Consider a thin slice of the first rod, of thickness dx1, that is located a distance x1 from the origin and a thin slice of the second rod, of thickness dx2, that is located a distance x2 from the origin. Enter an expression for the magnitude of the force dF12 that the thin slice of rod 1 exerts on the thin slice of rod 2, in terms of L1, L2, Q1, Q2, x1, x2, and the Coulomb constant k.

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Friction force is equal to the net force of the car.             fn                           |ff< ·                       |                     fgfg and fn cancel out.55 km/h is 15.2778 m/s. this is the initial velocityfinal velocity is 0 m/sfind acceleration using kinematic equationv² = v₀² + 2adsince v = 00 = v₀² + 2ada = -v₀²/(2d)a = -(15.2778 m/s)/(2 * 38m) a = -3.2418 m/s²find net force and you find friction.f = ma f = (1165 .2418 m/s²)f = 3776.7 n = ffw = ff · dw = (3776.7 n)(38 m)w = -1.4 × 10^5 jthe friction forces do 1.4 × 10^5 j of work on the car