. 10 × 10 one m / s. Therefore, in the horizontal direction the ball to do uniform deceleration, like \: z × 0.128s for the ball in the vertical direction only by gravity, from the beginning when the vertical velocity is zero, so the ball in the vertical direction of free fall. h ===: i × 10X (o .128): 0.08m.
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