反反变化问题

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What if you had a circuit with a current of 2 amps and two resistances in parallel of 10 and 20 ohms each? How could you find the circuit's voltage? After completing this Concept, you'll be able to solve real-world applications like this one using rational functions .
::如果你有一条电路,电流为2安普斯和两个阻力,每条电流平行10和20奥赫姆?你怎么能找到电路的电压?完成这个概念后,你就能用理性的功能解决真实世界的应用。

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Watch This
::观察这个

 

 

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Guidance
::指导指南指南指导指南

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We are going to investigate some problems that are described by rational functions. Our first example is one which is modeled by inverse variation , which are a special type of rational function. Many formulas in physics are described by inverse variation.
::我们将调查一些以理性函数描述的问题。我们的第一个例子是以反向变化为模型的,反向变化是一种特殊的理性功能。物理学的许多公式用反向变化来描述。

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Example A
::例A

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The frequency, $\begin{align*}f\end{align*}$ , of sound varies inversely with wavelength, $\begin{align*}\lambda\end{align*}$ . A sound signal that has a wavelength of 34 meters has a frequency of 10 hertz. What frequency does a sound signal of 120 meters have?
::音频的频率,f, 与波长反差, 。波长34米的音频信号频率为10赫兹。 120米的音频信号的频率是多少?

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Solution
::解决方案

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$$\begin{align*}\text{The inverse variation relationship is:} && f =\frac {k}{\lambda} \\ \\ \text{Plug in the values:} \ \lambda = 34 \ \text{and} \ f = 10: && 10=\frac {k}{34} \\ \\ \text{Multiply both sides by} \ 34: && k=340 \\ \\ \text{Thus, the relationship is given by:} && f=\frac {340}{\lambda} \\ \\ \text{Plug in} \ \lambda = 120 \ \text{meters:} && f=\frac {340}{120} \Rightarrow f=2.83 \ \text{Hertz}\end{align*}$$
::反差关系是:f=k+Plug, 值为:\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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Electrical circuits are commonplace is everyday life—for example, they’re in all the electrical appliances in your home. The figure below shows an example of a simple electrical circuit. It consists of a battery which provides a voltage ( $\begin{align*}V\end{align*}$ , measured in Volts, $\begin{align*}V\end{align*}$ ), a resistor ( $\begin{align*}R\end{align*}$ , measured in ohms, $\begin{align*}\Omega\end{align*}$ ) which resists the flow of electricity and an ammeter that measures the current ( $\begin{align*}I\end{align*}$ , measured in amperes, $\begin{align*}A\end{align*}$ ) in the circuit.
::电路是常见的日常生活,例如,电路存在于你家的所有电器中。下图显示了简单的电路的例子。它包括一个电池,提供电压(V,以伏尔特测量,V),一个抗电器(R,以ohms测量,___),以抗电流和测量电路电流(I,以安培尔测量,A)的仪表。

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Ohm’s Law gives a relationship between current, voltage and resistance. It states that
::Ohm’s Law给出了电流、电压和抵抗之间的关系。

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$$\begin{align*}I = \frac{V}{R}\end{align*}$$
::I=VR

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Your light bulbs, toaster and hairdryer are all basically simple resistors. In addition , resistors are used in an electrical circuit to control the amount of current flowing through a circuit and to regulate voltage levels. One important reason to do this is to prevent sensitive electrical components from burning out due to too much current or too high a voltage level. Resistors can be arranged in series or in parallel.
::您的灯泡、烤面包机和理发机基本上都是简单的阻力器。此外,电路还使用阻力器控制电流通过电路流的数量并调节电压水平。 这样做的一个重要原因是防止敏感电组件因电压水平过高或过高而燃烧。 阻力器可以按序列或平行排列。

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For resistors placed in a series:
::对于被排列在一系列抗体中的抗体:

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the total resistance is just the sum of the resistances of the individual resistors:
::完全抵抗只是个别抵抗者的抵抗的和:

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$$\begin{align*}R_{tot} = R_1 + R_2\end{align*}$$
::Rtot=R1+R2

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For resistors placed in parallel:
::对于平行的抵抗者:

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the reciprocal of the total resistance is the sum of the reciprocals of the resistances of the individual resistors:
::完全抵抗的对等性是个别抵抗者抵抗的对等性之和:

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$$\begin{align*}\frac{1}{R_c} = \frac{1}{R_1} + \frac{1}{R_2}\end{align*}$$
::1 Rc=1R1+1R2

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Example B
::例例BB

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Find the quantity labeled $\begin{align*}x\end{align*}$ in the following circuit.
::在下一条电路中查找标记为 x 的数量 。

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Solution
::解决方案

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$$\begin{align*}\text{We use the formula:} && I = \frac{V}{R} \\ \\ \text{Plug in the known values:} && I = 2 \ A, V = 12 \ V: \qquad \qquad 2 = \frac{12}{R} \\ \\ \text{Multiply both sides by} \ R: && 2R = 12 \\ \\ \text{Divide both sides by}\ 2: && R = 6 \Omega \quad \mathbf{Answer}\end{align*}$$
::我们使用的公式是:I=VRPLUG,在已知值中:I=2 A,V=12 V:2=12RMUTBY 以R:2R=12RMUTIBY 以R:2R=12DVIF 以2:R=6/28Answer为单位。

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Example C
::例例C

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Find the quantity labeled $\begin{align*}x\end{align*}$ in the following circuit.
::在下一条电路中查找标记为 x 的数量 。

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Solution
::解决方案

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$$\begin{align*}\text{Ohm's Law also tells us that:} && I_{total} = \frac{V_{total}}{R_{total}} \\ \\ \text{Plug in the values we know}, I = 2.5 \ A \ \text{and}\ E = 9 \ V: && 2.5 = \frac{9}{R_{tot}} \\ \\ \text{Multiply both sides by}\ R: && 2.5R_{tot} = 9 \\ \\ \text{Divide both sides by}\ 2.5: && R_{tot} = 3.6 \Omega \\ \\ \text{Since the resistors are placed in parallel,}\\ \text {the total resistance is given by:} && \frac{1}{R_{tot}} = \frac{1}{X} + \frac{1}{20} \\ \\ && \Rightarrow \frac{1}{3.6} = \frac{1}{X} + \frac{1}{20} \\ \\ \text{Multiply all terms by} \ 72X: && \frac{1}{3.6}(72X) = \frac{1}{X}(72X) + \frac{1}{20}(72X) \\ \\ \text{Cancel common factors:} && 20X = 72 + 3.6X \\ \\ \text{Solve:} && 16.4X = 72 \\ \\ \text{Divide both sides by}\ 16.4: && X = 4.39 \Omega \quad \mathbf{Answer}\end{align*}$$
::Ohm's Law 也告诉我们: 总计=我们所知道的数值中总总总总纯纯纯纯纯纯纯纯纯纯的值, I=2.5 A 和 E= 9 V:2.5= 9 RtutMultiply 双方以R: 2.5Rtot= 9Divide 双方以2.5: 2.5: Rtot= 3.6} 由于阻力器是平行的, 1Rtot=1X+120=1X=120x1X+120 ltiply 所有条件均以72X: 13.6(72X) +120(72X) Cancel 共同系数: 20X= 72+3.6XSolve: 16.4X=72Divide 双方以16.4: X=4.39/9Aswer提供总抗力。

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Watch this video for help with the Examples above.
::观看此视频, 帮助了解上面的例子 。

 

 

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Guided Practice
::实践指南 实践指南 实践指南 实践指南 实践指南 实践指南 实践指南

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Electrostatic force is the force of attraction or repulsion between two charges. The electrostatic force is given by the formula $\begin{align*}F = \frac{Kq_1 q_2}{d^2},\end{align*}$ where $\begin{align*}q_1\end{align*}$ and $\begin{align*}q_2\end{align*}$ are the charges of the charged particles, $\begin{align*}d\end{align*}$ is the distance between the charges and $\begin{align*}k\end{align*}$ is a proportionality constant . The charges do not change, so they too are constants; that means we can combine them with the other constant $\begin{align*}k\end{align*}$ to form a new constant $\begin{align*}K\end{align*}$ , so we can rewrite the equation as $\begin{align*}F = \frac{K}{d^2}\end{align*}$ .
::静电力是两种电荷之间的吸引力或击退力。静电力由公式F=Kq1q2d2提供,其中q1和q2是充电粒子的电荷,d是充电粒子的距离,d是充电量和k之间的距离是相称性常数。电荷不变,因此也是常数;这意味着我们可以将电力与其他恒定K结合,形成一个新的常数K,因此我们可以将方程式改写为F=Kd2。

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If the electrostatic force is $\begin{align*}F = 740\end{align*}$ Newtons when the distance between charges is $\begin{align*}5.3 \times 10^{-11}\end{align*}$ meters, what is $\begin{align*}F\end{align*}$ when $\begin{align*}d = 2.0 \times 10^{-10}\end{align*}$ meters?
::如果电阻力是F=740牛顿,当电荷的距离为5.3×10-11米时,当 d=2.0×10-10米时,F是多少?

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Solution
::解决方案

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$$\begin{align*}\text{The inverse variation relationship is:} && F=\frac {K}{d^2} \\ \\ \text{Plug in the values} \ F = 740 \ \text{and} \ d = 5.3\times10^{-11}: && 740=\frac {K}{\left ( {5.3\ \times \ 10^{-11}} \right )^2} \\ \\ \text{Multiply both sides by} \ (5.3\times10^{-11})^2: && K=740 \left ( {5.3\times 10^{-11}} \right )^2 \\ \\ && K = 2.08\times 10^{-18} \\ \\ \text{The electrostatic force is given by:} && F=\frac {2.08\ \times \ 10^{-18}}{d^2} \\ \\ \text{When} \ d = 2.0 \times 10^{-10}: && F=\frac {2.08\ \times \ 10^{-18}}{\left ( 2.0 \ \times \ 10^{-10} \right )^2} \\ \\ \text{Use scientific notation to simplify:} && F=52 \ \text{Newtons}\end{align*}$$
::逆差关系是:F=Kd2Plug F=740, d=53x10-10-11:740=K(5.3x10x10-11):K=740(5.3x10-11-11-11):K=740(5.3x10-10-11-11-11):K=2.08x10-18:K=2.08x10-18d2 when d=2.0x10-10:F=2.08x10-18(2.0x10-10)2 使用科学标记简化:F=52牛顿

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Explore More
::探索更多

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For 1-4, find the quantity labeled $\begin{align*}x\end{align*}$ in each of the following circuits.
::1-4,在以下每条电路中找到标记为x的数量。

1. 
2. 
3. 
4. 

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For 5-7, the intensity of light is inversely proportional to the square of the distance between the light source and the object being illuminated.
::就5-7而言,光的强度与光源与照亮的物体之间的距离平方成反比。

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5. A light meter that is 10 meters from a light source registers 35 lux. What intensity would it register 25 meters from the light source?
   ::光线计距光源登记簿35超载10米,光线计距光源登记簿25米,其强度是多少?
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6. A light meter that is registering 40 lux is moved twice as far away from the light source illuminating it. What intensity does it now register? (Hint: let $\begin{align*}x\end{align*}$ be the original distance from the light source.)
   ::登记在40卢布的光度计是距离光源点亮它的两倍。它现在登记了何种强度? (提示: x 是光源的原始距离 。)
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7. The same light meter is moved twice as far away again (so it is now four times as far from the light source as it started out). What intensity does it register now?
   ::同样的光计又移动了两倍远的距离(因此现在离光源的距离是最初的四倍 ) 。 它现在登记了多大的强度 ?

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8. Ohm’s Law states that current flowing in a wire is inversely proportional to the resistance of the wire. If the current is 2.5 Amperes when the resistance is 20 ohms, find the resistance when the current is 5 Amperes.
   ::《奥姆法》指出,电线中流的电流与电线的阻力成反比。 如果阻力为20奥姆时的电流为2.5安珀斯,当电流为5安珀斯时就能找到阻力。

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For 9-10, the volume of a gas varies directly with its temperature and inversely with its pressure. At 273 degrees Kelvin and pressure of 2 atmospheres, the volume of a certain gas is 24 liters.
::9-10年,气体的体积与其温度直接不同,与其压力反差,在273度开尔文和2度大气压力下,某一气体的体积为24升。

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9. Find the volume of the gas when the temperature is 220 Kelvin and the pressure is 1.2 atmospheres.
   ::当温度为220开尔文,压力为1.2个大气时,找到气体的体积。
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10. Find the temperature when the volume is 24 liters and the pressure is 3 atmospheres.
    ::当体积为24升 压力为3度大气时 找到温度

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For 11-13, the volume of a square pyramid varies jointly with the height and the square of the side length of the base. A pyramid whose height is 4 inches and whose base has a side length of 3 inches has a volume of $\begin{align*}12 \ in^3\end{align*}$ .
::11-13年,平方金字塔的体积与基数的高度和侧长的平方不同,高4英寸,基数为3英寸的平方金字塔的体积为12英寸。

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11. Find the volume of a square pyramid that has a height of 9 inches and whose base has a side length of 5 inches.
    ::找到一个方形金字塔的体积,其高度为9英寸,其底部侧长为5英寸。
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12. Find the height of a square pyramid that has a volume of $\begin{align*}49 \ in^3\end{align*}$ and whose base has a side length of 7 inches.
    ::找到一个平方金字塔的高度,其体积为49英寸3,其底部侧长为7英寸。
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13. A square pyramid has a volume of $\begin{align*}72 \ in^3\end{align*}$ and its base has a side length equal to its height. Find the height of the pyramid.
    ::方形金字塔的体积为72英寸3,其底部的侧长等于其高度。 查找金字塔的高度 。

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Answers for Explore More Problems
::探索更多问题的答案

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Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。