<span>If Jack is filing married-filing-separate he would report $76,000 gross income as head of household.</span>
Answer:
12.5752053801 m/s
No.
Explanation:
Q = Volume flow rate = 
d = Diameter of pipe = 9 cm
A = Area = 
Volume flow rate is given by
Velocity of fluid is 12.5752053801 m/s
The volume flow rate in m³/s is
The flow of fluid does not depend on the type of water used. Hence the answers would be same. If Q is constant v will be the same irrespective of the type of water used.
You would see mountains off in the distance as if the earth was actually flat. but it seems flat because its so big
In the writing of ionic chemical formulas the value of each ion's charge is crossed over in the crossover rule.
Rules for naming Ionic compounds
- Frist Rule
The cation (element with a negative charge) is written first in the name then the anion(element with a positive charge) is written second in the name.
- Second rule
When the formula unit contains two or more of the same polyatomic ion, that ion is written in parentheses with the subscript written outside the parentheses.
Example: Sodium carbonate is written as Na₂CO₃ not Na₂(CO)₃
- Third rule
If the cation is a metal ion with a fixed charge then the name of the cation will remain the same as the (neutral) element from which it is derived (Example: Na+ will be sodium).
If the cation is a metal ion with a variable charge, the charge on the cation is indicated using a Roman numeral, in parentheses, immediately following the name of the cation (example: Fe³⁺ = iron(III)).
- Fourth rule
If the anion is a monatomic ion, the anion is named by adding the suffix <em>-ide</em> to the root of the element name (example: F = Fluoride).
The oxidation state of each ion is also important, thus in the crossover rule, the value of each ion's charge is crossed over.
Learn more about chemical formulas here:
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<span>In this problem, we need to solve for Bubba’s mass. To do this, we let A be the area of the raft and set the weight of the displaced fluid with the raft alone as ρwAd1g and ρwAd2g with the person on the raft, </span>where ρw is the density of water, d1 = 7cm, and d2= 8.4 cm. Set the weight of displaced fluid equal to the weight of the floating objects to eliminate A and ρw then solve for m.
<span>ρwAd1g = Mg</span>
ρw<span>Ad2g = (M + m) g</span>
<span>d2∕d1 = (M + m)/g</span>
m = [(d2<span>∕d1)-1] M = [(8.4 cm/7.0 cm) - 1] (600 kg) =120 kg</span>
This means that Bubba’s mass is 120 kg.
