初中物理求过程
1)、p = mg/S = 2000 * 10/ ( 4 * 125 * 0.0001 ) = 400000 pa = 400kpa;
静止时对地面压强为 400kpa;
2)、匀加速度 a = V/t = (100 * 1000/3600)(m/s) / 10s = 25/9 m/s^2;
最大速度 V1 = at1 = 15 * 25/9 = 125/3 m/s;
匀速阶段位移 x2 = V1t2 = ( 40 - 15 ) * 125/3 = 3125/3 m;
匀速阶段拉力 F = 阻力 = 2000 * 0.1 * 10 = 2000N;
加速阶段发动机做功 W1 = Pt2 = 80 * 10^3( 40 - 15 ) = 2 * 10^6 J;
匀速阶段发动机做功 W2 = Fx2 = 2000 * 3125/3 = 6.25/3 * 10^6 J;
整个测试阶段发动机做功 W = W1 + W2 = 2 * 10^6 + 6.25/3 * 10^6 = 12.25/3 * 10^6 J;
3)、加速阶段位移 x1 = at1^2/2 = 15^2 * (25/9) /2 = 625/2 m;
匀减速度 a1 = V1/t3 = (125/3)/( 50-40 ) = 25/6 m/s^2;
减速阶段位移 x3 = a1t3^2/2 = 10^2 * (25/6) /2 = 625/3 m;
全过程平均速度 V = ( x1 + x2 + x3 )/50 = ( 625/2 + 3125/3 + 625/3 )/50 = 31.25 m/s
= 112.5 km/h 。
(2)解:以额定功率运动运动了40s,W=Pt=80000w×40s=3.6×106J
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