大学物理实验报告答案大全(实验数据及思考题答案全包括)
大学物理实验预习及思考题答案
大学物理实验预习及思考题答案实验一霍尔效应及其应用【预习思考题】1．列出计算霍尔系数、载流子浓度n、电导率σ及迁移率μ的计算公式，并注明单位。
霍尔系数，载流子浓度，电导率，迁移率。
2．如已知霍尔样品的工作电流及磁感应强度B的方向，如何判断样品的导电类型？以根据右手螺旋定则，从工作电流旋到磁感应强度B确定的方向为正向，若测得的霍尔电压为正，则样品为P型，反之则为N型。
3．本实验为什么要用3个换向开关？为了在测量时消除一些霍尔效应的副效应的影响，需要在测量时改变工作电流及磁感应强度B的方向，因此就需要2个换向开关；除了测量霍尔电压，还要测量A、C间的电位差，这是两个不同的测量位置，又需要1个换向开关。
总之，一共需要3个换向开关。
【分析讨论题】1．若磁感应强度B和霍尔器件平面不完全正交，按式（5.2-5）测出的霍尔系数比实际值大还是小？要准确测定值应怎样进行？若磁感应强度B和霍尔器件平面不完全正交，则测出的霍尔系数比实际值偏小。
要想准确测定，就需要保证磁感应强度B和霍尔器件平面完全正交，或者设法测量出磁感应强度B和霍尔器件平面的夹角。
2．若已知霍尔器件的性能参数，采用霍尔效应法测量一个未知磁场时，测量误差有哪些来源？误差来源有：测量工作电流的电流表的测量误差，测量霍尔器件厚度d的长度测量仪器的测量误差，测量霍尔电压的电压表的测量误差，磁场方向与霍尔器件平面的夹角影响等。
实验二声速的测量【预习思考题】1. 如何调节和判断测量系统是否处于共振状态？为什么要在系统处于共振的条件下进行声速测定？答：缓慢调节声速测试仪信号源面板上的“信号频率”旋钮，使交流毫伏表指针指示达到最大（或晶体管电压表的示值达到最大），此时系统处于共振状态，显示共振发生的信号指示灯亮，信号源面板上频率显示窗口显示共振频率。
在进行声速测定时需要测定驻波波节的位置，当发射换能器S1处于共振状态时，发射的超声波能量最大。
若在这样一个最佳状态移动S1至每一个波节处，媒质压缩形变最大，则产生的声压最大，接收换能器S2接收到的声压为最大，转变成电信号，晶体管电压表会显示出最大值。
大学物理实验思考题答案（Coll...
大学物理实验思考题答案（College physics experiment; thinkingquestion; answer）1. if the time can be measured as a swing cycle value? Why?A: no.. Because of a large random error measurement, multiple measurements can reduce random error.2. a radius of less than under the radius of disk disk in disk, and the center line, the discussion of three line pendulum cycle and no-load cycle is compared to the increase and the decrease or not? Explain the reason.Answer: quality when two disks for uniform distribution, compared with no-load, swing cycle will be reduced. Because if the two disc as a radius equal to the original lower disc, the disc inertia is less than I0 quality equal to that of the same diameter, according to formula (3-1-5), T0 will reduce the swing period.3. three line pendulum by air damping in the swing amplitude, more and more small, whether it will change the cycle? Great influence on the measurement result? Why?Answer: the decreasing cycle has little influence on the measurement results, because the measurement time is short.2 measurement of elastic modulus of metal wires1. what are the advantages of optical lever, how to improve the sensitivity of the optical lever measurement?A: the advantage is that it can measure the small length variation. To improve the vertical distance magnification is increasing the scale distance of D or properly reduce the optical lever before and after the foot of B, can improve the sensitivity, because the magnification of the optical lever is 2D/b.2. what is the disparity, how to judge and eliminate the disparity?Answer: eyes at the eyepiece to move up and down, if the scale level telescope crosshair scale and relative displacement, this phenomenon is called parallax, fine focusing handwheel can eliminate parallax.3. why use by experimental data processing method?Answer: the difference method is a basic method of experimental data processing, the essence is to make full use of the data from the experiment, is to reduce the random error, data averaging effect. Because of some experimental data, if the simple average value of each measurement, the measured values will eliminate all intermediate, only through two readings, equal to the actual single measurement. In order to maintain the advantages of multiple measurements, the general variable interval changes, the data is divided into two groups, two groups of successive difference and then calculate the average value of the difference.In experiment three, the statistical regularity of random error1. what is the histogram? What is the normal distribution curve of the two? And what is the relationship between the difference?Answer: to do n repeated measurements of a physical quantity under the same conditions, a series of measurements, find the maximum and minimum values, and then determine an interval, which contains all the measurement data, the interval is divided into several regions and the statistical measurement results appear in the M frequency of each interval. In order to measure the data as the abscissa and the frequency M as the ordinate, draw the area between the frequency and the corresponding height can be a rectangle, namely statistical histogram.If the number of measurements of the interval more smaller, while histogram will be close to a smooth curve, when the distribution of n tends to infinity is called the normal distribution, the distribution curve of normal distribution curve.2. if the measured data set, its discrete degree than the data in the table, which is S (x) is relatively large, the average value of the cycle is very different?Answer: (not a big gap, according to the characteristics, statistical regularity of random error. We know that when the number of measurements is large, the measurement data and the average error of plus or minus almost offset each other, and the algebraic error tends to zero.Measurement of thermal experiment four equivalent electric heating methodWhat are the conditions of the experiment and 1. must be the basic method? Sources of systematic error may have?Answer: the experimental condition is the system with no external large heat exchange, and the system (i.e. water) should be possible in the quasi-static process. The experimental method of electrothermal method.The main source of system error is a system of heat dissipation, and temperature correction often cannot fully compensate for the heat dissipation effect of the measurement. Not even other sources may be the temperature of the water, instead of using the local temperature temperature. The water temperature and the ambient temperature difference is large, so as to give the final temperature error correction. The temperature measurement error, power quality and other physical quantity.2. try a qualitative description of the following occurs during the experiment, the experimental result is too large or too small?(1) when mixing water is spilled;Answer: the experimental results will be too small. The water was spilled, namely water quality decrease, in the calculation of the mechanical equivalent of heat, but also to the quality of said horizontal water calculation, that the water quality does not change, but because of the water quality of waterreducing, heating, electric heating power in the same system, rising temperature than water when the temperature will not rise high water and not spill at the same heating power, should rise T degrees, and the water spilled after rising temperature should be T+ degrees T. Use Q = (cimiT), and the denominator bigger J smaller.The uneven stirring (2);Answer: between the position of J is too large, too small by the resistance wire and the thermometer into the distance and. Far away from the resistance wire, temperature display system ratio, uniform system of low temperature, T uniform temperature of the system, value should be T- T display thermometer, calculated by J=A/ theta, the denominator is smaller, J is bigger; from the resistance wire is near. The value should be T+ T display thermometer thus, the denominator bigger, smaller J.The (3) measured at room temperature is high or low.Answer:0. Theta is at room temperature, if the measured values measured by the delta theta, delta theta theta temperature value is 0+; the low delta theta theta, the measured temperature value 0- delta theta, in the calculation of temperature loss, dTi=k (Ti- 0), K with independent volume (k at room temperature and room temperature the initial temperature and cooling), only the cooling temperature and the cooling time, measured at room temperature is too high, dTi=k[Ti- (0+ delta theta theta)],second temperature in the loss of dTi is less than the actual value, T seconds at the end of the Ti= k[Ti- (sigma delta temperature loss 0+ Delta theta theta)]. This value is less than the actual value, because the temperature of heat dissipation caused by loss of delta Ti=Tf+ Tf "Tf", the temperature corrected: Tf = Tf - Delta Ti, Delta Ti is negative, when the measured value is less than the actual temperature, the smaller absolute value of delta Ti: Tf "=Tf+, Ti, Tf 8." smaller, smaller Delta T (which is T = Tf - Tf, Tf: early warming initial value),J, J and smaller.Experiment seven(1) the current correction table, if it is found that the modified milliammeters readings are always higher than standard gauge readings, the shunt resistance should be adjusted big or small? Why? Answer: should be small. Let the same standard reading meter circuit, which keeps the loop current constant, the shunt resistance value decreased after will receive more current, so that the current flowing through the modified table header modification decreases, the reading is reduced. (2) correction voltage meter, if it is found that the modification of the voltage meter reading is always lower than the standard gauge readings, pressure resistance should be adjusted in big or small? Why? Answer: should be small. Let the same standard reading meter circuit, which is added in the modified Ammeter voltage constant. Small resistance, total resistance modification table is reduced, through the modified milliammeter increases, which also increased reading. (3)prove that using an ohmmeter measuring resistance, if the head pointer is refers to the scale in the center of the dial, then the ohmmeter indication why is exactly equal to the internal resistance of the ohmmeter value. Answer: a head pointer full scale current is Ig, the head pointer refers to the center of the dial in the circuit when the current is I, according to the problem, when the internal table is Rg, resistance to be measured at Rx; according to the working principle of an ohmmeter, when the measured resistance Rx = 0,. That is, it can be Rx = Rg. So, the ohmmeter displays readings is the internal resistance of the ohmmeter.The principle and application of [eight] experimental oscilloscopeOneWhy can display analog oscilloscope signal path of rapid change? Answer: in the analog oscilloscope vertical deflection plate plus the observed signal voltage in the horizontal deflection plate plus the sawtooth wave (time linear change) signal voltage, so the tube axis of oscilloscope is equivalent to the Cartesian coordinate time axis, after a period of a sawtooth signal, electron beam at in the screen tube painted on the observed signal waveform of a trajectory. When a periodic sawtooth wave signal is greater than or equal to the cycle signal and its phase locked (synchronous), electron beam in the display screen tube painted on the same trajectory observed signal waveform, due to persistence of vision and screen the human eye afterglow, can be observed signal waveform. (2) in this experiment, to observe the pattern of Li Saru, why not longtime stable graphics? Answer: because the CH1 and CH2 inputs are two completely irrelevant signals, the phase difference between them is difficult to maintain a constant, so they don't have a long time stable waveform. (3) assume that the input is a sinusoidal signal in the oscilloscope Y axis, horizontal scanning frequency used for 120Hz, there are three stable complete sine wave on the screen, then what is the frequency of the signal? Whether this is a good method for the measurement of the signal frequency? Why? Answer: the frequency of the input signal is 360Hz. This method is not good because the measurement of the signal frequency, frequency accuracy by this method of measuring low. (4) the scanning frequency oscilloscope is far greater than or much less than the frequency of a sinusoidal signal input, the graphics on the screen is what? Answer: the scanning frequency is far less than the frequency of sinusoidal signal input, a graph is dense sine wave; scanning frequency is far higher than the frequency of sinusoidal signal input, the signal waveform of a cycle will be decomposed into several sections, graphic display will become mesh cross line.Experiment seven(1) the current correction table, if it is found that the modified milliammeters readings are always higher than standard gauge readings, the shunt resistance should be adjusted big or small? Why?Answer: should be small. Let the same standard reading meter circuit, which keeps the loop current constant, the shunt resistance value decreased after will receive more current, so that the current flowing through the modified table headermodification decreases, the reading is reduced.(2) correction voltage meter, if it is found that the modification of the voltage meter reading is always lower than the standard gauge readings, pressure resistance should be adjusted in big or small? Why?Answer: should be small. Let the same standard reading meter circuit, which is added in the modified Ammeter voltage constant. Small resistance, total resistance modification table is reduced, through the modified milliammeter increases, which also increased reading.(3) prove that using an ohmmeter measuring resistance, if the head pointer is refers to the scale in the center of the dial, then the ohmmeter indication why is exactly equal to the internal resistance of the ohmmeter value.Answer: a head pointer full scale current is Ig, the head pointer refers to the center of the dial in the circuit when the current is I, according to the problem, when the internal table is Rg, resistance to be measured at Rx; according to the working principle of an ohmmeter, when the measured resistance Rx = 0,. That is, it can be Rx = Rg. So, the ohmmeter displays readings is the internal resistance of the ohmmeter.The principle and application of [eight] experimental oscilloscope1. analog oscilloscope can display the signal of high speed track why change?Answer: in the analog oscilloscope vertical deflection plate plus the observed signal voltage in the horizontal deflection plate plus the sawtooth wave (time linear change) signal voltage, so the tube axis of oscilloscope is equivalent to the Cartesian coordinate time axis, after a period of a sawtooth signal, electron beam at in the screen tube painted on the observed signal waveform of a trajectory. When a periodic sawtooth wave signal is greater than or equal to the cycle signal and its phase locked (synchronous), electron beam in the display screen tube painted on the same trajectory observed signal waveform,Because of the persistence of vision and screen the human eye afterglow, can be observed signal waveform.(2) in this experiment, to observe the pattern of Li Saru, why not long time stable graphics?Answer: because the CH1 and CH2 inputs are two completely irrelevant signals, the phase difference between them is difficult to maintain a constant, so they don't have a long time stable waveform.(3) assume that the input is a sinusoidal signal in the oscilloscope Y axis, horizontal scanning frequency used for 120Hz, there are three stable complete sine wave on the screen, then what is the frequency of the signal? Whether this is a good method for the measurement of the signal frequency? Why?Answer: the frequency of the input signal is 360Hz. This methodis not good because the measurement of the signal frequency, frequency accuracy by this method of measuring low.(4) the scanning frequency oscilloscope is far greater than or much less than the frequency of a sinusoidal signal input, the graphics on the screen is what?Answer: the scanning frequency is far less than the frequency of sinusoidal signal input, a graph is dense sine wave; scanning frequency is far higher than the frequency of sinusoidal signal input, the signal waveform of a cycle will be decomposed into several sections, graphic display will become mesh cross line.。
大学物理实验报告答案
大学物理实验报告答案大学物理实验报告答案1.伏安法测电阻实验目的( 1 ) 利用伏安法测电阻。
( 2 ) 验证欧姆定律。
( 3 ) 学会间接测量量不确定度的计算；进一步掌握有效数字的概念。
实验方法原理根据欧姆定律，I U R = ，如测得U 和I 则可计算出R 。
值得注意的是，本实验待测电阻有两只，一个阻值相对较大，一个较小，因此测量时必须采用安培表内接和外接两个方式，以减小测量误差。
实验装置待测电阻两只，0 ～5 m A 电流表1 只，0 －5 V 电压表1 只，0 ～5 0 m A 电流表1 只，0 ～1 0 V 电压表一只，滑线变阻器1 只，D F 1 7 3 0 S B 3 A 稳压源1 台。
实验步骤本实验为简单设计性实验，实验线路、数据记录表格和具体实验步骤应由学生自行设计。
必要时，可提示学生参照第 2 章中的第 2 .4 一节的有关内容。
分压电路是必须要使用的，并作具体提示。
( 1 ) 根据相应的电路图对电阻进行测量，记录U 值和I 值。
对每一个电阻测量3 次。
( 2 ) 计算各次测量结果。
如多次测量值相差不大，可取其平均值作为测量结果。
( 3 ) 如果同一电阻多次测量结果相差很大，应分析原因并重新测量。
数据处理测量次数1 2 3 U 1 / V5 . 46 . 9 8 . 5 I 1 / m A 2 . 0 0 2 . 6 0 3 . 2 0 R 1 / Ω 27 0 0 2 6 5 4 2 6 5 6 测量次数1 2 3 U 2 / V 2 . 08 2 . 2 2 2 . 5 0 I 2 / m A 3 8 . 0 4 2 . 0 4 7 . 0 R 2 / Ω 5 4 . 7 5 2 .9 5 3 . 2 ( 1 ) 由% . m a x 5555 1111 ×××× ==== U U ∆ ，得到, . V U 1 5 1 5 1 5 1 5 0000 1111 ==== ∆ V U 0 7 5 0 7 5 0 7 5 0 7 5 0000 2222 . ==== ∆ ；( 2 ) 由% . m a x 5555 1111 ××××==== I I ∆ ，得到, . m A I 0 7 5 0 7 5 0 7 5 0 7 5 0000 1111 ==== ∆ m A I 7 5 7 5 7 5 7 5 0000 2222 . ==== ∆ ；( 3 ) 再由2222 2222 33 33 33 33 ) ( ) ( I I V U R u R ∆ ∆ ++++ ==== ，求得Ω Ω 1111 1 0 1 0 1 0 1 0 9999 2222 1111 1111 ==== ×××× ==== R R u u , ；( 4 ) 结果表示Ω ± = Ω × ± = ) 1 4 4 ( , 1 0 ) 0 9 . 0 9 2 . 2 ( 2 3 1 R R 实验目的( 1 ) 了解分光计的原理和构造。
大学物理实验思考题答案
力学和热学电磁学近代物理是否可以测摆动一次的时间作周期值？为什么？答：不可以。
因为一次测量随机误差较大，多次测量可减少随机误差。
将一半径小于下圆盘半径的圆盘，放在下圆盘上，并使中心一致，讨论此时三线摆的周期和空载时的周期相比是增大、减小还是不一定？说明理由。
答：当两个圆盘的质量为均匀分布时，与空载时比较，摆动周期将会减小。
因为此时若把两盘看成为一个半径等于原下盘的圆盘时，其转动惯量I0小于质量与此相等的同直径的圆盘，根据公式T0三线摆在摆动中受空气阻尼，振幅越来越小，它的周期是否会变化？对测量结果影响大吗？为什么？答：周期减小，对测量结果影响不大，因为本实验测量的时间比较短。
金属丝弹性模量的测量光杠杆有什么优点，怎样提高光杠杆测量的灵敏度?答：优点是：可以测量微小长度变化量。
提高放大倍数即适当地增大标尺距离或适当地减小光杠杆前后脚的垂直距离，可以提高灵敏度，因为光杠杆的放大倍数为何谓视差，怎样判断与消除视差?答：眼睛对着目镜上、下移动，若望远镜十字叉丝的水平线与标尺的刻度有相对位移，这种现象叫视差，细调调焦手轮可消除视差。
为什么要用逐差法处理实验数据?答：逐差法是实验数据处理的一种基本方法，实质就是充分利用实验所得的数据，减少随机误差，具有对数据取平均的效果。
因为对有些实验数据，若简单的取各次测量的平均值，中间各测量值将全部消掉，只剩始末两个读数，实际等于单次测量。
为了保持多次测量的优越性，一般对这种自变量等间隔变化的情况，常把数据分成两组，两组逐次求差再算这个差的平均值。
实验三，随即误差的统计规律什么是统计直方图什么是正态分布曲线两者有何关系与区别?答：对某一物理量在相同条件下做次重复测量，得到一系列测量值，找出它的最大值和最小值，然后确定一个区间，使其包含全部测量数据，将区间分成若干小区间，统计测量结果出现在各小区间的频数，以测量数据为横坐标，以频数为纵坐标，划出各小区间及其对应的频数高度，则可得到一个矩形图，即统计直方图。
[实用参考]大学物理实验课思考题参考答案
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大学物理实验思考题参考答案目录一、转动惯量：二、伏安法与补偿法三、混沌思考题四、半导体PN结五、地磁场六、牛顿环七、麦克尔逊干涉仪八、全息照相九、光电效应十、声速测量十一、用电位差计校准毫安表十二、落球法测量液体的黏度十三、电子束偏转与电子比荷测量十四、铁磁材料磁化特性研究十P五、光栅衍射十六、电桥十七、电位差计十八、密立根油滴十九、模拟示波器二十、金属杨氏摸量二十一、导热系数二十二、分光计二十三、集成霍尔传感器特性与简谐振动一、转动惯量：1、由于采用了气垫装置，这使得气垫摆摆轮在摆动过程中受到的空气粘滞阻尼力矩降低至最小程度，可以忽略不计。
但如果考虑这种阻尼的存在，试问它对气垫摆的摆动（如频率等）有无影响？在摆轮摆动中，阻尼力矩是否保持不变？答：如果考虑空气粘滞阻尼力矩的存在，气垫摆摆动时频率减小，振幅会变小。
（或者说对频率有影响，对振幅有影响）在摆轮摆动中，阻尼力矩会越变越小。
2、为什么圆环的内、外径只需单次测量？实验中对转动惯量的测量精度影响最大的是哪些因素？答：圆环的内、外径相对圆柱的直径大很多，使用相同的测量工具测量时，相对误差较小，故只需单次测量即可。
（对测量结果影响大小）实验中对转动惯量测量影响最大的因素是周期的测量。
（或者阻尼力矩的影响、摆轮是否正常、平稳的摆动、物体摆放位置是否合适、摆轮摆动的角度是否合适等）3、试总结用气垫摆测量物体转动惯量的方法有什么基本特点？答：原理清晰、结论简单、设计巧妙、测量方便、最大限度的减小了阻尼力矩。
二、伏安法与补偿法1、利用补偿法测量电阻消除了伏安法的系统误差，还可能存在的误差包括：读数误差、计算产生的误差、仪器误差、导线阻值的影响等或其他。
2、能利用电流补偿电路对电流表内接法进行改进：三、混沌思考题1、有程序（各种语言皆可）、K 值的取值范围、图+5分有程序没有K 值范围和图+2分只有K 值范围+1分有图和K 值范围+2分2、（1）混沌具有内在的随机性：从确定性非线性系统的演化过程看，它们在混沌区的行为都表现出随机不确定性。
大学物理实验报告及答案共30页word资料
大学物理实验报告答案大全（实验数据及思考题答案全包括）伏安法测电阻实验目的(1) 利用伏安法测电阻。
(2) 验证欧姆定律。
(3) 学会间接测量量不确定度的计算；进一步掌握有效数字的概念。
U实验方法原理根据欧姆定律，R =，如测得U 和I 则可计算出R。
值得注意的是，本实验待测电阻有两只，I一个阻值相对较大，一个较小，因此测量时必须采用安培表内接和外接两个方式，以减小测量误差。
实验装置待测电阻两只，0～5mA 电流表1 只，0－5V 电压表1 只，0～50mA 电流表1 只，0～10V 电压表一只，滑线变阻器1 只，DF1730SB3A 稳压源1 台。
实验步骤本实验为简单设计性实验，实验线路、数据记录表格和具体实验步骤应由学生自行设计。
必要时，可提示学生参照第2 章中的第2.4 一节的有关内容。
分压电路是必须要使用的，并作具体提示。
(1) 根据相应的电路图对电阻进行测量，记录U 值和I 值。
对每一个电阻测量3 次。
(2) 计算各次测量结果。
如多次测量值相差不大，可取其平均值作为测量结果。
(3) 如果同一电阻多次测量结果相差很大，应分析原因并重新测量。
(1) 由∆U =U max ，得到 1 2 ；(2) 由∆I = I max ×1.5% ，得到∆I 1 = 0.075mA,∆I 2 = 0.75mA；(3) 再由u= R ( ∆U )2 + (∆I ) 2，求得u= 9 ×101Ω, u= 1Ω；R 3V 3I R1 R2(4) 结果表示R1 = (2.92 ± 0.09) ×10光栅衍射实验目的(1) 了解分光计的原理和构造。
(2) 学会分光计的调节和使用方法。
Ω, R2= (44 ±1)Ω(3) 观测汞灯在可见光范围内几条光谱线的波长实验方法原理若以单色平行光垂直照射在光栅面上，按照光栅衍射理论，衍射光谱中明条纹的位置由下式决定： (a + b) sin ψk =dsin ψk =±k λ 如果人射光不是单色，则由上式可以看出，光的波长不同，其衍射角也各不相同，于是复色光将被分解，而在中央 k =0、 ψ =0 处，各色光仍重叠在一起，形成中央明条纹。
大学物理实验报告范文(含答案)
大学物理实验报告范文(含答案)
实验目的
探究{实验内容}对{实验目标}的影响。
实验器材
- {器材1}
- {器材2}
- {器材3}
实验步骤
1. 设置实验装置，确保器材摆放正确。
2. 测量并记录初始值。
3. 对实验进行{操作1}，并记录相应数据。
4. 对实验进行{操作2}，并记录相应数据。
5. 对实验进行{操作3}，并记录相应数据。
6. 分析实验数据，得出结论。
实验结果
以下是实验中所测得的数据和结果：
经过数据分析，我们发现{结论}。
实验结论
根据实验结果和分析，我们得出以下结论：
- {结论1}
- {结论2}
- {结论3}
总结
通过这次实验，我们深入了解了{实验内容}对{实验目标}的影响。
实验结果验证了相关理论，并且经过数据分析，我们得出了一些有意义的结论。
同时，我们也发现了实验中的一些问题和改进的可能性。
附录
以下是实验过程中的一些额外信息和实验数据的详细记录：
实验过程记录
- 步骤1：...
- 步骤2：...
- 步骤3：...
实验数据详细记录
- 数据记录1：...
- 数据记录2：...
- 数据记录3：...。
大学物理实验习题包括答案整理版
第一部分：基本实验基础1．（直、圆）游标尺、千分尺的读数方法。
答： P462．物理天平1．感量与天平敏捷度关系。
天平感量或敏捷度与负载的关系。
答：感量的倒数称为天平的敏捷度。
负载越大，敏捷度越低。
2．物理天平在称衡中，为何要把横梁放下后才能够增减砝码或挪动游码。
答：保护天平的刀口。
3．检流计1．哪些用途使用时的注意点如何使检流计很快停止振荡答：用途：用于鉴别电路中两点能否相等或检查电路中有无轻微电流经过。
注意事项：要加限流保护电阻要保护检流计，随时准备松开按键。
很快停止振荡：短路检流计。
4．电表量程如何选用量程与内阻大小关系答：先预计待丈量的大小，选稍大批程试测，再采纳适合的量程。
电流表：量程越大，内阻越小。
电压表：内阻=量程×每伏欧姆数5．万用表不一样欧姆档测同一只二极管正向电阻时，读测值差别的原由答：不一样欧姆档，内阻不一样，输出电压随负载不一样而不一样。
二极管是非线性器件，不一样欧姆档测，加在二极管上电压不一样，读测值有很大差异。
6．信号发生器功率输出与电压输出的差别答：功率输出：能带负载，比方能够给扬声器加信号而发声音。
电压输出：实现电压输出，接上的负载电阻一般要大于50Ω。
比方不可以够此后输出口给扬声器加信号，即带不动负载。
7．光学元件光学表面有尘埃，可否用手帕擦试答：不可以够8．箱式电桥倍率的选择方法。
答：尽量使读数的有效数字位数最大的原则选择适合的倍率。
9．逐差法什么是逐差法，其长处答：把丈量数据分红两组，每组相应的数据分别相减，而后取差值的均匀值。
长处：每个数据都起作用，表现多次丈量的长处。
10．杨氏模量实验1．为何各长胸怀用不一样的量具测答：恪守偏差均分原理。
2．测钢丝直径时，为安在钢丝上、中、下三部位的互相垂直的方向上各测一次直径，而不是在同一部位采样数据答：钢丝不行能到处均匀。
3．钢丝长度是杨氏模量仪上下两个螺丝夹之间的长度仍是上端螺丝夹到挂砝码的砝码钩之间的长度答：前者4．采纳光放大方法测钢丝的细小伸长量时要测望远镜到标尺之间的距离L，请问， L 是指平面镜镜面到望远镜旁标尺的距离仍是指平面镜镜面到望远镜物镜之间的距离答：前者5．一定预加砝码使钢丝拉直，你能用什么方法判断需预加几个砝码答：用图示法。
大学物理实验报告预习题与思考题部分答案(周岚)
实验一物体密度的测定【预习题】1．简述游标卡尺、螺旋测微器的测量原理及使用时的注意事项。
答：（1）游标卡尺的测量原理及使用时的注意事项：游标卡尺是一种利用游标提高精度的长度测量仪器，它由主尺和游标组成。
设主尺上的刻度间距为y，游标上的刻度间距为x，x比y略小一点。
一般游标上的n个刻度间距等于主尺上(n-1)个刻度间距，即yn(-=。
由此可知，游标上的刻度间距与主尺上nx)11,这就是游标的精度。
刻度间距相差n1,即主尺上49mm与教材P33图1-2所示的游标卡尺精度为mm50游标上50格同长，如教材图1-3所示。
这样，游标上50格比主尺上50格（50mm）少一格（1mm），即游标上每格长度比主尺每格少1÷50 = 0.02(mm)，所以该游标卡尺的精度为0.02mm。
使用游标卡尺时应注意：①一手拿待测物体，一手持主尺，将物体轻轻卡住，才可读数。
②注意保护量爪不被磨损，决不允许被量物体在量爪中挪动。
③游标卡尺的外量爪用来测量厚度或外径，内量爪用来测量内径，深度尺用来测量槽或筒的深度，紧固螺丝用来固定读数。
（2）螺旋测微器的测量原理及使用时的注意事项：螺旋测微器又称千分尺，它是把测微螺杆的角位移转变为直线位移来测量微小长度的长度测量仪器。
螺旋测微器主要由固定套筒、测量轴、活动套筒(即微分筒)组成。
如教材P24图1-4所示，固定套管D上套有一个活动套筒C(微分筒)，两者由高精度螺纹紧密咬合，活动套筒与测量轴A相联，转动活动套筒可带动测量轴伸出与缩进，活动套筒转动一周(360)，测量轴伸出或缩进1个螺距。
因此，可根据活动套筒转动的角度求得测量轴移动的距离。
对于螺距是0.5mm 螺旋测微器，活动套筒C 的周界被等分为50格，故活动套筒转动1 格，测量轴相应地移动0.5/50=0.01mm,再加上估读，其测量精度可达到0.001 mm 。
使用螺旋测微器时应注意：①测量轴向砧台靠近快夹住待测物时，必须使用棘轮而不能直接转动活动套筒，听到“咯、咯”即表示已经夹住待测物体，棘轮在空转，这时应停止转动棘轮，进行读数，不要将被测物拉出，以免磨损砧台和测量轴。