一道物理题,明天要交,各位大侠帮帮忙,怎么做,要具体的计算过程
\u4e00\u9053\u7269\u7406\u9898\u76ee\uff0c\u8bf7\u5404\u4f4d\u5927\u4fa0\u5feb\u5e2e\u5e2e\u5fd9\u5206\u6790\u7269\u7406\u8fc7\u7a0b\u5bb9\u6613\u5f97\u5230:
0.2s\u5185,\u5f39\u7c27\u5904\u4e8e\u4f38\u957f\u72b6\u6001.0.2s\u65f6,\u5f39\u7c27\u5728\u539f\u957f.0.2s\u540e,\u5f39\u7c27\u4e0e\u7269\u4f53\u5206\u79bb.
\u5f00\u59cb\u5f39\u7c27\u4f38\u957f\u91cfx=F/k=mg/k=0.15m
\u7531s=(1/2)at^2=0.15\u5f97
a=12m/s^2
F\u7684\u6700\u5c0f\u503c\u51fa\u73b0\u5728\u5f00\u59cb\u65f6\u523b(\u5f39\u7c27\u5f39\u529b\u6700\u5927)
Fmin=m2a=10.5*12=126N
0.2s\u540e,F\u4fdd\u6301\u4e0d\u53d8,\u63d0\u4f9b\u52a0\u901f\u5ea6
Fmax-m2g=m2a
Fmax=m2a+m2g=231N
1\u7535\u8def\u8fde\u63a5\u65b9\u6cd5\uff1a
\u5f85\u6d4b\u7535\u963bRx\u3001\u7535\u538b\u6052\u5b9a\u4e0d\u53d8\u4f46\u672a\u77e5\u7684\u7535\u6e90\u3001\u5df2\u77e5\u7535\u963bR\uff0c\u4e09\u8005\u4e32\u8054\u4e3a\u95ed\u5408\u56de\u8def\u3002\u7535\u538b\u8868\u5e76\u8054\u5728\u5df2\u77e5\u7535\u963bR\u7684\u4e24\u7aef\uff0c\u6d4b\u5df2\u77e5\u7535\u963bR\u7684\u4e24\u7aef\u7684\u7535\u538b\u3002\u5f00\u5173\u5e76\u8054\u5728\u5f85\u6d4b\u7535\u963bRx\uff0c\u7528\u4e8e\u77ed\u8def\u5f85\u6d4b\u7535\u963bRx\u3002
2\u3001\u6d4b\u91cf\u65b9\u6cd5\uff1a
\u5148\u95ed\u5408\u5f00\u5173\uff0c\u77ed\u8def\u5f85\u6d4b\u7535\u963bRx\u3002\u6b64\u65f6\u7535\u538b\u8868\u8bfb\u6570\u4e3a\u7535\u538b\u7535\u538b\uff0c\u4e5f\u662f\u5df2\u77e5\u7535\u963bR\u4e24\u7aef\u7684\u7535\u538b\uff0c\u8bb0\u4e3aV1
\u518d\u65ad\u5f00\u5f00\u5173\uff0c\u6b64\u65f6\u7535\u538b\u8868\u8bfb\u6570\u4e3a\u5df2\u77e5\u7535\u963bR\u4e24\u7aef\u7684\u7535\u538b\uff0c\u8bb0\u4e3aV2
\u5219Rx/(V1-V2)=R/V2
Rx=R*(V1-V2)/V2
\u7b2c\u4e8c\u9898\u867d\u7136\u4f60\u6ca1\u6709\u6807\u660eABCD\u5404\u70b9\u7684\u4f4d\u7f6e\uff0c\u4f46\u662f\u6211\u5206\u6790\u5e94\u8be5\u662f\u9009\u62e9\uff08D \u7535\u8def\u7684C\u3001D\u4e24\u70b9\u95f4\u5bfc\u7ebf\u65ad\u8def\uff09\uff0c\u56e0\u4e3aABC\u4e09\u4e2a\u9009\u9879\u5747\u4f1a\u5bf9\u7535\u706f\u9020\u6210\u5f71\u54cd\uff0c\u867d\u7136\u56fe\u4e2d\u6ca1\u6709\u6807\u660eC\u3001D\u4e24\u70b9\u7684\u4f4d\u7f6e\uff0c\u4f46\u662f\u5c31\u53ea\u5269\u8fd9\u4e2a\u9009\u9879\u80fd\u9009\u4e86\u3002
2)杯子对桌面的压力:F=mg+F水=22N 压强: p=F/s=3,6Kkpa
水对杯子一定要先求压强P=水的密度*g*深度30㎝,然后用这个压强去诚意杯子的底面积。不能用水中作为压力,因为不是柱形容器,水对杯子底的压力不等于水的重力。
杯子对桌面的要先求压力,就等于杯子和水的总重22牛,然后用这个压力除以杯子的底面积。因为固体放在水平面上静止,对固体的压力就等于固体的重力。
6750/1.2=5625 5625开平方=75
绛旓細1锛夋按瀵规澂搴曠殑鍘嬪姏锛欶姘=20N,鍚戜笅 鍘嬪己锛歱=F/s=3,3Kpa 2)鏉瓙瀵规闈㈢殑鍘嬪姏锛欶=mg+F姘=22N 鍘嬪己锛 p=F/s=3,6Kkpa
绛旓細姘存《搴曢儴鐨勫帇寮烘槸pgh p=1000kg/m^3 g=10N/kg h=15/1000m 姘存《搴曢儴鐨勫帇寮哄氨鏄150pa 搴曢潰绉槸200/10^4m^2 姘存《搴曢儴鍘嬪姏灏辨槸150*200/10000=30N 姘存《瀵规闈㈢殑鍘嬪姏灏辨槸40+20=60N 鍘嬪己灏辨槸F/S=60*10000/200=3000pa
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绛旓細I=U/(R1+R2)4=U*R2/(R1+R2) (1)U2'=U*0.5R2/(R1+0.5R2) (2)(1)/(2)2/U2'=(R1+0.5R2)/(R1+R2)U2'>2V