In other words, if you stacked one marshmallow on top of another and stacked them up how many would it take to stack a line all the way to the moon?
211435
Approximately 38 billon marshmallows\n\nThe average distance between Earth and Moon is approximately 30 times Earth's diameter.\n\nEarth Mean diameter is 12,742 km \nMoon Mean diameter is 3,476 km\n\nWhen the Moon is at apogee, it is 11% farther from Earth than it is at perigee. \n\nPerigee: 363,300 km \nApogee:  405,500 km \n\nMean distance between Earth and Moon is therefore = 384,400 km = 384,400,000 meters\n\nIf the size of average Marshmallow is 1 cm = 0.01 meter\n\nNumber of Marshmallows between Earth and Moon =\n\n384,400,000 / 0.01 = 384,400,000 x 100 = 38,440,000,000 \n\n~= 38 billion
515937
Mass of moon\n-------------------- ------------------- =\nMass of earth's atmosphere\n\n73 800 000 000 000 000 000\n-------------------- ------------------- =\n5 700 000 000 000 000\n\n≈13000 (nearest thousand)
645627
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
493536
Let's do a thought experiment.Imagine that you had a giant compass and drew a circle about 12 feet in diameter on your school parking lot. You are going to be the earth so you stand in the center where the giant stick pin of the compass went. Pick a friend to be the moon. She is going to walk along the path of the  circle and you are going to face her at all times and observe carefully what you see. First have her look in one direction as she marches along the path - like maybe looking at the school building or maybe facing north. Make sure her entire body faces the same way at all times. In other words, she moves in a circle but does not rotate her body. Notice that you see her from the front at times, then the left side, back, right and front again. Now. Compare this to the moon. If the moon didn't rotate we would see all sides of it during the course of its almost one month journey around the earth. In other words, if the moon did NOT rotate we WOULD be able to see the back side.  Now, have your friend do the same as before but this time have her face you at all times so you cannot see her back - only her front . Now. Ask her what she saw in addition to you and she will say that part of the time she saw the school house and part of the time she was even facing away from it and finally ended up facing the school again. She faced first north, then east, then south, then west and finally north again. In other words, she turned in a complete circle while her face was turned toward you at all times. This is what the moon does. One face of the moon is turned toward the earth at all times. In order for that to be true it must rotate, just like your friend. The secret is that it rotates on its axis at the exact same speed as it revolves in its path about the earth. The reason for this behavior is that the moon is captured by the earth's gravity.
8940
Provided you mix the same amount of each liquid.
645938
in order to solve this problem, you need to work part (b) first before (a)\n\nb): you know gravity (g=32.2 ft/sec) acts opposite the rocket, so calculate, using:\n\nv_final = v_initial - 2*g*y_1\n\nyou know that the final velocity is zero (max height) and you are given the inital velocity. the negative sign is to account for the opposite direction of gravity. move the equation around to solve for y_1 (height from tower to max. height). then add 100 feet (to account for the tower) and you have your answer.\n\na): since you know your maximum height (from part b), substitute that answer for h. [*NOTE: remember to subtract 100 feet from your answer in part b, because t=0 at the top of the tower] You now have a quadratic equation and solve for t using the quadratic formula (you should know what that is). your answer will be the answer!\n\nThis prob is easy only because it worries upto the max height. if we had to worry about the falling down part, well thats another story.\n\n** i already worked out this prob, so if you want to ask more questions/verify your answer, just post an additional response and ill modify my question to answer it. GL!
429077
ok 1.plz spess equation right\nnxt\n2. 6284000 X .01= 62840\n\n3. it would end up being well over 10 mil.
584081
You could do a base and acid mix and explain the reaction, such as vinegar and baking soda.\n\nAlso the erupting marshmallow mixture is pretty cool... and edible.\n\nCheck out this link for a bunch of cool activities you could do.\nhttp://www.calstatela.edu/dept/chem/chem2/LACTE/K12.html
417501
i guess...but the moon has to be off good size..and the "miimoons" has to be really small...smaller than the moon...
473855
let the amount of pizzas equal x\nron makes x/30 pizzas per minute and jess makes x/40 pizzas per minute \ntogether they make:    x/40+x/30 pizzas per minute\n                                   =7x/120 pizzas per minute\ntherefore to make x pizzas they will take 17.14 (=120/7) minutes\nThey make 7 times the amount needed in 120 minutes therefore we must divide the 120 by 7.
481594
What type of flow charts are you referring to?  They d not react with each other. They do not dissole in each other.  With some flour and perhaps some baking soda you could make a cake.
368161
So, you would set up two equations and set them equal to each other...first you would have to find the amount of time it would take the mass to fall on earth...then set the equation from earth with the found time against the equation from planet Zog with its gravity as the only variable.\n\nI'll figure it out and give you more of an answer.\n\nalright...ready?\n\nI was stupid...just use this equation:\n\nY = Yo + Vo t + .5 a t^2\n\nwhere Y is the final height (zero), Yo is the initial height, Vo is the initial velocity (zero), a is the acceleration of gravity on the planet Zog, and t is the time it takes for the mass to hit the ground.
139272
24 men have orbited the moon, 12 have walked on its surface\n\n3 men orbited:\nApollo 8\nApollo 10\nApollo 13\n\n3 men orbited, 2 of whom landed:\nApollo 11\nApollo 12\nApollo 14\nApollo 15\nApollo 16\nApollo 17\n\nThe above math shows 3*9=27 men, but 3 got to orbit twice (Jim Lovell, Gene Cernan, and John Young)
91120
It depends on the object.  You find the area of all the sides and then add them together.
432046
Assuming you somehow manage to get the ladder pass the centre and out the other side; could you throw a rope ladder 6,378km towards space?\n\nThen the next hurdle; have you ever tried to walk feet-first up a ladder? Because that's what you'll be doing once you've got past the centre of the Earth.
219490
It is theoretically possible, but very, very, very difficult!\nL2 is roughly 1,500,00 km (or a million miles) from the earth, on the far side from the sun.  So when the moon is on the sun side of the earth, with it's dark face toward the earth (a new moon), the distance from the L2 point to the moon would be 1.5M km (to earth), plus 375,000 km  (to the moon), then bounce back to earth, for a total of 2.25 million km (or 1.5M miles).  Since the moon only reflects about 1/10 of the light that hits it (its albedo is .12), the light source at L2 would have to be 10 times as powerful as if you were reflecting off a mirror (or 6-8 times what it would be off a good screen).\n So let's figure how powerful that light would have to be.  Let's say that you want to just barely be able to see this projection, say, about as well as you could see a candle on a perfectly dark night at a distance of a kilometer. Well, light power varies inversely as the square of the distance.  So (simplifying a lot to just get a rough idea) that light would have to be 2.25M, squared, times 10 (for the moon's reflectivity) candlepower to hit your eye with that same power.  So you'd need a light of 50,625,000,000,000,000,000 candlepower.  By one account, the most powerful light currently existing on earth is about 1,000,000,000 (one billion) candlepower.  \n I've simplified quite a bit; you could reduce the power requirements many orders of magnitude by focusing the beam or using a laser, but the answer still comes out to be that you'd need a light millions of times more powerful than the brightest lighthouse.  And you'll need thousands of square kilometers of solar cells to power this super projector.  And thousands of rockets to get all this equipment out to L2!\n\n So to answer your question - is it THEORETICALLY possible?  Sure!  But there is a huge world of difference between theory and practical application on this one!
534096
Since the Moon's gravity is only one sixth that of the Earth, much less fuel is needed to launch from the Moon than from the Earth, assuming the mass launched from each is the same.\nOf course the Moon would not be visible.  It does not make its own light, because its just a big rocky ball.