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递等式计算 828-828÷23 18÷45+40×0.35 (0.52+)÷-
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递等式计算.200÷[172-72÷25]
常用的食品冷藏温度和冷冻温度分别是
0~15℃和-23~-12℃
0-10℃和-20~-18℃
0~8℃和-18~-12℃
0~5℃和-18~-10℃
0~15℃2和-23~-18℃
递等式计算 35÷[﹣×3]1÷××161.2÷0.75+7.2÷2.490720÷72﹣23×
用递等式计算能简算的要简算.
计算题有变量定义的伪指令如下NUMSDW18DUP4DUP523VARDB’HOWAREYOU!’0
递等式计算.能用简便算法的用简便方法计算
用递等式计算.1﹣×45%218%÷32%+18%
已知都是正实数函数的图象过02点则的最小值是
常用的食品冷藏温度和冷冻温度分别是
0~15℃和-23~-12℃
0~10℃和-20~-18℃
0~8℃和-18~-12℃
0~5℃和-18~-10℃
0~15℃和-23~-18℃
递等式计算.329÷[22-9+6]
用递等式计算写出主要的计算过程能简便计算的要用简便方法计算
中闪点易燃液体的分类判据是
0℃≤闪点<23℃
0℃≤闪点<28℃
-18℃≤闪点<23℃
-18℃<闪点≤23℃
递等式计算. 5.4÷3.94+6.86
24.00分递等式计算怎样计算简便就怎样算
解不等式组并在数轴上表示不等式组的解集.
用递等式计算能简算的要简算.
递等式计算.
常用的食品冷藏温度和冷冻温度分别是
0~15℃和-12~-23℃
0~8℃和-12~-18℃
0~10℃和-18~-20℃
0~5℃和-10~-18℃
0~15℃和-18~-23℃
计算.
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如果将两只标有“2.5V0.3A”的相同灯泡串联在2.5V电压的电路中则每只灯泡的额定功率是W若灯丝的电阻不变每只灯泡的实际功率是W.
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请你在图中的方框内设计一个“病房呼叫电路”电路图病人在病床按下开关医护人员在值班室能够听到铃声并通过灯泡的发光可以判断哪个病床在呼叫两个病床即可.
小明看到闪电后12.3s又听到雷声已知声音每3s传播的距离约为1000m光速约为3×108m/s于是他用12.3除以3很快估算出闪电发生位置到他的距离为41km.假设声速或光速发生以下变化这种估算方法不再适用的是
如图所示电源电压恒定R1R2为定值电阻R2=20Ω灯泡L标有“6V3W”字样. 1当SS1S2都闭合时灯泡L正常发光电流表示数为A电路消耗的总功率为W 2当S闭合S1S2断开时电压表示数为2V则R1的阻值为Ω通电1minR1产生的热量为J.
如图所示电路电源电压恒为4.5V定值电阻R0的阻值为10Ω滑动变阻器的最大阻值为30Ω电阻箱Rx最大阻值为999.9Ω电流表量程为0~0.6A电压表量程为0~3V.闭合开关S下列说法正确的是
在综合实践活动中科技小组设计了一个由压敏电阻控制的报警电路如图所示电源电压恒为18V电阻箱最大阻值为999.9Ω.报警器电阻不计通过的电流达到或超过10mA会报警超过20mA会损坏.压敏电阻Rx在压力不超过800N的前提下其阻值随压力F的变化规律如下表所示. 1为不损坏元件报警电路允许消耗的最大功率是多少 2在压敏电阻Rx所受压力不超过800N的前提下报警电路所选滑动变阻器的最大阻值不得小于多少 3现要求压敏电阻受到的压力达到或超过200N时电路报警按照下列步骤调试此报警电路 ①电路接通前滑动变阻器滑片P置于b端根据实验要求应将电阻箱调到一定的阻值这一阻值为Ω ②将开关向端填数字闭合调节直至 ③保持将开关向另一端闭合报警电路即可正常使用. 4对3中已调试好的报警电路现要求压敏电阻受到的压力达到或超过700N时电路报警若电源电压可调其它条件不变则将电源电压调为V即可.
如图所示太阳光跟地面成60°角晓彤想用一个平面镜把太阳光竖直反射到井底请画出平面镜的位置.
有两只阻值未知的定值电阻R1R2. 1图甲是用伏安法测R1阻值的实物电路电源电压恒为3V滑动变阻器最大阻值为10Ω. ①甲图中有一根导线连接错误请在该导线上画“×”并在图上改正所画的导线用实线且不能交叉. ②改正电路后闭合开关发现无论怎样移动滑动变阻器滑片两电表均无示数其原因可能是填字母. A.滑动变阻器断路B.R1断路C.R1短路 ③故障排除后正确操作实验器材移动滑片当电压表示数为1.2V时电流表示数如图乙所示则待测电阻R1=Ω. ④现有以下三组数据分析可知不可能通过以上实验获得的数据有填序号. 2图丙是能巧测R2阻值的实验电路图.图中R为电阻箱R0为定值电阻阻值未知.要求仅利用电阻箱读数表达R2的阻值请在空白处填上适当内容. ①将开关接a调节电阻箱和滑动变阻器滑片P至适当位置记下 ②将开关接b调节保持不变记下 ③则R2=.
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完成图中的光路.
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