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10027945
11.09.2019 •
Mathematics
Vector c has a magnitude of 22.2 m and points in the −y‑ direction. vectors a and b both have positive y‑ components, and make angles of α=41.9° and β=28.2° with the positive and negative x- axis, respectively. if the vector sum a+b+c=0 , what are the magnitudes of a and b?
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Ответ:
The magnitude of A is 17.46 m and B is 1.50 m
Step-by-step explanation:
If the vector sum A+B+C =0, then the sum of the projection of the vector in axes x- is zero and the sum of the projection of the vector in the axes y- is also zero.
Ax+Bx+Cx = 0
Ay+By+Cy = 0
|Ax| = cos 41.9 * |A|
|Ay| = sin 41.9 * |A|
|Bx| = cos 28.2 * |B|
|By| = sin 28.2 * |B|
|Cx| = 0
|Cy| = 22.2
Ax+Bx+Cx = 0
|Ax|-|Bx|+0 =0
the vector Ax is in the positive direction of the x- axes and Bx in the negative direction and C do not have a component in the x- axes
cos 41.9 * |A| - cos 28.2 * |B| = 0 (I)
Ay+By+Cy = 0
|Ay|+|By|-|Cy|=0
the vector Ay and By are the positive direction of the y- axes and Cy in the negative direction
sin 41.9 * |A| + sin 28.2 * |B| - 22 =0 (II)
Now we have a system of 2 (I and II) equations and 2 variables (|A| and |B|)
cos 41.9 * |A| - cos 28.2 * |B| = 0
sin 41.9 * |A| + sin 28.2 * |B| = 22
cos 41.9 * |A| = cos 28.2 * |B|
|A| = cos 28.2 * |B| / cos 41.9
sin 41.9 * |A| + sin 28.2 * |B| = 22
sin 41.9 * cos 28.2 * |B| / cos 41.9 + sin 28.2 * |B| = 22
tg 41.9 * cos 28.2 * |B| + sin 28.2 * |B| = 22
(tg 41.9 * cos 28.2 + sin 28.2) * |B| = 22
|B| = 22 / (tg 41.9 * cos 28.2 + sin 28.2)
|B| = 17.46
|A| = 1.50
Ответ:
32,058,508 bacteria at 10:42am
Step-by-step explanation:
10am: 8,014,627
10:21: 8,014,627 × 2 (doubled) = 16,029,254
10:42: 16,029,254 × 2 (doubled) = 32,058,508