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Answer:
- F1.x ≈ -28.93
- F1.y ≈ 34.47
- F2.x = 70
- F2.y = 0
- (F1+F2).x ≈ 41.07
- (F1+F2).y ≈ 34.47
- |F1+F2| ≈ 53.62
- ∠(F1+F2) ≈ 40.0°
Explanation:
A suitable calculator can show you the vector components and their resultant in polar or rectangular format. (See attached.) 2D vectors are conveniently treated as complex numbers, which is why the y-component values are shown as imaginary.
(The 50° angle measured from the -x axis is equivalent to 130° measured from the +x axis, which is the reference we're using here.)
If you'd like to compute the vector components by hand, they are ...
(x, y) = magnitude×(cos(angle), sin(angle))
This notation is sometimes abbreviated <em>magnitude cis angle</em>, a reference to the complex number form x+yi.
Answer:
A) approximate alkalinity = 123.361 mg/l
B) exact alkalinity = 124.708 mg/l
Explanation:
Given data :
A) determine approximate alkalinity first
Bicarbonate ion = 120 mg/l
carbonate ion = 15 mg/l
Approximate alkalinity = ( carbonate ion ) * 50/30 + ( bicarbonate ion ) * 50/61
= 15 * (50/30) + 120*( 50/61 ) = 123.361 mg/l as CaCO3
B) calculate the exact alkalinity of the water if the pH = 9.43
pH + pOH = 14
9.43 + pOH = 14. therefore pOH = 14 - 9.43 = 4.57
[OH^- ] = 10^-4.57 = 2.692*10^-5 moles/l
[ OH^- ] = 2.692*10^-5 * 179/mole * 10^3 mg/g = 0.458 mg/l
[ H^+ ] = 10^-9.43 * 1 * 10^3 = 3.7154 * 10^-7 mg/l
therefore the exact alkalinity can be calculated as
= ( approximate alkalinity ) + ( [ OH^- ] * 50/17 ) - ( [ H^+ ] * 50/1 )
= 123.361 + ( 0.458 * 50/17 ) - ( 3.7154 * 10^-7 * 50/1 )
= 124.708 mg/l
I don’t know if I’m right but I’m guessing B