Answer:
<em><u>The final atmospheric pressure is 5.19 · 10⁴ Pa</u></em>
Step-by-step explanation:
Assuming that the temperature of the air does not change, we can use Boyle's law, which states that for a gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume. In formula,
pV = const.
where p is the gas pressure and V is the volume
The equation can also be rewritten as
p₁ V₁ = p₂ V₂
where in our problem we have:
p₁ = 1.03 · 10₅ Pa is the initial pressure (the atmospheric pressure at sea level)
V₁ = 90.0L is the initial volume
p₂ is the final pressure
V₂ = 175.0L is the final volume
Solving the equation for p2, we find the final pressure:
p₂ = p₁ v₁ divided by V₂ = (1.01 · 10⁵)(90.0) divided by 175.0 = 5.19 · 10⁴ Pa
Answer:
Over 2 oz. Class A-II felony
Step-by-step explanation:
One would need to carefully weigh the material.*
0.06 kg ≈ 2.12 oz
Note that .06 kg is 1 significant figure, so this rounds to 2 oz (1 significant figure). Given the precision of the reported weight, there is insufficient precision to say the amount is actually over 2 oz.
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If you take the numbers at face value, the suspect is in possession of over 2 oz, so will be charged with a Class A-II felony.
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* 2.00 ounces translates to about 0.0567 kg, which rounds to 0.06 kg.
The Pythagoras theorem states that
the sum of squares of the shorter sides (legs) of a right triangle equals the square of the third side.
A corollary from the same theorem helps us solve this problem:
If the sum of the squares of the shorter sides of a triangle is greater than the square of the third side, the included angle is acute. ..... (case 1)
Conversely, if the sum of the squares of the shorter sides of a triangle is less than the square of the third side, the triangle is obtuse. .....(case 2)
Here we have
6^2+10^2 = 36+100=136 <12^2=144
Therefore case 2 applies, and the triangle is obtuse.
x=6x-2 This would be your answer, hope this helps if you need any further explanation feel free to let me know and I would be happy to help you out.