第 1 頁:試題 |
第 8 頁:答案 |
26.解:⑴柑橘排水質量:
m排=m總-m剩=360g-240g=120 g ····································· (1分)
⑵由 ρ=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)
⑵當開關S閉合、S0斷開時,電熱飲水機只有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)
當開關S、S0閉合時,電熱飲水機處于加熱狀態(tài)。此時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)
⑶方法一:電熱飲水機處于加熱狀態(tài)時的總電阻為:
R總=U2/P總=(220V)2/920W=1210/23Ω≈ 52.6Ω ······· (1分)
實際加熱功率:P實=U實2/R總
=(198V)2/(1210/23)Ω=745.2W ············· (1分)
[或P實=U實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2/P實 ···································· (1分)
實際加熱功率:P實=P額×U實2 /U2
=920 ×(198/220)2=745.2 W ···················· (1分)
方法三:實際加熱功率:P實=P1實+P2實=U實2/R1+U實2/R2 ···· (1分)
=(198V)2/55Ω+(198V)2/1210Ω
=745.2 W ········································· (1分)
28.解:1 m3可燃冰轉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)動機的實際功率: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)動機效率:
η=W有 /Q=2.952×109 J/5.904×109J=50% ······················· (1分)
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