As shown in the figure, on the horizontal plane, there are two smooth metal guides without resistance, which are parallel and fixed. The spacing D is 0.5m. The right end is connected with a small bulb l with resistance of 4 Ω through a wire. There is a uniform vertical magnetic field in the rectangular area of cdfe. The length of CE is 2m. The change of magnetic induction intensity B with time t in the area of cdfe is shown in the figure. When t = 0s, a metal bar with resistance of 1 Ω acts under constant force F When the metal rod moves from ab position to ef position, the brightness of the small bulb does not change______ N. The mass of the metal bar is______  kg．

As shown in the figure, on the horizontal plane, there are two smooth metal guides without resistance, which are parallel and fixed. The spacing D is 0.5m. The right end is connected with a small bulb l with resistance of 4 Ω through a wire. There is a uniform vertical magnetic field in the rectangular area of cdfe. The length of CE is 2m. The change of magnetic induction intensity B with time t in the area of cdfe is shown in the figure. When t = 0s, a metal bar with resistance of 1 Ω acts under constant force F When the metal rod moves from ab position to ef position, the brightness of the small bulb does not change______ N. The mass of the metal bar is______  kg．

Within 0-4s, the induced electromotive force generated in the circuit is e = △ B △ TS = 0.5 × 0.5 × 2V = 0.5V, and the induced current is I = ER + r = 0.54 + 1A = 0.1A. When the metal rod moves from ab position to ef position, the brightness of the small bulb does not change, which indicates that the induced electromotive force generated by the metal rod entering the magnetic field after 4S is equal to that generated in 0-4s; F = bil = 2 × 0.1 × 0.5N = 0.1N when the metal rod just enters the magnetic field, it is obtained by E = BLV, v = EBL = 0.52 × 0.5m/s = 0.5m/s, v = at, a = FM, v = fmtm = FTV = 0.1 × 40.5kg = 0.8kg

As shown in the figure, the distance between two parallel long straight metal guides fixed in the same horizontal plane is D, and the right end is connected with a resistance of R. the whole device is in a uniform magnetic field with vertical upward magnetic induction intensity of B. a conductor rod AB with mass m (uniform mass distribution) is placed perpendicular to the guide rail and keeps good contact with the two guide rails. The dynamic friction between the rod and the guide rail is stable The factor is μ. Under the action of constant force F, which is horizontal to the left and perpendicular to the rod, the speed of the rod just reaches the maximum when it moves along the guide rail from rest (the rod always keeps vertical with the guide rail in the process of movement). If the resistance of the rod connecting to the circuit is r, the resistance of the guide rail is ignored, and the acceleration of gravity is g
A. The maximum velocity of the rod is f − μ MGRC. The sum of work done by constant force F and friction force is equal to the change of rod kinetic energy. D. the sum of work done by constant force F and Ampere force is equal to the change of rod kinetic energy

A. Let the maximum velocity of the rod be V, then the ampere force on the rod is fa = b2l2vr + r = b2d2vr + R, and the force on the rod is balanced, then f = FA + μ mg, and the solution is v = (f − μ mg) (R + R) b2d2

In the series circuit, the two constant resistors are 1 ohm and 2 ohm respectively. Calculate the resistance of the circuit (excluding the wire resistance)
I know the ratio is 1:1, but what is the resistance of the whole circuit

In a series circuit, the total resistance is equal to the sum of the partial resistances
So, r = R1 + R2 = 1 + 2 = 3 (Euro)
A: slightly