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海狗油及ω-3不饱和脂肪酸的分析、分离的研究
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海狗油及ω-3不饱和脂肪酸的分析、分离的研究
官方公共微信如图所示，两平行导轨间距L＝0.1 m，足够长光滑的倾斜部分和粗糙的水平部分圆滑连接，倾斜部分与水平面的夹角θ＝30°，垂直斜面方向向上的磁场磁感应强度B＝0.5 T，水平部分没有磁场．金属棒ab质量m＝0.005 kg、电阻r＝0.02 Ω，运动中与导轨始终接触良好，并且垂直于导轨．电阻R＝0.08 Ω，其余电阻不计．当金属棒从斜面上离地高h＝1.0 m以上的任何地方由静止释放后，在水平面上滑行的最大距离x都是1.25 m．取g＝10 m/s2，求：(1)金属棒在斜面上的最大速度；(2)金属棒与水平面间的动摩擦因数；(3)从高度h＝1.0 m处滑下后电阻R上产生的热量．-乐乐题库
& 法拉第电磁感应定律知识点 & “如图所示，两平行导轨间距L＝0.1 m，...”习题详情
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如图所示，两平行导轨间距L＝0.1 m，足够长光滑的倾斜部分和粗糙的水平部分圆滑连接，倾斜部分与水平面的夹角θ＝30°，垂直斜面方向向上的磁场磁感应强度B＝0.5 T，水平部分没有磁场．金属棒ab质量m＝0.005 kg、电阻r＝0.02 Ω，运动中与导轨始终接触良好，并且垂直于导轨．电阻R＝0.08 Ω，其余电阻不计．当金属棒从斜面上离地高h＝1.0 m以上的任何地方由静止释放后，在水平面上滑行的最大距离x都是1.25 m．取g＝10 m/s2，求：(1)金属棒在斜面上的最大速度；(2)金属棒与水平面间的动摩擦因数；(3)从高度h＝1.0 m处滑下后电阻R上产生的热量．&
本题难度：一般
题型：解答题&|&来源：2014-江苏省启东中学高二下学期第二次月考（实验班）物理试卷
分析与解答
习题“如图所示，两平行导轨间距L＝0.1 m，足够长光滑的倾斜部分和粗糙的水平部分圆滑连接，倾斜部分与水平面的夹角θ＝30°，垂直斜面方向向上的磁场磁感应强度B＝0.5 T，水平部分没有磁场．金属棒ab质量m＝0.0...”的分析与解答如下所示：
(1)到达水平面之前已经开始匀速运动，设最大速度为v，感应电动势E＝BLv&&&&&&感应电流I＝安培力F＝BIL&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&匀速运动时，mgsin θ＝F&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&解得v＝1.0 m/s&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&(2)滑动摩擦力f＝μmg&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&金属棒在摩擦力作用下做匀减速直线运动，有f＝ma&&&&&&&&&&&&&&&&&&&&&&金属棒在水平面做匀减速直线运动，有v2＝2ax&&&&&&&&&&&&&&&&&&&&&&&&&&解得μ＝0.04&&&&&&& (用动能定理同样可以得分)(3)下滑的过程中，由动能定理可得：mgh－W＝mv2&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&安培力所做的功等于电路中产生的焦耳热W＝Q&&&&&&&&&&&&&&&&&&&&&&&电阻R上产生的热量：QR＝Q&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&联立解得：QR＝3.8×10－2J
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如图所示，两平行导轨间距L＝0.1 m，足够长光滑的倾斜部分和粗糙的水平部分圆滑连接，倾斜部分与水平面的夹角θ＝30°，垂直斜面方向向上的磁场磁感应强度B＝0.5 T，水平部分没有磁场．金属棒ab质量...
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经过分析，习题“如图所示，两平行导轨间距L＝0.1 m，足够长光滑的倾斜部分和粗糙的水平部分圆滑连接，倾斜部分与水平面的夹角θ＝30°，垂直斜面方向向上的磁场磁感应强度B＝0.5 T，水平部分没有磁场．金属棒ab质量m＝0.0...”主要考察你对“法拉第电磁感应定律”
等考点的理解。
因为篇幅有限，只列出部分考点，详细请访问。
法拉第电磁感应定律
与“如图所示，两平行导轨间距L＝0.1 m，足够长光滑的倾斜部分和粗糙的水平部分圆滑连接，倾斜部分与水平面的夹角θ＝30°，垂直斜面方向向上的磁场磁感应强度B＝0.5 T，水平部分没有磁场．金属棒ab质量m＝0.0...”相似的题目：
下列有关电磁感应的说法中正确的是（　　）当回路里的磁通量发生变化时，回路中一定产生感应电动势，形成感应电流闭合回路中的感应电流激发的磁场的方向总是与产生感应电流的原磁场方向相反变压器是利用互感原理来工作的在通、断电自感现象中，流过电感线圈的电流发生突变
（8分）如图所示，一个100匝的圆形线圈（图中只画了2匝），面积为200cm2，线圈的电阻为1Ω，在线圈外接一个阻值为4Ω的电阻和一个理想电压表。线圈放入方向垂直线圈平面指向纸内的匀强磁场中，磁感强度随时间变化规律如B－t图所示，求：（1）t＝3s时穿过线圈的磁通量。（2）t＝5s时，电压表的读数。&&&&
（1）如图所示，有一面积为S=100cm2金属环如图甲所示，电阻为R=0.1Ω．环中磁场变化规律如图乙所示，且磁场方向垂直环面向里，在0～2s的时间内，求①环中感应电流的大小和方向②通过金属环的电荷量的多少（2）一矩形线圈与外电阻相接后，感应电流随时间变化的图象如图丙所示，若线圈的面积为200cm2，回路总电阻为10Ω，求线圈所处匀强磁场的磁感应强度．
“如图所示，两平行导轨间距L＝0.1 m，...”的最新评论
该知识点好题
1一矩形线圈位于一随时间t变化的匀强磁场内，磁场方向垂直线圈所在的平面（纸面）向里，如图1所示．磁感应强度B随t的变化规律如图2所示．以I表示线圈中的感应电流，以图1中线圈上箭头所示方向的电流为正，则以下的I-t图中正确的是（　　）
2（2012o北京）物理课上，老师做了一个奇妙的“跳环实验”．如图所示，她把一个带铁芯的线圈、开关和电源用导终连接起来后，将一金属套环置于线圈上，且使铁芯穿过套环．闭合开关的瞬间，套环立刻跳起．某同学另找来器材再探究此实验．他连接好电路，经重复试验，线圈上的套环均末动．对比老师演示的实验，下列四个选项中，导致套环未动的原因可能是（　　）
3将闭合多匝线圈置于仅随时间变化的磁场中，线圈平面与磁场方向垂直，关于线圈中产生的感应电动势和感应电流，下列表述正确的是（　　）
该知识点易错题
1（2007o宁夏）电阻R、电容C与一线圈连成闭合回路，条形磁铁静止于线圈的正上方，N极朝下，如图所示．现使磁铁开始自由下落，在N极接近线圈上端的过程中，流过R的电流方向和电容器极板的带电情况是（　　）
2如图所示，A是长直密绕通电螺线管．小线圈B与电流表连接，并沿A的轴线OX从D点自左向右匀速穿过螺线管A．能正确反映通过电流表中电流，随X变化规律的是（　　）
3（2002o天津）图中MN、GH为平行导轨，AB、CD为跨在导轨上的两根横杆，导轨和横杆均为导体．有匀强磁场垂直于导轨所在的平面，方向如图．用I表示回路中的电流．（　　）
欢迎来到乐乐题库，查看习题“如图所示，两平行导轨间距L＝0.1 m，足够长光滑的倾斜部分和粗糙的水平部分圆滑连接，倾斜部分与水平面的夹角θ＝30°，垂直斜面方向向上的磁场磁感应强度B＝0.5 T，水平部分没有磁场．金属棒ab质量m＝0.005 kg、电阻r＝0.02 Ω，运动中与导轨始终接触良好，并且垂直于导轨．电阻R＝0.08 Ω，其余电阻不计．当金属棒从斜面上离地高h＝1.0 m以上的任何地方由静止释放后，在水平面上滑行的最大距离x都是1.25 m．取g＝10 m/s2，求：(1)金属棒在斜面上的最大速度；(2)金属棒与水平面间的动摩擦因数；(3)从高度h＝1.0 m处滑下后电阻R上产生的热量．”的答案、考点梳理，并查找与习题“如图所示，两平行导轨间距L＝0.1 m，足够长光滑的倾斜部分和粗糙的水平部分圆滑连接，倾斜部分与水平面的夹角θ＝30°，垂直斜面方向向上的磁场磁感应强度B＝0.5 T，水平部分没有磁场．金属棒ab质量m＝0.005 kg、电阻r＝0.02 Ω，运动中与导轨始终接触良好，并且垂直于导轨．电阻R＝0.08 Ω，其余电阻不计．当金属棒从斜面上离地高h＝1.0 m以上的任何地方由静止释放后，在水平面上滑行的最大距离x都是1.25 m．取g＝10 m/s2，求：(1)金属棒在斜面上的最大速度；(2)金属棒与水平面间的动摩擦因数；(3)从高度h＝1.0 m处滑下后电阻R上产生的热量．”相似的习题。韩医生:您好,我女儿杨瑾钰今年七周岁(出生日期),现读一年级
日快速散瞳数据
VOD -0.5OUS+1.25M*73→0.8 ;
VOS-1.00OUS+1.75M*84→0.8
日复光数据
VOD +0.5OUS-1.25M*163→1.0 ;
VOS+0.75OUS-1.75M*174→0.9+
日裸眼视力是:VOD0.6,
日快速散瞳,挂不到您的号,检查数据是
VOD -0.5OUS-1.25M*165→0.9+ ;
-1.75M*175→0.9
日复光数据
VOD -0.5OUS-1.25M*165→0.9+ ;
-1.75M*175→0.9+
请问 (1)现在她需要配眼镜吗? (2)配眼镜能治疗好散光吗?还是只是临时改善视力?
(3)配了眼镜她永远得戴眼镜了是吗?不配眼镜近视与散光会越来越严重吗?
(4)如果要配眼镜是不是拿日复光的数据去配?
(5)配了眼镜需要时时戴着吗?还是去上学戴着就可以?
望回复!非常感谢!-医生在线咨询-就医160网
韩医生:您好,我女儿杨瑾钰今年七周岁(出生日期),现读一年级
日快速散瞳数据
VOD -0.5OUS+1.25M*73→0.8 ;
VOS-1.00OUS+1.75M*84→0.8
日复光数据
VOD +0.5OUS-1.25M*163→1.0 ;
VOS+0.75OUS-1.75M*174→0.9+
日裸眼视力是:VOD0.6,
日快速散瞳,挂不到您的号,检查数据是
VOD -0.5OUS-1.25M*165→0.9+ ;
-1.75M*175→0.9
日复光数据
VOD -0.5OUS-1.25M*165→0.9+ ;
-1.75M*175→0.9+
请问 (1)现在她需要配眼镜吗? (2)配眼镜能治疗好散光吗?还是只是临时改善视力?
(3)配了眼镜她永远得戴眼镜了是吗?不配眼镜近视与散光会越来越严重吗?
(4)如果要配眼镜是不是拿日复光的数据去配?
(5)配了眼镜需要时时戴着吗?还是去上学戴着就可以?
望回复!非常感谢!
【病情状况】韩医生:您好,我女儿杨瑾钰今年七周岁(出生日期),现读一年级
日快速散瞳数据
VOD -0.5OUS+1.25M*73→0.8 ;
VOS-1.00OUS+1.75M*84→0.8
日复光数据
VOD +0.5OUS-1.25M*163→1.0 ;
VOS+0.75OUS-1.75M*174→0.9+
日裸眼视力是:VOD0.6,
日快速散瞳,挂不到您的号,检查数据是
VOD -0.5OUS-1.25M*165→0.9+ ;
-1.75M*175→0.9
日复光数据
VOD -0.5OUS-1.25M*165→0.9+ ;
-1.75M*175→0.9+
请问 (1)现在她需要配眼镜吗? (2)配眼镜能治疗好散光吗?还是只是临时改善视力?
(3)配了眼镜她永远得戴眼镜了是吗?不配眼镜近视与散光会越来越严重吗?
(4)如果要配眼镜是不是拿日复光的数据去配?
(5)配了眼镜需要时时戴着吗?还是去上学戴着就可以?
望回复!非常感谢!
1、如果上课看得清，可以暂时不配眼镜。2、配眼镜不能治疗好她的近视和散光。只是帮助她看得清楚。3、配了眼镜在18岁前要戴着，18岁后可以选择激光手术摘掉眼镜。近视会越来越严重，加重速度与眼疲劳有关。4、是的。5、最好时时戴着。
好的，非常感谢韩医生，那我先不给配眼镜，半年后再去验光对吗？
如果开学后不影响上课，先不配镜。要尽量不看电视、电脑、手机、ipad等电子产品。滴假性近视的眼药水。4－6个月复查验光。
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预约挂号 随时随地CHAPTER 1 - BASIC TERMS AND CALCULATIONS
It is important to be able to measure and calculate surface areas. It might be necessary to calculate, for example, the surface area of the cross-section of a canal or the surface area of a farm.
This Section will discuss the calculation of some of the most common surface areas: the triangle, the square, the rectangle, the rhombus, the parallelogram, the trapezium and the circle (see Fig. 1a).
Fig. 1a. The most common surface areas
The height (h) of a triangle, a rhombus, a parallelogram or a trapezium, is the distance from a top corner to the opposite side called base (b). The height is always perpe in other words, the height makes a &right angle& with the base. An example of a right angle is the corner of this page.
In the case of a square or a rectangle, the expression length (1) is commonly used instead of base and width (w) instead of height. In the case of a circle the expression diametre (d) is used (see Fig. 1b).
Fig. 1b. The height (h), base (b), width (w), length (1) and diametre (d) of the most common surface areas
The surface area or surface (A) of a triangle is calculated by the formula:
A (triangle) = 0.5 x base x height = 0.5 x b x h ..... (1)
Triangles can have many shapes (see Fig. 2) but the same formula is used for all of them.
Fig. 2. Some examples of triangles
Calculate the surface area of the triangles no. 1, no. 1a and no. 2
Triangles no. 1 and no. 1a:
base = 3 cmheight = 2 cm
A = 0.5 x base x height= 0.5 x 3 cm x 2 cm = 3 cm2
Triangle no. 2:
base = 3 cmheight = 2 cm
A = 0.5 x 3 cm x 2 cm = 3 cm2
It can be seen that triangles no. 1, no. 1a and no. 2 h the shapes of the triangles are different, but the base and the height are in all three cases the same, so the surface is the same.
The surface of these triangles is expressed in square centimetres (written as cm2). Surface areas can also be expressed in square decimetres (dm2), square metres (m2), etc...
Calculate the surface areas of the triangles nos. 3, 4, 5 and 6.
Triangle no. 3:
base = 3 cmheight = 2 cm
A = 0.5 x base x height= 0.5 x 3 cm x 2 cm = 3 cm2
Triangle no. 4:
base = 4 cmheight = 1 cm
A = 0.5 x 4 cm x 1 cm = 2 cm2
Triangle no. 5:
base = 2 cmheight = 3 cm
A = 0.5 x 2 cm x 3 cm = 3 cm2
Triangle no. 6:
base = 4 cmheight = 3 cm
A = 0.5 x 4 cm x 3 cm = 6 cm2
The surface area or surface (A) of a square or a rectangle is calculated by the formula:
A (square or rectangle) = length x width = l x w ..... (2)
In a square the lengths of all four sides are equal and all four angles are right angles.
In a rectangle, the lengths of the opposite sides are equal and all four angles are right angles.
Fig. 3. A square and a rectangle
Note that in a square the length and width are equal and that in a rectangle the length and width are not equal (see Fig. 3).
Calculate the surface areas of the rectangle and of the square (see Fig. 3).
length = 2 cmwidth = 2 cm
A = length x width= 2 cm x 2 cm = 4 cm2
Rectangle:
length = 5 cmwidth = 3 cm
A = length x width= 5 cm x 3 cm = 15 cm2
Related to irrigation, you will often come across the expression hectare (ha), which is a surface area unit. By definition, 1 hectare equals 10 000 m2. For example, a field with a length of 100 m and a width of 100 m2 (see Fig. 4) has a surface area of 100 m x 100 m = 10 000 m2 = 1 ha.
Fig. 4. One hectare equals 10 000 m2
The surface area or surface (A) of a rhombus or a parallelogram is calculated by the formula:
A (rhombus or parallelogram) = base x height = b x h ..... (3)
In a rhombus the lengths of all
none of the ang opposite sides run parallel.
In a parallelogram the lengths of the oppo none of the ang opposite sides run parallel (see Fig. 5).
Fig. 5. A rhombus and a parallelogram
Calculate the surface areas of the rhombus and the parallelogram (see Fig. 5).
base = 3 cmheight = 2 cm
A = base x height= 3 cm x 2 cm = 6 cm2
Parallelogram:
base = 3.5 cmheight = 3 cm
A = base x height= 3.5 cm x 3 cm = 10.5 cm2
The surface area or surface (A) of a trapezium is calculated by the formula:
A (trapezium) = 0.5 (base + top) x height = 0.5 (b + a) x h ..... (4)
The top (a) is the side opposite and parallel to the base (b). In a trapezium only the base and the top run parallel.
Some examples are shown in Fig. 6:
Fig. 6. Some examples of trapeziums
Calculate the surface area of trapezium no. 1.
Trapezium no. 1:
base = 4 cmtop = 2 cmheight = 2 cm
A = 0.5 x (base x top) x height= 0.5 x (4 cm + 2 cm) x 2 cm= 0.5 x 6 cm x 2 cm = 6 cm2
Calculate the surface areas trapeziums nos. 2, 3 and 4.
Trapezium no. 2:
base = 5 cmtop = 1 cmheight = 2 cm
A = 0.5 x (base + top) x height= 0.5 x (5 cm + 1 cm) x 2 cm= 0.5 x 6 cm x 2 cm = 6 cm2
Trapezium no. 3:
base = 3 cmtop = 1 cmheight = 1 cm
A = 0.5 x (3 cm + 1 cm) x 2 cm= 0.5 x 4 cm x 2 cm = 4 cm2
Trapezium no. 4:
base = 2 cmtop = 4 cmheight = 2 cm
A = 0.5 x (2 cm + 4 cm) x 2 cm= 0.5 x 6 cm x 2 cm = 6 cm2
Note that the surface areas of the trapeziums 1 and 4 are equal. Number 4 is the same as number 1 but upside down.
Another method to calculate the surface area of a trapezium is to divide the trapezium into a rectangle and two triangles, to measure their sides and to determine separately the surface areas of the rectangle and the two triangles (see Fig. 7).
Fig. 7. Splitting a trapezium into one rectangle and two triangles. Note that A = A1 + A2 + A3 = 1 + 6 + 2 = 9 cm2
The surface area or surface (A) of a circle is calculated by the formula:
A (circle) = 1/4 (p x d x d) = 1/4 (p x d2) = 1/4 (3.14 x d2) ..... (5)
whereby d is the diameter of the circle and p (a Greek letter, pronounced Pi) a constant (p = 3.14). A diameter (d) is a straight line which divides the circle in two equal parts.
Fig. 8. A circle
Circle: d = 4.5 cm
A = 1/4 (p x d²)
= 1/4 (3.14 x d x d)= 1/4 (3.14 x 4.5 cm x 4.5 cm)= 15.9 cm2
Calculate the surface area of a circle with a diameter of 3 m.
Circle: d = 3 m
A = 1/4 (p x d²) = 1/4 (3.14 x d
= 1/4 (3.14 x 3 m x 3 m) = 7.07 m2
i. Units of length
The basic unit of length in the metric system is the metre (m). One metre can be divided into 10 decimetres (dm), 100 centimetres (cm) or 1000 millimetres (mm); 100 m equals to 1 hectometre (hm); while 1000 m is 1 kilometre (km).
1 m = 10 dm = 100 cm = 1000 mm0.1 m = 1 dm = 10 cm = 100 mm0.01 m = 0.1 dm = 1 cm = 10 mm0.001 m = 0.01 dm = 0.1 cm = 1 mm
1 km = 10 hm = 1000 m0.1 km = 1 hm = 100 m0.01 km = 0.1 hm = 10 m0.001 km = 0.01 hm = 1 m
ii. Units of surface
The basic unit of area in the metric system is the square metre (m), which is obtained by multiplying a length of 1 metre by a width of 1 metre (see Fig. 9).
Fig. 9. A square metre
1 m2 = 100 dm2 = 10 000 cm2 = 1 000 000 mm20.01 m2 = 1 dm2 = 100 cm2 = 10 000 mm20.0001 m2 = 0.01 dm2 = 1 cm2 = 100 mm20.000001 m2 = 0.0001 dm2 = 0.01 cm2 = 1 mm2
1 km2 = 100 ha2 = 1 000 000 m20.01 km2 = 1 ha2 = 10 000 m20.000001 km2 = 0.0001 ha2 = 1 m2
1 ha = 100 m x 100 m = 10 000 m2
This Section explains how to apply the surface area formulas to two common practical problems that will often be met in the field.
The most common shape of a canal cross-section is a trapezium or, more truly, an &up-side-down& trapezium (see Fig. 10).
The area (A B C D), hatched on the above drawing, is called the canal cross-section and has a trapezium shape (compare with trapezium no. 4). Thus, the formula to calculate its surface is similar to the formula used to calculate the surface area of a trapezium (formula 4):
Surface area of the canal cross-section = 0.5 (base + top line) x canal depth = 0.5 (b + a) x h ..... (6)
base (b) = bottom width of the canal
top line (a) = top width of the canal
canal depth (h) = height of the canal (from the bottom of the canal to the top of the embankment)
Suppose that the canal contains water, as shown in Fig. 11.
The area (A B C D), hatched on the above drawing, is called the wetted canal cross-section or wetted cross-section. It also has a trapezium shape and the formula to calculate its surface area is:
Surface area of the wetted canal cross-section = 0.5 (base + top line) x water depth = 0.5 (b + a1) x h1 ..... (7)
base (b) = bottom width of the canal
top line (a1) = top width of the water level
water depth (h1) = the height or depth of the water in the canal (from the bottom of the canal to the water level).
Calculate the surface area of the cross-section and the wetted cross-section, of the canal shown in Fig. 12 below.
Canal cross-section:
base (b) = 1.25 mtop line (a) = 3.75 mcanal depth (h) = 1.25 m
A = 0.5 x (b + a) x h= 0.5 x (1.25 m + 3.75 m) x 1.25 m= 3.125 m2
Canal wetted cross-section:
base (b) = 1.25 mtop line (a1) = 3.25 mwater depth (h1) = 1.00 m
A = 0.5 x (b + a1) x h= 0.5 x (1.25 m + 3.25 m) x 1.00 m= 2.25 m2
It may be necessary to determine the surface area of a farmer's field. For example, when calculating how much irrigation water should be given to a certain field, the size of the field must be known.
When the shape of the field is regular and has, for example, a rectangular shape, it should not be too difficult to calculate the surface area once the length of the field (that is the base of its regular shape) and the width of the field have been measured (see Fig. 13).
Fig. 13. Field of regular shape
Length of the field = 50 mWidth of the field = 30 m
A = length x width (formula 2)= 50 m x 30 m = 1500 m2
What is the area of the same field, expressed in hectares?
Section 1.1.2 explained that a hectare is equal to 10 000 m. Thus, the formula to calculate a surface area in hectares is:
In this case: area of the field in
More often, however, the field shape is not regular, as shown in Fig. 14a.
Fig. 14a. Field of irregular shape
In this case, the field should be divided in several regular areas (square, rectangle, triangle, etc.), as has been done in Fig. 14b.
Fig. 14b. Division of irregular field into regular areas
Surface area of the square: As = length x width = 30 m x 30 m = 900 m2Surface area of the rectangle: Ar = length x width = 50 m x 15 m = 750 m2Surface area of the triangle: At = 0.5 x base x height = 0.5 x 20 m x 30 m = 300 m2Total surface area of the field: A = As + Ar + At = 900 m2 + 750 m2 + 300 m2 = 1950 m2
A volume (V) is the content of a body or object. Take for example a block (Fig 15). A block has a certain length (l), width (w) and height (h). With these three data, the volume of the block can be calculated using the formula:
V (block) = length x width x height = l x w x h ..... (9)
Fig. 15. A block
Calculate the volume of the above block.
length = 4 cmwidth = 3 cmheight = 2 cm
V = length x width x height= 4 cm x 3 cm x 2 cm= 24 cm3
The volume of this block is expressed in cubic centimetres (written as cm). Volumes can also be expressed in cubic decimetres (dm3), cubic metres (m3), etc.
Calculate the volume in m3 of a block with a length of 4 m, a width of 50 cm and a height of 200 mm.
All data must be converted in metres (m)
length = 4 mwidth = 50 cm = 0.50 mheight = 200 mm = 0.20 m
V = length x width x height= 4 m x 0.50 m x 0.20 m= 0.40 m3
Calculate the volume of the same block, this time in cubic centimetres (cm3)
All data must be converted in centimetres (cm)
length = 4 m = 400 cmwidth = 50 cmheight = 200 mm = 20 cm
V = length x width x height= 400 cm x 50 cm x 20 cm= 400 000 cm3
Of course, the result is the same: 0.4 m3 = 400 000 cm3
The basic unit of volume in the metric system is the cubic metre (m3) which is obtained by multiplying a length of 1 metre, by a width of 1 metre and a height of 1 metre (see Fig. 16).
Fig. 16. One cubic metre
1 m3 = 1.000 dm3 = 1 000 000 cm3 = 1 000 000 000 mm30.001 m3 = 1 dm3 = 1 000 cm3 = 1 000 000 mm30.000001 m3 = 0.001 dm3 = 1 cm3 = 1 000 mm30. m3 = 0.000001 dm3 = 0.001 cm3 = 1 mm3
1 dm3 = 1 litre
1 m3 = 1000 litres
Suppose a one-litre bottle is filled with water. The volume of the water is thus 1 litre or 1 dm3. When the bottle of water is emptied on a table, the water will spread out over the table and form a thin water layer. The amount of water on the table is the same as the amount of water th being 1 litre.
The volume of wa only the shape of the &water body& changes (see Fig. 17).
Fig. 17. One litre of water spread over a table
A similar process happens if you spread irrigation water from a storage reservoir over a farmer's field.
Suppose there is a reservoir, filled with water, with a length of 5 m, a width of 10 m and a depth of 2 m. All the water from the reservoir is spread over a field of 1 hectare. Calculate the water depth (which is the thickness of the water layer) on the field, see Fig. 18.
The formula to use is:
..... (10)
As the first step, the volume of water must be calculated. It is the volume of the filled reservoir, calculated with formula (9):
Volume (V) = length x width x height = 5 m x 10 m x 2 m = 100 m3
As the second step, the thickness of the water layer is calculated using formula (10):
Surface of the field = 10 000 m2
Volume of water = 100 m3
d = 0.01 m
A water layer 1 mm thick is spread over a field of 1 ha. Calculate the volume of the water (in m3), with the help of Fig. 19.
Fig. 19. One millimetre water depth on a field of one hectare
The formula to use is:
Volume of water (V) = Surface of the field (A) x Water depth (d) ..... (11)
Surface of the field = 10 000 m2Water depth = 1 mm =1/1 000 = 0.001 m
Formula: Volume (m³)
= surface of the field (m²) x water depth (m)
V = 10 000 m2 x 0.001 mV = 10 m3 or 10 000 litres
The flow-rate of a river, or of a canal, is the volume of water discharged through this river, or this canal, during a given period of time. Related to irrigation, the volume of water is usually expressed in litres (l) or cubic metres (m3) and the time in seconds (s) or hours (h). The flow-rate is also called discharge-rate.
The water running out of a tap fills a one litre bottle in one second. Thus the flow rate (Q) is one litre per second (1 l/s) (see Fig. 20).
Fig. 20. A flow-rate of one litre per second
The water supplied by a pump fills a drum of 200 litres in 20 seconds. What is the flow rate of this pump?
The formula used is:
..... (12a)
Volume of water: 200 lTime: 20 s
The unit &litre per second& is commonly used for small flows, e.g. a tap or a small ditch. For larger flows, e.g. a river or a main canal, the unit &cubic metre per second& (m3/s) is more conveniently used.
A river discharges 100 m3 of water to the sea every 2 seconds. What is the flow-rate of this river expressed in m3/s?
The formula used is:
..... (12b)
Volume of water: 100 m3Time: 2 s
The discharge rate of a pump is often expressed in m3 per hour (m3/h) or in litres per minute (l/min).
..... (12c)
..... (12d)
NOTE: Formula 12a, 12b, 12c and 12 only the units change
In relation to agriculture, the words percentage and per mil will be met regularly. For instance &60 percent of the total area is irrigated during the dry season&. In this Section the meaning of the words &percentage& and &per mil& will be discussed.
The word &percentage& means literally &per hundred&; in other words one percent is the one hundredth part of the total. You can either write percent, or %, or 1/100, or 0.01.
Some examples are:
5 percent = 5% =5/100 = 0.0520 percent = 20% = 20/100= 0.2025 percent = 25% = 25/100 = 0.2550 percent = 50% = 50/100 =0.50100 percent = 100% = 100/100 = 1150 percent = 150% = 150/100 = 1.5
How many oranges are 1% of a total of 300 oranges? (see Fig. 21)
Fig. 21. Three oranges are 1% of 300 oranges
1% of 300 oranges = 1/100 x 300 = 3 oranges
6% of 100 cows
6/100 x 100 = 6 cows
15% of 28 hectares
15/100 x 28 = 4.2 ha
80% of 90 irrigation projects
80/100 x 90 = 72 projects
150% of a monthly salary of $100
150/100 x 100 = 1.5 x 100 = $150
0.5% of 194.5 litres
0.5/100 x 194.5 = 0.005 x 194.5 = 0.9725 litres
The word &per mil& means literally &per thousand&; in other words one per mil is one thousandth part of the total.
You can either write: per mil, or ‰, or 1/1000, or 0.001.
Some examples are:
5 per mil = 5‰ =5/50 per mil = 50‰ = 50/1000 = 0.05095 per mil = 95‰ = 95/1000 = 0.095
How many oranges are 4‰ of 1000 oranges? (see Fig. 22)
Fig. 22. Four oranges are 4‰ of 1000 oranges
4‰ of 1000 oranges = 4/1000 x 1000 = 4 oranges
10‰ = 1%
because 10‰ = 10/1000 = 1/10 = 1%
3‰ of 3 000 oranges
3/1000 x 3 000 = 9 oranges
35‰ of 10 000 ha
35/1000 x 10 000 = 350 ha
0.5‰ of 750 km2
0.5/1000 x 750 =0.375 km2
A graph is a drawing in which the relationship between two (or more) items of information (e.g. time and plant growth) is shown in a symbolic way.
To this end, two lines are drawn at a right angle. The horizontal one is called the x axis and the vertical one is called the y axis.
Where the x axis and the y axis intersect is the &0& (zero) point (see Fig. 23).
The plotting of the information on the graph is discussed in the following examples.
Fig. 23. A graph
Suppose it is necessary to make a graph of the growth rate of a maize plant. Each week the height of the plant is measured. One week after planting the seed, the plant measures 2 cm in height, two weeks after planting it measures 5 cm and 3 weeks after planting the height is 10 cm, as illustrated in Fig. 24a.
These results can be plotted on a graph. The time (in weeks) will be ind 2 cm on the axis represents 1 week. The plant height (in centimetres) will be ind 1 cm on the axis represents 1 cm of plant height.
After 1 week the height is 2 this is indicated on the graph with A; after 2 weeks the height is 5 cm, see B, and after 3 weeks the height is 10 cm, see C, as shown in Fig. 24b.
At planting (Time = 0) the height was zero, see D.
Now connect the crosses (see Fig. 24c) with a straight line. The line indicates the grow this is the height increase over time.
Fig. 24b. Growth rate of a maize plant
It can be seen from the graph that the plant is growing faster and faster (during the first week 2 cm and during the third week 5 cm); the line from B to C is steeper than the line from D to A.
From the graph can be read what the height of the plant was after, say 2 1/2 see the dotted line (Fig. 24c). Locate on the horizontal axis 2 1/2 weeks and follow the dotted line upwards until the dotted line crosses the graph. From this crossing follow the dotted line to the left until the vertical axis is reached. Now take the reading: 7.5 cm, which means that the plant had a height of 7.5 cm after 2 1/2 weeks. This height has not been measured in reality, but with the graph the height can be determined anyway.
What was the height of the plant after 1 1/2 weeks?
The height of the plant after 1 1/2 weeks was 3.5 cm (see Fig. 24c).
Fig. 24c. Graph of the growth rate of a maize plant
Another example to illustrate how a graph should be made is the variation of the temperature over one full day (24 hours). Suppose the outside temperature (always in the shade) is measured, with a thermometer, every two hours, starting at midnight and ending the following midnight.
Suppose the following results are found:
Temperature (°C)
On the x axis indicate the time in hours, whereby 1 cm on the graph is 2 hours. On the y axis indicate the temperature in degrees Celsius (°C), whereby 1 cm on the graph is 5°C.
Now indicate (with crosses) the values from the table (above) on the graph paper and connect the crosses with straight dotted lines (see Fig. 25a).
At this stage, if you look attentively at the graph, you will note that there is a very abrupt change in its shape around the sixteenth hour. The outside temperature seems to have fallen from 28°C to 2°C in two hours time! That does not make sense, and the reading of the thermometer at the sixteenth hour must have been wrong. This cross cannot be taken in consideration for the graph and should be rejected. The only dotted line we can accept is the straight one in between the reading at the fourteenth hour and the reading at the eighteenth hour (see Fig. 25b).
In reality the temperature will change more gradually than indicate that is why a smooth curve is made (continuous line). The smooth curve represents the most realistic approximation of the temperature over 24 hours (see Fig. 25c).
From the graph it can be seen that the minimum or lowest temperature was reached around 4 o'clock in the morning and was about 6°C. The highest temperature was reached at 4 o'clock in the afternoon and was approximately 29°C.
What was the temperature at 7, 15 and 23 hours? (Always use the smooth curve to take the readings).
ANSWER (see Fig. 25c)
Temperature at 7 hours: 10°CTemperature at 15 hours: 29°CTemperature at 23 hours: 17°C
1) Calculate the surface areas of the following triangles:
a. height = 6 cm, base = 12 cm = = = A = .....cm2b. height = 22 cm, base = 48 cm = = = A = .....cm2c. height = 16 cm, base = 24 cm = = = A = .....cm2d. height = 0.8 m, base = 0.3 m = = = A = .....m2
2) Calculate the surface areas of the following trapeziums:
a. height = 12 cm, base = 52 cm, top = 16 cm = = = A = .....cm2b. height = 20 cm, base = 108 dm, top = 16 cm = = = A = .....cm2c. height = 0.3 m, base = 1.8 m, top = 1.5 m = = = A = .....m2
3) Calculate the cross-section of the canal when given:
height = 1 mtop width = 2.6 mbottom width = 1.2 m
4) Calculate the wetted cross-section when in addition to 3) is given that the water height is 0.8 m and the top width of the water surface is 2.32 m.
5) A rectangular field has a length of 120 m and a width of 85 m. What is the area of the field in hectares?
a. 25% of 1820 metres = .....metresb. 13% of 971 cm = .....cmc. 83% of 8000 apples = .....applesd. 7‰ of 18 060 metres = .....metrese. 13% of 26 hectares = .....hectaresf. 1.5‰ of 28 000 metres = .....metres
7) Calculate the volume of the following blocks, when given:
a. length = 75 cm, width = 3 m, height = 6 cm = = = V = .....m3b. length = 0.5 cm, width = 1 dm, height = 20 cm = = = V = .....m3c. length = 15 cm, width = 2 dm, height = 0.5 m = = = V = .....litres
8) Calculate the volume of water (in m3) on a field, when given: the length = 150 m, the width = 56 m and the water layer = 70 mm.
9) Calculate the minimum depth of a reservoir, which has: a length of 15 m and a width of 10 m and which has to provide 50 mm water for a field of 175 m long and 95 m wide.
10) Make a graph of the monthly rainfall over a period of 1 year, when given:
Rain (mm/month)
a. A = 0.5 x b x h = 0.5 x 6 cm x 12 cm = 36 cm2b. A = 0.5 x 22 cm x 48 cm = 528 cm2c. A = 0.5 x 16 cm x 24 cm = 192 cm2d. A = 0.5 x 0.8 m x 0.3 m = 0.12 m2
a. A = 0.5 x (b + a) x h = 0.5 x (52 cm + 16 cm) x 12 cm = 408 cm2b. A = 0.5 x (108 cm + 16 cm) x 20 cm = 1240 cm2c. A = 0.5 x (1.8 m + 1.5 m) x 0.3 m = 0.495 m2
3) A = 0.5 x (b + a) x h = 0.5 x (1.2 m + 2.6 m) x 1 m = 1.9 m2
4) A = 0.5 x (b + a1) x h1 - 0.5 x (1.2 m + 2.32 m) x 0.8 m = 1.408 m2
5) Area of the field in square metres = l (m) x w (m) = 120 m x 85 m = 10 200 m2
a. 1820 m x 25/100 = 455 mb. 971 cm x 13/100 = 126.23 cmc. 8000 apples x 83/100 = 6640 applesd. 18 060 m x 7/1000 = 126.42 me. 26 ha x 13/100= 3.38 haf. 28 000 m x 1.5/1000 = 42 m
a. V = l x w x h = 0.75 m x 3 m x 0.06 m = 0.135 m3b. V = 0.005 m x 0.10 m x 0.20 m = 0.0001 m3c. V = 1.5 dm x 2 dm x 5 dm = 15 dm3 = 15 litres
8) V = l x w x h = 150 m x 56 m x 0.070 m = 588 m3
Volume of water required on a field: V = length of field (m) x width of field (m) x thickness of water layer (m) = 175 m x 95 m x 0.050 m = 831.25 m3
Volume of the reservoir: V = 831.25 m3 = length of reservoir (m) x width of reservoir (m) x depth of reservoir (m)}