人类遵循地质构造规律开展的实践活动是A. 在背斜构造地区找石油B. 在向斜构造盆地找天然气C. 在断裂带上修建大型水库D. 沿地下断层线开挖地下隧道

黄土和古土壤的测试指标中，指示东亚冬季风势力强弱的是（ ）。A. 粒度B. 磁化率C. 碳酸盐氧同位素组成D. 有机碳含量

升交点的赤径Ω,轨道的倾角 i ,确定了卫星轨道平面与地球体之间的相对定向。()A. 正确B. 错误

正射影像特点:每个像素都具有(),逐像素纠正(看不到建筑物()面),可()。A. 经纬度坐标,立,三维浏览B. 三维坐标值,顶,量测C. 经纬度坐标,顶,测量D. 经纬度坐标,立,量测

42.国家建立地图审核制度,向社会公开的所有地图-|||-均应当报送有审核权的测绘地理信息主管部门审-|||-核,方可向社会公布o-|||-A正确-|||-B错误

城市化现象是社会经济发展和工业化的必然产物，是社会进步的表现，据此回答1~2题。 (1) 关于城市化的叙述，正确的有 (2) 衡量城市化水平最重要的标志是 A.城市化指农村人口向城市聚集，不包括乡村变为城市地区的过程； B.一个国家或地区的城市化水平可以体现社会经济发展水平； C.城市是区域经济的发展中心，但对周边地区的经济发展影响不大； D.城市化必然将扩大城乡之间的差距

岩石受力发生破坏后，破裂两侧岩块发生显著位移的构造称之为( ）。A. 褶皱B. 节理C. 断层D. 构造体系

黄土和以下哪两项并称全球变化研究的三大支柱?A. 化石B. 极地冰芯C. 深海沉积D. 湖湘沉积

第一次课后作业答案:选择题: 1 B 2 C 3 D 5 D3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 105页参考答案4-1:b 4-2: c 4-4: a 4-5: a3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 不同温度的S值也可以直接用饱和水表查得。计算结果是0.3363-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 6-43-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 在总压101.33kPa、温度350.8K下,苯(1)-正己烷(2)形成x1=0.525的恒沸混合物。此温度下两组分的蒸汽压分别是99.4kPa和97.27kPa,液相活度系数模型选用Margules方程,气相服从理想气体,求350.8K下的气液平衡关系3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 和3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 的函数式。

具有规则的几何形状,纹理平滑细腻,地块边界多有路、渠、田间防护林网等。山区、半山区耕地多为不规则的几何形状,一般都有农村道路连接。色调随土壤、湿度、农作物种类、及生长季节不同而变化。通过这些标志一般解译地类为()A. 草地B. 耕地C. 林地D. 园地