7. -Congratulations to you ! I hear you got the first place in the English Speech Contest.
-_______ .
A. Don’t mention it. I think that’s only me past. B. Thanks. But I think I could have done better. C. There’s nothing to cheer for. D. No, no. It’s a piece of cake.
6. ______ the Atlantic Ocean crosses the equator, the trade winds cause a flow of water to the west.
A. That B. When C. Where D. Though
5. ______ that the South African writer John Coetzee won the Nobel Prize in Literature for 2003.
A. They’er reported B. He’s reported C. We’er reported D. It’s reported
4. -You've made great progress in your studies of English, haven't you?
-Yes, but much .
A. remains to do B. is remained to do
C. remains to be done D. is remained to be done
3. ______ is known to us all is that China has launched Shenzhou V spaceship successfully.
A. That B. What C. It D. As
2.He hasn't come yet.What do you consider _______ to him?
A.happens B.has happened C.happening D.to happen
1. Unfortunately, when I dropped in, Doctor Li _____ for Beijing to join in the fight again H1N1, so we only had time for a few words.
A. just left B. has just left C. is just leaving D. was just leaving
17.如图14所示,A、B为两块平行金属板,A板带正电、B板带负电。两板之间存在着匀强电场,两板间距为d、电势差为U,在B板上开有两个间距为L的小孔。C、D为两块同心半圆形金属板,圆心都在贴近B板的O′处,C带正电、D带负电。两半圆形金属板间的距离很近,两板间的电场强度可认为大小处处相等,方向都指向O′。且两板末端的中心线正对着B板上的小孔。半圆形金属板两端与B板的间隙可忽略不计。现从正对B板小孔紧靠A板的O处由静止释放一个质量为m、电量为q的带正电微粒(微粒的重力不计),问:
(1)微粒穿过B板小孔时的速度多大?
(2)为了使微粒能在CD板间运动而不碰板,CD板间的电场强度大小应满足什么条件?
(3)从静止释放开始,微粒通过半圆形金属板间的最低点P点的时间?
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命题:刘立军 审题:孙智勇
温州中学2009学年第一学期期中考试
16.如图13所示,半径R=0.80m的
光滑圆弧轨道竖直固定,过最低点的半径OC处于竖直位置.其右方有底面半径r=0.2m的转筒,转筒顶端与C等高,下部有一小孔,距顶端h=0.8m.转筒的轴线与圆弧轨道在同一竖直平面内,开始时小孔也在这一平面内的图示位置。今让一质量m=0.1kg的小球自A点由静止开始下落后打在圆弧轨道上的B点,但未反弹,在瞬问碰撞过程中,小球沿半径方向的分速度立刻减为零,而沿切线方向的分速度不变.此后,小球沿圆弧轨道滑下,到达C点时触动光电装置,使圆筒立刻以某一角速度匀速转动起来,且小球最终正好进入小孔。已知A、B到圆心O的距离均为R,与水平方向的夹角均为θ=30°,不计空气阻力,g取l0m/s2.求:
(1)小球到达C点时对轨道的压力 FC;
(2)转筒转动的角速度ω大小.
15.在如图12所示电路中,电源电动势为ε= 6V,内阻不计,小灯L上标有“6V,0.3A”字样,滑动变阻器R1的阻值范围是0-20Ω,电阻R2上标有“15Ω,4A”字样,电流表的量程为0-0.6A。甲、乙两同学在讨论滑动变阻器功率的取值范围时,产生了分歧。甲同学认为:由于电流表允许通过的最大电流为0.6A,所以通过R1的最大电流为
I1m = IAm–IL = 0.6A–0.3A = 0.3A,
这时滑动变阻器R1两端的电压为 U1m =ε–I1m R2 = 6V–0.3×15V = 1.5V
因此,滑动变阻器的最大功率为 P1m = I1m U1m = 0.3×1.5W = 0.45W。
乙同学不同意甲同学的看法,他认为滑动变阻器的功率决定于通过它的电流和它两端电压的乘积,即P1 = I1 U1,电流最大时功率未必最大,只有电流、电压的乘积最大时,功率才最大。
你认为甲、乙两位同学中,哪位同学的看法正确?如果你认为甲同学正确,请简述他正确的理由;如果你认为乙同学正确,请求出滑动变阻器R1的最大功率P1m?
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