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8. What a table ! I’ve never seen such a thing before .It is it is long.

 A. half not as wide as               B. wide not as half as

 C. not half as wide as               D. as wide as not half

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7. The children loved their day trip, and they enjoyed the horse ride ___

A. most            B. more           C. less              D. little

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6. To all the famous artists’ surprise, the unknown woman’s two  paintings are also on show in the art exhibition.

A. little blue oil      B. blue little oil     C. oil blue little      D. little oil blue

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5. -- What do you think of chemistry?

--In my opinion, chemistry is ______ physics

A. subject so difficult as                B. as difficult a subject as

C. as a difficult subject as             D. difficult as a subject as

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4. Anybody in our class is _________ of working out the puzzle in ten minutes

A. likely            B. probable       C. capable         D. unable

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3. I know this is not quite the right word, but I can’t be bothered to think of _____

A. a better          B. a best          C. the better        D. the good

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2. He’s feeling rather _______ with himself after the examination.

A. pleasing         B. pleased        C. pleasant         D. pleasurable

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1. -- Do you think we need so much money?

-- Yes, you know we’ll have to buy ________ equipment

A. much            B. many           C. less              D. vast

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12.

图2-2-27

如图2-2-27所示,AB两物体叠放在水平地面上,已知AB的质量分别均为mA=10 kg,mB=20 kg,AB之间,B与地面之间的动摩擦因数均为μ=0.5.一轻绳一端系住物体A,另一端系于墙上,绳与竖直方向的夹角为37°,今欲用外力将物体B匀速向右拉出,求所加水平力F的大小,并画出AB的受力分析图.(取g=10 m/s2,sin 37°=0.6,cos 37°=0.8)

解析:AB的受力分析如右图所示

A应用平衡条件

Tsin 37°=Ff1μFN1

Tcos 37°+FN1mAg

联立①、②两式可得:FN1==60 N

Ff1μFN1=30 N

B用平衡条件

FFf1′+Ff2Ff1′+μFN2Ff1+μ(FN1+mBg)=2Ff1+μmBg=160 N

答案:160 N 图见解析

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11.

图2-2-26

如图2-2-26所示,在倾角为37°的固定斜面上静置一个质量为5 kg的物体,物体与斜面间的动摩擦因数为0.8.求:

(1)物体所受的摩擦力;(sin 37°=0.6,cos 37°=0.8)

(2)若用原长为10 cm,劲度系数为3.1×103 N/m的弹簧沿斜面向上拉物体,使之向上匀速运动,则弹簧的最终长度是多少?(取g=10 m/s2)

解析:

图a

(1)物体静止在斜面上受力分析如图a所示,则物体受到的静摩擦力Ffmgsin 37°

代入数据得Ff=5×10×sin 37° N=30 N,摩擦力方向为沿斜面向上

(2)当物体沿斜面向上被匀速拉动时,如图b所示,弹簧拉力设为F,伸长量为x,则Fkx

图b

Fmgsin 37°+F

Fμmgcos 37°

弹簧最终长度ll0+x,由以上方程解得l=12 cm.

答案:(1)30 N,方向沿斜面向上

(2)12 cm

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