请以资深中国初中物理教师的身份，讲解试卷图片上第21题，要求讲解过程图文并茂，用不同的颜色标注，语言风格风趣幽默，简单易懂。我将提供题目和答案---Here is the extraction of the content from the image: **General Information:** * Title/Section: 五、综合应用题 (Five, Comprehensive Application Questions) * Score Distribution: (本题共 2 小题, 第 20 题 8 分, 第 21 题 9 分, 共 17 分) (Total 2 questions, Question 20 is 8 points, Question 21 is 9 points, total 17 points) * Page Footer: 物理试卷(A) 第 6 页(共 6 页) (Physics Test Paper (A) Page 6 (Total 6 pages)) **Problem 20:** * Question Stem: 20. 如图 15 所示是一款用于物流搬运的电动机器人,它能够代替人工进行货物装卸,自动化程度高,节省人力成本。该机器人在某次作业时叉起质量为 900 kg 的货物,沿水平仓库地面匀速前进 40 m,用时 20 s,然后在 10 s 内将货物沿竖直方向匀速提升 2 m。货物底面积为 1 m²,与两个叉臂的总接触面积为 0.1 m²,g 取 10 N/kg。 (As shown in Figure 15, this is an electric robot used for logistics handling. It can replace manual labor for cargo loading and unloading, with high automation and cost savings. During a certain operation, this robot lifted a 900 kg cargo, moved forward uniformly 40 m along the horizontal warehouse floor, taking 20 s. Then it lifted the cargo uniformly 2 m vertically in 10 s. The cargo's bottom area is 1 m², and the total contact area with the two forks is 0.1 m². g is taken as 10 N/kg.) * Sub-questions: (1) 货物对叉臂的压强是多少? (1) What is the pressure exerted by the cargo on the forks? (2) 机器人沿水平地面匀速前进的速度为多大? 匀速前进过程中叉臂对货物的支持力做了多少功? (2) What is the speed of the robot moving uniformly along the horizontal ground? How much work is done by the forks' support force on the cargo during the uniform horizontal movement? (3) 机器人竖直提升货物的功率是多少? (3) What is the power of the robot lifting the cargo vertically? (4) 若竖直提升货物的过程中消耗了 0.008 kW·h 的电能,则该过程中机器人做功的效率是多少? (4) If 0.008 kW·h of electrical energy is consumed during the vertical lifting process, what is the efficiency of the robot's work in this process? * Figure 15 Description: * Type: Image of a physical object (Electric Robot). * Main elements: Shows a robot with two forks at the front, designed for lifting and moving objects. The robot appears to have wheels or tracks at the bottom and a control panel or main body. * Label: 图 15 (Figure 15) **Problem 21:** * Question Stem: 21. 小明家新购买的电热水壶(如图 16 甲),其部分参数如图 16 乙 所示。 (21. Xiao Ming's family recently purchased an electric kettle (as shown in Figure 16 甲), and some of its parameters are shown in Figure 16 乙.) * Figure 16 Description: * Type: Composite figure. * Part 甲 (Figure 16 甲): Image of a physical object (Electric Kettle). Shows a typical electric kettle with a handle, spout, lid, and base. * Part 乙 (Figure 16 乙): Table of parameters. * Label: 乙 * Table Header/Rows: * 型号: xxx (Model: xxx) * 额定电压: 220 V (Rated Voltage: 220 V) * 额定功率: 1000 W (Rated Power: 1000 W) * 额定容量: 2.0 L (Rated Capacity: 2.0 L) * Label: 图 16 (Figure 16) * Sub-questions: (1) 电热水壶的电源线粗而短,因此电阻较\_\_\_\_\_\_\_\_\_\_\_\_,\_\_\_\_\_\_\_\_\_\_\_\_产生热量较\_\_\_\_\_\_\_\_\_\_\_\_,使用时更加安全。 (1) The power cord of the electric kettle is thick and short, so the resistance is relatively \_\_\_\_\_\_\_\_\_\_\_\_, \_\_\_\_\_\_\_\_\_\_\_\_ generates relatively \_\_\_\_\_\_\_\_\_\_\_\_ heat, making it safer to use. (Fill in the blanks) (2) 小明在壶中加入额定容量的初温为 25 ℃ 的水,把电热水壶插进插座,闭合开关,突然家里的用电器都不工作了,请写出可能的两个原因:①\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_;\_ ②\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_。 (2) Xiao Ming added water of rated capacity with an initial temperature of 25 ℃ into the kettle, plugged the electric kettle into the socket, and closed the switch. Suddenly, none of the household appliances worked. Please write down two possible reasons: ①\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_;\_\_ ②\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_。 (Fill in the blanks) (3) 排除故障后,小明继续实验,将壶中水加热到沸腾,从理论上讲,需要多长时间?实际所用时间大于理论时间,可能是由什么原因造成的?请写出两条。已知烧水 时气压为 1 个标准大气压,$\rho_{水} = 1.0\times10^{3}$ kg/m$^{3}$, $c_{水} = 4.2\times10^{3}$ J/(kg·℃)。 (3) After troubleshooting, Xiao Ming continued the experiment, heating the water in the kettle to boiling. Theoretically, how long does it take? The actual time taken is longer than the theoretical time, what could be the reasons for this? Please write two reasons. Given that the atmospheric pressure during boiling is 1 standard atmospheric pressure, $\rho_{水} = 1.0\times10^{3}$ kg/m$^{3}$, $c_{水} = 4.2\times10^{3}$ J/(kg·℃). **Options Section (Part above Problem 20):** * This section appears to be related to a previous question (possibly related to evaluating solutions or options). It shows a table header with columns "得分" (Score) and "评卷人" (Grader). * Below this table header, there are four options labeled A, B, C, D: * A. 两种方案都能 (Both options are possible) * B. 两种方案都不能 (Neither option is possible) * C. A 方案能, B 方案不能 (Option A is possible, Option B is not possible) * D. A 方案不能, B 方案能 (Option A is not possible, Option B is possible) * This section is separate from the main problems 20 and 21. **Extraction Content:** **Calculations/Explanations (Likely from previous questions):** 匀速前进过程中臂对货物的支持力与运动方向垂直, 不做功。(1分) (3)机器人提升货物做功 W=Gh=9000 N x 2 m = 1.8 x 10⁴ J 机器人提升货物的功率 P = W/t = 1.8x10⁴ J / 10 s = 1.8x10³ W (2分) (4)机器人做功的效率 η = W有用 / W总 x 100% = 1.8x10⁴ J / 0.008x3.6x10⁶ J x 100% = 62.5% (2分) **Question 21:** 21. (1) 小 少 (2分, 每空 1分) (2) ①电热水壶内部电路短路; ②电热水壶使用后家中用电器总功率过大 (2分) (3)壶中水的质量 m=ρ水V=1.0x10³ kg/m³ x 2.0x10⁻³ m³ = 2 kg 将壶中水加热到沸腾, 水吸收热量 Q=cm(t₂-t₁) = 4.2x10³ J/(kg・℃) x 2 kg x (100 ℃ - 25 ℃) = 6.3x10⁵ J (2分) 电流做功 W=Pt=Q, 可得需要时间 t = Q/P = 6.3x10⁵ J / 1000 W = 630 s (2分) 实际加热时间大于理论时间的原因可能是: (1)加热时有热量散失, 加热效率低于 100%; (2)家庭电路实际电压低于 220 V; (3)电热水壶额定功率低于 1000 W。(1分) **Chart/Diagram Description:** * **Type:** Circuit diagram. * **Main Elements:** * **Components:** A battery (labeled with positive and negative terminals, voltage possibly labeled 2.4 V nearby), a switch (labeled S), an ammeter (labeled A inside a circle), a light bulb (represented by a circle with an X inside), and a voltmeter (labeled V inside a circle). * **Connections:** The battery, switch, ammeter, and light bulb are connected in series, forming a single loop. The voltmeter is connected in parallel across the light bulb. Lines represent wires connecting the components. * **Labels and Annotations:** Labels include S (switch), A (ammeter), V (voltmeter), possibly voltage value near the battery (e.g., 2.4 V), and other unclear labels around the components or measurements. **Other Text:** 物理试卷(A)答案 第 2 页 (Physics Test (A) Answer Page 2)