You need to find the mass of water in the pool.
Find the volume (10 x 4 x 3) = 120 m3
Water has a density of 1000g/m3,so 120 m3 = 120 x 1000 = 120 000 kg
[delta]H = 4.187 x 120 000 x 3.4 (and the units will be kJ)
You then use the heat of combustion knowing that each mole of methane
releases 891 kJ of heat so if you divide 891 into the previous answer,
you will get the number of moles of CH4
The trickiest part of this problem was making sure where the Yakima Valley is.
OK so it's generally around the city of the same name in Washington State.
Just for a place to work with, I picked the Yakima Valley Junior College, at the
corner of W Nob Hill Blvd and S16th Ave in Yakima. The latitude in the middle
of that intersection is 46.585° North. <u>That's</u> the number we need.
Here's how I would do it:
-- The altitude of the due-south point on the celestial equator is always
(90° - latitude), no matter what the date or time of day.
-- The highest above the celestial equator that the ecliptic ever gets
is about 23.5°.
-- The mean inclination of the moon's orbit to the ecliptic is 5.14°, so
that's the highest above the ecliptic that the moon can ever appear
in the sky.
This sets the limit of the highest in the sky that the moon can ever appear.
90° - 46.585° + 23.5° + 5.14° = 72.1° above the horizon .
That doesn't happen regularly. It would depend on everything coming
together at the same time ... the moon happens to be at the point in its
orbit that's 5.14° above ==> (the point on the ecliptic that's 23.5° above
the celestial equator).
Depending on the time of year, that can be any time of the day or night.
The most striking combination is at midnight, within a day or two of the
Winter solstice, when the moon happens to be full.
In general, the Full Moon closest to the Winter solstice is going to be
the moon highest in the sky. Then it's going to be somewhere near
67° above the horizon at midnight.
Just do what u would do if u were at a stop sign
Answer:
Work done = 35467.278 J
Explanation:
Given:
Height of the cone = 4m
radius (r) of the cone = 1.2m
Density of the cone = 600kg/m³
Acceleration due to gravity, g = 9.8 m/s²
Now,
The total mass of the cone (m) = Density of the cone × volume of the cone
Volume of the cone = 
thus,
volume of the cone = 
= 6.03 m³
therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg
The center of mass for the cone lies at the 
times the total height
thus,
center of mass lies at, h' = 
Now, the work gone (W) against gravity is given as:
W = mgh'
W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J
Answer:
i don't know if this is good for you but
Explanation:
ignoring frictional air resistance (drag) the speed on return is the same as when it left the ground (5 m/s but in the opposite direction).
Note: this points out a good reason for not firing live bullets into the air..they will return somewhere and at the same speed.
However, if you take into account the atmospheric drag the reurn speed will be somewhat smaller (but in the case of a bullet, probably still lethal.) Drag depends on many factors and is difficult to calculate.
