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Anna Mae O.

Licensed Chemical Engineer

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Trigonometry

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Question:

The time in an analog clock reads 2:27. The minute hand is 8 inches long and the hour hand is 6.3 inches long. If the hour hand is positioned an inch above the minute hand, what is the shortest distance between the tips of the hands of the clock?

Anna Mae O.

Answer:

First, let's assume that the minute hand and the hour hand lies on the same plane: The minute hand moves at a rate of $$360^{o} per$$ $$ 60$$ $$ minutes$$ or $$6^{o}/min$$ The hour hand moves at a rate of $$360^{o} per$$ $$ 12$$ $$ hours$$ or $$0.5^{o}/min$$ The angle formed by the two hands is: $$ \theta = (6^{o}/min)(27 min)$$ - $$(0.5^{o}/min)(2x60+27)min$$ $$\theta=88.5^{o}$$ The apparent distance(x) between the tips of the two hands can be solved using cosine law: $$x^{2}=h^{2} +m^ {2}-2 h m cos(\theta)$$; where h is the length of the hour hand and m is the length of the minute hand $$x^{2}=6.3^{2} +8^ {2}-2 h m cos(88.5)$$ $$x=10.05 in $$ Now let's consider the fact the hands are not really coplanar. To visualize, consider a right triangle having the legs equal to x (the apparent distance) and y = 1 in (the vertical distance between the hands) in and hypotenuse z (the true distance between the tips of the hands). The true distance (z) can be calculated using pythagorean theorem: $$z^2=x^2+y^2$$ $$z^{2}= (10.05 in)^{2}+ (1in)^{2}$$ $$z=10.10 in$$ Therefore, the distance between the tips of the two hands is 10.10 inches.

Chemistry

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Question:

Does a precipitate form when 0.10 L of 8.0x10 ^-3 M Pb(NO3)2 is added to 0.40 L of 5.0x10^-3 M Na2SO4? Ksp of PbSO4 is 6.4x10^-7

Anna Mae O.

Answer:

Pb(NO3)2 + Na2SO4 ------> PbSO4(s) + NaNO3(aq) PbSO4 will precipitate if Q=[Pb2+][SO4-] is greater than Ksp=6.4x10^{-7}. $$[Pb^{2+}] = \frac{(0.1 L)(8.0x10^{-3}\frac {mol PbSO4}{L}) (\frac {mol Pb^{2+}}{mol PbSO4})}{0.1 L + 0.40 L}$$ $$[Pb^{2+}]= 1.60 x 10^{-3} M$$ $$[SO_{4}^{2-}] = \frac{(0.4 L)(5.0x10^{-3}\frac {mol Na_{2}SO4}{L}) (\frac {mol SO_{4}^{2-}}{mol Na_{2}SO4})}{0.1 L + 0.40 L}$$ $$[SO_{4}^{2-}]=4.0x10^{-3}M$$ $$Q=[Pb^{2+}][SO_{4}^{2-}]$$ $$Q=(1.60 x 10^{-3} M)(4.0x10^{-3}M)$$ $$Q=6.4x10^{-6} > K_{sp}$$ Since Q> Ksp then PbSO4 will precipitate.

Algebra

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Question:

A merchant has three items on sale, namely a radio for P50, a clock for P30 and a flashlight for P1. At the end of the day, he sold a total of 100 of the three items and has taken exactly P1000 on the total sales. If the number of radios sold is 4 times that of the clocks, how many radios, clocks, and flashlights had the merchant sold?

Anna Mae O.

Answer:

Let x = the number of radios sold 50x = total sales from sold radios y = the number of clocks sold 30y = total sales from sold clocks z= the number of flashlights sold z = total sales from sold flashlights Radios sold is 4 times that of the clocks: x=4y -----> (equantion 1) A total of 100 of theitems were sold: x + y + z = 100 -----> (equantion 2) The total sales is worth P1000, hence: 50x + 30y + z = 1000 -----> (equantion 3) Substitute Equation 1 to 2: 4y + y +z =100 5y + z= 100 z= 100- 5y -----> equation 4 Substitute Equation 1 to 3: 50(4y) +30y + z = 1000 200y + 30y +z = 1000 230y + z = 1000 Substitute equation 4: 230y + (100-5y) =1000 225y = 900 y = 4 From equation 1: x = 4y = 4(4) =16 From equation 4: z = 100 - 5y =100 - 5 (4) = 80 Therefore the merchant was able to sell 16 radios, 4 clocks and 80 flashlights.

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