2012廣西省桂林市中考物理試題及答案

　26.解：⑴柑橘排水質(zhì)量：

　　m排=m總-m剩=360g-240g=120 g ····································· (1分)

　�、朴伞ˇ�=m/V 可得 ····································································· (1分)

　　柑橘的體積：V橘 =V排=m排/ρ水=120 g/1.0 g/cm3=120 cm3 (2分)

　　柑橘的密度：ρ橘=m橘/V橘=114g/120 cm3=0.95g/cm3 ············ (1分)

　　⑶偏小 ··························································································· (1分)

　　27.解：⑴由　P=UI　可得 ······································································· (1分)

　　正常加熱狀態(tài)下的電流：I= P/U = 920W/220V ≈ 4.2A　 ··············· (1分)

　�、飘�(dāng)開關(guān)S閉合、S0斷開時(shí)，電熱飲水機(jī)只有R2工作，處于保溫狀態(tài)。

　　由　P=UI　I=U/R　可得 ························································ (1分)

　　電阻R2消耗的功率：P2=U2/R2=(220V)2/1210Ω=40W ····· (1分)

　　當(dāng)開關(guān)S、S0閉合時(shí)，電熱飲水機(jī)處于加熱狀態(tài)。此時(shí)R1消耗的電功率為：

　　P1 =P總- P2=920 W- 40W=880 W ··························· (1分)

　　則有：R1=U2/P1=(220V)2/880 W=55Ω ··························· (1分)

　�、欠椒ㄒ唬弘姛犸嬎畽C(jī)處于加熱狀態(tài)時(shí)的總電阻為：

　　R總=U2/P總=(220V)2/920W=1210/23Ω≈ 52.6Ω ······· (1分)

　　實(shí)際加熱功率：P實(shí)=U實(shí)2/R總

　　=(198V)2/(1210/23)Ω=745.2W ············· (1分)

　　[或P實(shí)=U實(shí)2/R總=(198V)2/52.6Ω≈745.3 W ]

　　方法二：電路電阻不變，可得：

　　R總=U2/P總=U實(shí)2/P實(shí) ···································· (1分)

　　實(shí)際加熱功率：P實(shí)=P額×U實(shí)2 /U2

　　=920 ×(198/220)2=745.2 W ···················· (1分)

　　方法三：實(shí)際加熱功率：P實(shí)=P1實(shí)+P2實(shí)=U實(shí)2/R1+U實(shí)2/R2 ···· (1分)

　　=(198V)2/55Ω+(198V)2/1210Ω

　　=745.2 W ········································· (1分)

　　28.解：1 m3可燃冰轉(zhuǎn)化生成的甲烷氣體完全燃燒放出熱量：

　　Q=q甲烷V

　　=3.6×107J/ m3×164 m3=5.904×109 J ··································· (1分)

　　由題知： t=1640min=9.84×104s

　　由　P=W / t　可得 …………………………………………………………(1分)

　　發(fā)動(dòng)機(jī)的實(shí)際功率：P=Q / t=5.904×109 J/9.84×104s=6.0×104 W (2分)

　　由題知：36km/h=10m/s

　　由　v=s/t 可得 ·········································································· (1分)

　　公交車行駛路程：s=v t=10m/s×9.84×104s=9.84×105 m ················ (1分)

　　車勻速行駛，有：

　　F牽=f=0.05 G車=0.05mg=0.05×6000 kg×10N/kg=3×103N (1分)

　　由W=Fs可得牽引力做的有用功： ·················································· (1分)

　　W有=F牽s=3×103N×9.84×105 m=2.952×109 J ··················· (1分)

　　發(fā)動(dòng)機(jī)效率：

　　η=W有 /Q=2.952×109 J/5.904×109J=50% ······················· (1分)

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