(1) Only turn on the S 1, R 1 = 6 access circuit, and the current value through R 1 = 6V/6ohm =1a.
(2) Only when S2 is closed, the resistor R 1 = 6 and the resistor R2 = 4 are connected in series, and the current value is = 6V/(6Ω+4Ω) = 0.6A. During the power-on period 10s, the electric heat generated by the current through R2 is Q2 = I2R2T = 0.6A * 0.6A * 4Ω *.
(3) According to P=UI, the power supply voltage U is constant, the current I is the largest when the total resistance is the smallest, and the electric power consumed by the whole circuit is the largest. When S 1S3 of the three switches is closed and S2 is open, the two resistors are connected in parallel, with the minimum total resistance and the maximum total power.
The maximum electric power at this time = UU/R1+UU/R2 = = (6+9) W.
20\ As shown in Figure 18A, the bottom area of horizontally rotating flat-bottomed cylindrical container A is 200cm? Cube B does not absorb water and weighs 10cm. It is placed at the bottom of the container. One end of the thin line with neglected mass and volume is fixed at the bottom of the container, and the other end is fixed at the center of the bottom of the box. The length of the thin line is l = 5 cm, and the known density of water is 1.0× 10? kg/m？ . Q:
In Figure A (1), the pressure of wood blocks on the bottom of the container = 5n/(0.10m * 0.1m) = 500 Pa.
(2) slowly add water to container a, and stop adding water when the thin line Zhang Liwei 1N is reached, as shown in figure 18b. At this time, the buoyancy of wood block B is = tension weighting force = 1N+5N=6N.
Immersion volume =6N/ (water density * g) = 6 * 10 (-4) m3,
3. When the line is not broken, the immersion depth of the block = immersion volume divided by the block bottom area = 6 *10 (-4) m3/(0.10m * 0.1m) = 6 *10 (-2) m =
Total volume of water =200cm? *5cm+(200cm？ - 10cm * 10cm)* 6cm = 1600 cm3
After cutting the thin line connected with B in Figure 18B, when the pulley is stationary, it floats on the water, and the buoyancy is equal to 5N gravity.
Immersion volume =5N/ (water density * g) = 5 * 10 (-4) m3, = 500 cm3.
Water depth = (1600+500) cm3/200 cm2 =10.5cm = 0.105m.
Pressure of water at the bottom of the container = density of water *g* depth of water = 1050pa.
It is difficult to do the problem. If you understand, please adopt it.