2012广西省桂林市中考物理试题及答案
2012广西省桂林市中考物理试题及答案来源：考试吧（Exam8）2012-7-318:31:26【考试吧：中国教育培训第一门户】模拟考场专业生产尼龙输送带价格低廉.国际质量体系认证,尼龙输送带价格优惠与科研机构建立技术合作交流0312-83..鼓浪屿旅馆—厦门鑫雅琪公馆.鼓浪屿鑫雅琪公馆视野开阔奢华享受躺着即可饱览海景您绝佳的度假之地百度推广第1页：试题第8页：答案26.解：⑴柑橘排水质量：m排=m总-m剩=360g-240g=120g·····································(1分)⑵由ρ=m/V可得·····································································(1分)柑橘的体积：V橘=V排=m排/ρ水=120g/1.0g/cm3=(2分)柑橘的密度：ρ橘=m橘/V橘=114g/=0.95g/cm3············(1分)⑶偏小···························································································(1分)27.解：⑴由P=UI可得·······································································(1分)正常加热状态下的电流：I=P/U=920W/220V≈4.2A···············(1分)⑵当开关S闭合、S0断开时，电热饮水机只有R2工作，处于保温状态。由P=UII=U/R可得························································(1分)电阻R2消耗的功率：P2=U2/R2=(220V)2/1210Ω=40W·····(1分)当开关S、S0闭合时，电热饮水机处于加热状态。此时R1消耗的电功率为：P1=P总-P2=920W-40W=880W···························(1分)则有：R1=U2/P1=(220V)2/880W=55Ω···························(1分)⑶方法一：电热饮水机处于加热状态时的总电阻为：R总=U2/P总=(220V)2/920W=1210/23Ω≈52.6Ω·······(1分)实际加热功率：P实=U实2/R总=(198V)2/(1210/23)Ω=745.2W·············(1分)[或P实=U实2/R总=(198V)2/52.6Ω≈745.3W]方法二：电路电阻不变，可得：R总=U2/P总=U实2/P实····································(1分)实际加热功率：P实=P额×U实2/U2=920×(198/220)2=745.2W····················(1分)方法三：实际加热功率：P实=P1实+P2实=U实2/R1+U实2/R2····(1分)=(198V)2/55Ω+(198V)2/1210Ω=745.2W·········································(1分)28.解：1m3可燃冰转化生成的甲烷气体完全燃烧放出热量：Q=q甲烷V=3.6×107J/m3×164m3=5.904×109J···································(1分)由题知：t==9.84×104s由P=W/t可得…………………………………………………………(1分)发动机的实际功率：P=Q/t=5.904×109J/9.84×104s=6.0×104W(2分)由题知：36km/h=10m/s由v=s/t可得··········································································(1分)公交车行驶路程：s=vt=10m/s×9.84×104s=9.84×105m················(1分)车匀速行驶，有：F牵=f=0.05G车=0.05mg=0.05××10N/kg=3×103N(1分)由W=Fs可得牵引力做的有用功：··················································(1分)W有=F牵s=3×103N×9.84×105m=2.952×109J···················(1分)发动机效率：η=W有/Q=2.952×109J/5.904×109J=50%·······················(1分)
页:
[1]