王斌打算将800元钱存人银行。有两种储蓄办法：一种是存两年期的，年利率是3.06%；另一种先存一年期的，年利率是2.52

(4/5)/(1/4)x1+1/5=16/5+1/5=17/5 (千米)
王斌家到图书馆有17/5千米

20 June 2014
Dear Wangbin,
Thank you very much for taking me with you on that splendid outing to London. It was the first time that I had seen the Tower or any of the other famous sights. If I'd gone alone, I couldn't have seen nearly as much, because I wouldn't have known my way about.
The weather was splendid on that day, which I thought was rare. I still remember some people told me that in Britain there was weather and no climate. During the same day, it might snow in the morning, rain at noon, shine in the afternoon and be windy before the night falls. So I think I was lucky.
I think the river trip was the best thing of all. London really came alive for me as we saw it from the Thames during that wonderful journey down to Greenwich. It was all tremendously exciting － a day that I shall never forget.
Thank you for giving me such a great birthday treat.

Yours sincerely,
Lihua

Hi!Liming.l am wangbin.if you have something of lerning is understand.l hope help you.l think you shoud knowe how to lern . Do not think trouble or hard .you can ask for better student tant something of lernning.we should find good way to lern.you can study for your student of room.l hope my idear can helpe you .

解题思路：闭合电路的部分导体在磁场中做切割磁感线运动时，就会产生感应电流，感应电流的方向和导体的运动方向、磁场方向有关；感应电流的大小和线圈的匝数、磁场的强弱、线圈的转速有关．

由于铁线圈的电阻大于铜线圈的电阻，因此把铜线圈换成铁线圈会增大电路中的电阻，从而减小电路中的电流，使灯泡发光更暗，故A不符合题意；
因为影响感应电流的大小的因素有线圈的匝数、磁场的强弱、线圈的转速，因此可以通过增加线圈匝数、换用磁性较强的磁体、提高线圈的转速来增大感应电流的大小；故BCD符合题意．
故选BCD．

点评：
本题考点： 电磁感应．

考点点评： 此题考查了影响感应电流大小的因素，是一道基础题．

王斌是我（知心）的朋友,我生活上遇到困难时,他总会（热心）帮助

Wang Bin walks to school at 7：10.

解题思路：（1）根据题意求出变阻器接入电路中的电阻，根据电阻的串联和欧姆定律的应用求出电路中的电流即为电流表的示数，再利用欧姆定律求出变阻器两端的电压即为电压表的示数；
（2）当电压表的示数为3V时，该装置所能测量的直径最大，根据串联电路各处的电流相等和串联电路的电压特点得出等式即可求出变阻器接入电路中的最大阻值，进一步根据题意求出该装置所能测量的直径范围；
（3）要使刻度之间等距只需电压表的示数与被测电阻的阻值成正比即可．

（1）被测直径是5cm时，R_PB=5×4Ω=20Ω，则
电路中的电流I=[U
R0+RPB=
4V/20Ω+20Ω]=0.1A，
电压表的示数U_PB=IR_PB=0.1A×20Ω=2V；
（2）电压表的示数为3V时，该装置所能测量的直径最大，
∵串联电路各处的电流相等，
∴
U′PB
R′PB=
U−U0
R0，即
3V
R′PB=[4V−3V/20Ω]，
解得：R_PB′=60Ω，
故d=[60Ω/4Ω/cm]=15cm，
所以该装置所能测量的直径范围是0～15cm；
（3）当定值电阻与Rab的最大阻值串联，电压表测滑动变阻器下端接入电路中电阻时，电路中的电流不变，根据欧姆定律可知电压表的示数与变阻器下方电阻成正比，改进的电路图如下图所示：

答：（1）被测直径是5cm时，电路中的电流为0.1A，电压表的读数为2V；
（2）该装置所能测量的直径范围是0～15cm；
（3）如上图所示．

点评：
本题考点： 欧姆定律的应用；串联电路的电流规律；电路图及元件符号；串联电路的电压规律．

考点点评： 本题考查了串联电路的特点和欧姆定律的应用，根据题意得出球的直径与接入电路中电阻的关系以及会判断该装置所能测量的直径范围是本题的关键，难点是根据电压表的示数与变阻器接入电路的电阻成正比设计电路．

(94-2)*5-94*4=84

1000*2*0.0355 =71
1000*1*0.0275+1000=1027.5
1027.5*0.0275约等于28.26
28.26小于71
答：第一种
71-28.26=42.74
答：多42.74

第一种,百分之百对,有三种方法,但有一种你自己去思考哦,加我分吧!：
1.第一种：800*0.0306*2=48.96
第二种：第一年800*0.02=16
第二年（800+16）=16.32
16+16.32=32.32
2.存两年的利息高.
存两年：800×3.06%×2＝48.96元
先存一年：800×2%＋（800×2%＋800）×2%＝32.32元
3.王斌打算将800元存入银行.有两种储办法、一种是存两年期的、年利是3.06%、另一种先存一年期的、年利率是2.52%,第一年到期后再把本金和税后利息合在一起再存一年,利息税率为5%.算一算,哪一种存款方式获得的税后利息多一些?多多少元?

第一种储蓄方法税后利息800*4.68%*2*（1-5%）=71.14元第二种储蓄方法税后利息：第一年年末本息和：800+800*4.41%*（1-5%）=833.52元第二年年末税后利息：833.52*4.41%*（1-5%）=34.92元税后利息共计833.52+34.92-800...

Wang Bin and his family are all very happy.

设:王斌跑步上篮投中X次,没有投中Y次
X+Y=10 (1)
3X-Y+5=27 (2)
(这是一个方程组,我已经标上(1)式和(2)式了,下面,我用代入法给你解这个方程组)
由(1)得Y=10-X(3)
(3)代入(2)得3X-(10-X)+5=27
3X-10+X+5=27
4X-5=27
4X=32
X=8
把X=8代入(1)得8+Y=10
Y=2
斌跑步上篮10次,投中8次,未投中2次

Wang Bing has had a headache and cough since last night. Then he went to see the doctors. The doctor asked:"What's wrong with you?" Wang Bing answered,"I got wet in the rain last night and then I had a bad cold." So the doctor examined him carefully and took his temperature. After that, the doctor said to Wang Bing,"There is nothing serious. Take this medicine twice a day and drink more water. I think you'll be better soon." Wang Bing thanked the doctor and went home.

Liu Tao gave a football to Wang Bin as a birthday gift yesterday.Today,they want to play football in the park.There is a……后面是什么?

解题思路：分析表中数据，弹簧受到的拉力为0时，弹簧伸长为0，当弹簧受到的拉力成倍增加时，弹簧的伸长也相应的成倍增大，根据这个规律得出实验结论．

根据表格数据可以看出，弹簧受到的拉力越大，弹簧伸长的长度越长，弹簧伸长的长度与弹簧受到的拉力之比是定值，即在一定条件下，弹簧伸长的长度与它所受的拉力成正比．
故答案为：弹簧的伸长与所受拉力大小成正比．

点评：
本题考点： 探究弹簧测力计原理的实验．

考点点评： 本题既考查了学生分析实验数据的能力，又考查了学生应用获取的信息解决问题的能力，是中考探究实验考查的重点问题．

I‘d like to introduce my penfriend to you.His name is WangBing and his English name is Tommy.He studies in class 2 at The Second Middle School .He was given birth to on the 6th of June.He lives in Nanjing,JiangSu province.Besides,He is thirteen years of age.Not only does he enjoy music,but he also loves playing basketball.He prefers English rather than any other subjects.

设王斌数学x分
(78+82+80+60+x)÷5=x+4
（300+x）÷5=x+4
60+1/5x=x+4
4/5x=56
x=70
x+4=70+4=74
答：平均成绩是74分