Well, I'm assuming that everything aside form watermelon is extraneous information. Knowing that the Sun to Earth volume ratio is about 1,300,000 : 1 and the average mass of a watermelon is 5 KG while the Earth' density is 5.52 g/cm3
The density of an average watermelon can be obtained using density formula( D = . The average diameter is about 25 cm, radius 12.5 cm. Volume of a sphere (assuming these are theoretical watermelons) is 4/3pir^3. (4/3)(pi)(12.5^3) = 8181.23 cm^3
using this, we can obtain that 5,000/ 8181.23 cm^3 = .6112 g/cm^3
.6112g/1cm^3 = 5.52g/cm^3
we can conclude that the volume of the Earth is approximately 1.08 * 10^12km^3. Now, using Earth volume, we can conclude that the Earth to Sun ratio is 1 : 333,000 (Volume ratio divided by the density)
5.9736 * 10^24 Kg * 333, 000 (our ratio), we can see that the sun is
....1.9892 * 10 ^ 30 Kg
Holy ****. I came pretty damn close. Although, I did need to know ahead of time information that they hadn't given you that I researched
(1 : 1,300,000 and 5.52 on top of some ratio estimates that I rounded according to their true nature due to my somewhat flawed astronomy skills)
Now, we must consider that I basically already took known facts such as just about eveyr single ratio in there and some facts about watermelons. THe only reason my answer came so close was because half of the things in there were already accepted givens. I just made sure my answers looked like them.
**anonymous rolled a random image posted in comment #657 at spongebob ** Do you even know what OC means?
You didn't make that image. Therefore not OC. Therefore shut the **** up.
in notes (x-h)^2+(y-k)^2=r^2
homework- a water main broke at an intersection causing water to spill out at 5t feet with t being the number of seconds. a runner is running 6 miles away towards the interseciton at a rate of 17 feet per second,(a) at what time does the runners feet get wet (b) say the runner is 6 miles east and 5000 feet north he is running west what time will the water get his feet wet.