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ira [324]
3 years ago
15

Two vectors are presented as a=3.0i +5.0j and b=2.0i+4.0j find (a) a x b, ab (c) (a+b)b and (d) the component of a along the dir

ection of
b.
Physics
1 answer:
Svet_ta [14]3 years ago
3 0
Let's ask this question step by step:
 Part A) 
 a x b = (3.0i + 5.0j) x (2.0i + 4.0j) = (12-10) k = 2k
 ab = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
 Part (c)
 (a + b) b = [(3.0i + 5.0j) + (2.0i + 4.0j)]. (2.0i + 4.0j)
 (a + b) b = (5.0i + 9.0j). (2.0i + 4.0j)
 (a + b) b = 10 + 36
 (a + b) b = 46
 Part (d)
 comp (ba) = (a.b) / lbl
 a.b = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
 lbl = root ((2.0) ^ 2 + (4.0) ^ 2) = root (20)
 comp (ba) = 26 / root (20)
 answer
 2k
 26
 46
 26 / root (20)
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Explanation:

= 5.75 × 10-6

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(scientific e notation)

= 5.75 × 10-6

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Answer:

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Explanation:

For this problem to be easier to calculate, we can represent the triangle as a right triangle whose right angle is located at the origin of a coordinate system. (See picture attached).

With this disposition of the triangle, we can start finding our integral. The hydrostatic force can be set as an integral with the following shape:

\int\limits^a_bγhxdy

we know that γ=62.5 lb/ft^{3}

from the drawing, we can determine the height (or depth under the water) of each differential area is given by:

h=8-y

x can be found by getting the equation of the line, which we'll get by finding the slope of the line and using one of the points to complete the equation:

m=\frac{y_{2}-y_{1}}{x_{2}-x{1}}

when substituting the x and y-values given on the graph, we get that the slope is:

m=\frac{0-6}{7-0}=-\frac{6}{7}

once we got this slope, we can substitute it in the point-slope form of the equation:

y_{2}-y_{1}=m(x_{2}-x_{1})

which yields:

y-6=-\frac{6}{7}(x-0)

which simplifies to:

y-6=-\frac{6}{7}x

we can now solve this equation for x, so we get that:

x=-\frac{7}{6}y+7

with this last equation, we can substitute everything into our integral, so it will now look like this:

\int\limits^6_0{(62.5)(8-y)(-\frac{7}{6}y+7)}\,dy

Now that it's all written in terms of y we can now simplify it, so we get:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)}dy

we can now proceed and evaluate it.

When using the power rule on each of the terms, we get the integral to be:

62.5[\frac{7}{18}y^{3}-\frac{49}{6}y^{2}+56y]^{6}_{0}

By using the fundamental theorem of calculus we get:

62.5[(\frac{7}{18}(6)^{3}-\frac{49}{6}(6)^{2}+56(6))-(\frac{7}{18}(0)^{3}-\frac{49}{6}(0)^{2}+56(0))]

When solving we get:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)  } \, dy = 7875 lb

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