Since velocity is a speed and a direction, there are only two ways for you to accelerate: change your speed or change your direction—or change both. If you're not changing your speed and you're not changing your direction, then you simply cannot be accelerating—no matter how fast you're going.
Recall the definition of the cross product with respect to the unit vectors:
i × i = j × j = k × k = 0
i × j = k
j × k = i
k × i = j
and that the product is anticommutative, so that for any two vectors u and v, we have u × v = - (v × u). (This essentially takes care of part (b).)
Now, given a = 8i + j - 2k and b = 5i - 3j + k, we have
a × b = (8i + j - 2k) × (5i - 3j + k)
a × b = 40 (i × i) + 5 (j × i) - 10 (k × i)
… … … … - 24 (i × j) - 3 (j × j) + 6 (k × j)
… … … … + 8 (i × k) + (j × k) - 2 (k × k)
a × b = - 5 (i × j) - 10 (k × i) - 24 (i × j) - 6 (j × k) - 8 (k × i) + (j × k)
a × b = - 5k - 10j - 24k - 6i - 8j + i
a × b = -5i - 18j - 29k
Answer:
21 m/s.
Explanation:
The computation of the wind velocity is shown below:
But before that, we need to find out the angles between the vectors
53° - 35° = 18°
Now we have to sqaure it i.e given below
v^2 = 55^2 + 40^2 - 2 · 55 · 40 · cos 18°
v^2 = 3025 + 1600 - 2 · 55 · 40 · 0.951
v^2 = 440.6
v = √440.6
v = 20.99
≈ 21 m/s
Hence, The wind velocity is 21 m/s.
I belive it could be 6.5 but I could be wrong
Answer:
What is the correct path of sperm cells through the male reproductive system?
Epididymis, seminiferous tubules, urethra, vas deferens
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Urethra, seminiferous tubules, epididymis, vas deferens
Seminiferous tubules, vas deferens, epididymis, urethra
Hope this helps :)
Have a great day !
5INGH
Explanation:
