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匿名使用者 發問時間: 科學其他:科學 · 1 0 年前

有關物理向量的計算問題

1. A golfer putting on a green , requires three strokes to hole the ball

During the first putt , the ball rolls 5 m due east

for the second putt , the travels 2.1m at an angle of 20度 north of east

The third putt is 0.5m due north

what displacement (magnitude and direction relative to due east) would has been needed to " hole the ball " on the very first putt ?

2. what is the value of each of the angles of a triangle whose sides are 95 , 150 , and 190 cm in lenght ?

( hint: consider using the law of cosines given in appendix E.)

3. The speed of an object and the direction in which it moves constitute a vector quantity known as the velocity

An ostrich is running at a speed of 17m/s in a direction of 68度 north of west

what is the magnitude of the ostrich's velocity component that is directed (a) due north (b) due west

如果方便的話麻煩予以解釋計算過程

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1 個解答

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    Lv 7
    1 0 年前
    最佳解答

    1. 某人在果嶺推桿三次進洞,第一次向東5m, 第二次20度偏北2.1m, 第三次向北0.5m, 如果要推桿一次就進洞, 則距離與方向為何?

    此題解答為三個向量和. 用law of cosines計算, 第一和第二個向量 的和, r=(52+2.12-2*5*2.1*cos(180-20))0.5=(25+4.41-(-19.7))0.5=7, 7/sin(160)=2.1/sin(q), q=5.88度, 加第三個向量 r=(72+0.52-2*7*0.5*cos(180-5.88))0.5=7.08m, 7.08/sin(180-5.88)=0.5/sin(q), q=4.03度, 故欲推一桿就進洞, 需向東偏北 5.88+4.03=9.91度, 推7.08公尺.

    2. 三邊長為 95, 150, 190 cm的三角形各角大小為何?

    用餘弦定律: 95邊對角: 952=1502+1902-2*150*190*cos(q), q=29.6o, 150邊對角: 1502=952+1902-2*190*95*cos(q), q=51.2o, 190邊對角: 1902=1502+952-2*150*95*cos(q), q=99.2o.

    3. 鴕鳥以17m/s的速率向西偏北68度奔跑, 問它(a)向北, (b)向西的速率分量.

    (a) 17*sin(68)=15.8 m/s, (b) 17*cos(68)=6.4 m/s

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