(3) v_2=90km/h=25m/s ,由P=W/t=(F_s)/t=Fv 得,F=P/(v_2)= (90000W)/(25m/s)=3600N ,由于货车匀速行驶,所以 f_m=F_L=3600N,货车与货物的总量G_B=(f_(Rt))/(0.1)=10f_B=36000N ,最多可装载的货物重 G_(40)=G_(32)=G=36000N=11000N=1.1*10^4N.相关推荐 1一...