how do you solve gravity using the 6.67 and a bunch of m's?
925552
What is " a bunch of m's"?\nThe 6.67 refers to the Gravitational constant: 6.67x10-8 dyne cm2 g-2.  \nAccording to the law of universal gravitation, the attractive force between two bodies is proportional to the product of their masses and inversely proportional to the square of the distance between them.\nThe formula is F=G (m1m2)/r2, where F is the force, G is the 6.67 business above, m1 and m2 are their masses (in grams) and r2 is the distance between them squared (in cms).
1093836
Centrifugal force and gravity.
921703
Just let the units tell you what to do.  H=mass* T*sp heat.  with T in K
34572
alright, well just convert the mN into N and divide by gravity... \nso we multiply be the conversion factor getting .00000671N. now F=ma=mg so m=F/g(mass equals Force divided by gravity)  so .00000671/9.8= 6.85 x 10^-7 kg
929849
solve three body problems
70889
convert to Newtons from mN, then divide by gravity because F=mg. so you get .00000780/9.8
967250
The gravitotional force F of two masses m1 and m2 separated by a distance r is given by the equation\n\nF = G * m1 * m2 / r^2\n\nwhere G is a physical constant with value 6.67*10^ -11 N*m^2/kg^2\n\nSInce the third object is placed midway then its distance from either m1 or m2 is 0.15 m.\n\nSubstituting we get\nF1=3.9*10^-5 N\nand\nF2=9,5*10^-5 N\n\nSince the two forces have opposite directions then \nFtotal = F1 - F2 = -5.6*10^-5 N having the direction of F2.\n\nIn order for total to be zero we have\nF1=F2 ->\nG*m1*m/x^2=G*m2*m/y^2 ->\nm1/x^2 = m2/y^2 ->\nm1*y^2=m2*x^2 (1)\n\nwhere x and y the diastances of m from m1 and m2 respectively.\n\nsince\n\nx+y=r=0.300 m\n\nit follows that y = 0.3-x (2)\nCombining (1) and (2) we have\n\nm1*(0.3-x)^2=m2*x^2 ->\n215*(0.09-0.6*x+x^2)=515x^2 ->\n300x^2+129x-19,35=0\n\nsolving this we get\n\nx1=0.12m\nand\nx2=-0.55 m which is of course rejected since it represents a point where there can be no equilibrium.
969486
I just helped someone else with their physics question, so I'm alittle tired of it.  But, I suggest you make an equation sheet...physics is all about equations.  So, if you have a piece of paper with your equations, all you have to do is look at the sheet and plug things in...
948476
I won't show in details since it's hard to type out the work, but you'll get the rest, here it goes:\n\na)\nWork = mgd  *d=change in height\nW = (10kg) (9.8m/ss) (sin15) (-7.5m) = -190J\n\nb)\nW=(friction coef)(natural force)(distance traveled)\nnatural force = (cos15)(10kg)(9.8m/ss)\nW= (0.4) (cos15) (10kg) (9.8m/ss) (-7.5m) = -280J\n\nc) \nW=Force x distance\nW=(100N)(7.5m)=750J\n\nd)\n(initial KE)+ (puller work)=(gravity work) + (frictional work) + (final KE)\n(final KE) - (initial KE) = 750J - 280J - 190J = 280J\n\ne)\n(final KE) - (initial KE) = 280J\n*solve for final v\nv = squareroot (2 x [280j+ (1/2)(10kg)(1.5squared)] / 10kg) =7.63m/s
750320
If it takes 6 secs to reach the max, it will take an additional 6 secs to reach your starting point.  In this case it is 2560.  Then it is a matter of finding the time from that point to going to the ground.\n1st lets correct the orginal equations (acceleration due to gravity is down/negative)\nh = -16t^2 + 192t + 2560\n2nd - set the equation to zero since we want to find the time to ground.\n0 = -16t^2 + 192t + 2560\n16t^2 - 192t = 2560\nt^2 - 12t = 160\n(t^2 - 12t + 36) - 36 = 160\n(t - 6)^2 = 196\nt - 6 = 14\nt = 20\nConclusion: 6 seconds to reach max ht of 3136, another 6 seconds to reach starting point of 2560 and another 8 seconds to hit the ground.
950533
You must be talking about the planet called Venus. \n\nto calculate the acceleration due to gravity, you have to use the following formula:\n\ng = GM/(r^2)\n\nwhere G is the universal constrant of gravitation = 6.673x10^-11\n\nyou know that:\n the mass (M) is 4.88 x 10^24 kg\n and the radius (r) is 6.06 x 10^6 m\n\nnow substitute into the equation. \n\ng = [(6.673x10^-11)(4.88 x 10^24)] / [(6.06 x 10^6)^2]\ng = (3.256x10^14) / (3.672x10^13)\ng = 8.867m/s/s\ng = 8.87m/s/s (thats metres per second squared)
700115
Very simple, in your case you only need to convert from N(Newton) to Kg(kilogram) so just divide by 9.8 (gravity constant)\n\nN/m2---- /9.8 ---> Kg/m2\nN/m2<-- X9.8 ---- Kg/m2
295799
k ill try to help u  even though i dont condone cheating. i just know how it is to not understand something and its due in a few hours and it sucks. just remember, nobody will be able to help u on the exam.\n\nso 1. \na. acceleration is change in velocity divided by time. \nsince it started from rest, initial velocity is 0 and then final velocity is given as 25.3 m/s. and the time is given as 0.59s. so to get acceleration   you would do  \na =  Vf-Vi / t\n  =  25.3 - 0 / 0.59\n  = 42.88......\n\nb. the force is equal to mass x acceleration. since you know the mass and acceleration, you just multiply them and you get F = 38421.69... N. \n\nc. you have to find the normal force in this case, which is the force that opposes gravity. like upward force. so in this case, gravity is acting on the driver and force of gravity is equal to mass x 9.8 = 666.4 N. and that means the normal force is 666.4N (since he is not moving up or down the total vertical force = 0)\n\n2. ok well in this case, you know the initial velocity is 0, the distance is 404m and the time is 4.944s. if its constant acceleration, you could assume that the average velocity is just Vi+Vf/2. so if thatz average velocity, then distance = average velocity x time\n     404 = 4.944s  x  (0+Vf)/2\n 404 x 2 / 4.944 = Vf\nVf = 163.43........m/s\n\nthe acceleration would just be Vf-Vi/t. and since you know Vf now  you just do   a = 163.43-0/4.944\na = 33.056....m/s^2\n\n3. ok well acceleration is 9.8m/s^2 (gravity)     distance is 9.7m. initial velocity is 0m/s. so this time, you have to solve 2 equations. the first one is the acceleration equation. a = Vf-Vi / t = Vf / t = 9.8\n9.8 = Vf/t\nVf = 9.8t\nand the other one is the distance = t  x  Vi+Vf/2   average velocity one. \n9.7 = t x Vf / 2\n19.4 = Vf x t\n19.4 = (9.8t) x t\n19.4 = 9.8t^2\nt^2 = 1.979...\nt = 1.406....s\n\nnow that you know the time, you find the Vf by putting it in any of those 2 equations. Vf = 9.8t = 9.8 (1.406)\nVf = 13.788... m/s\n\npart b. so in this case, you use the initial velocity as the one u just found (the velocity of the swimmer at the surface of the water)   so Vi = 13.788m/s\nand Vf = 0    since they come to a stop. and d = 2m\nso you have to find the acceleration. first you need to find the time by using the average velocity equation.\nd = t x Vi+Vf/2\n2m = t x (0+13.788)/2\n4 = 13.788t\nt = 0.29s\nnow that you know the time, you have to find acceleration\na = Vf-Vi/t\n  = 0-13.788 / 0.29\na = -47.5286m/s^2    its negative because its downwards and up is positive in our case. so the force = mass x acceleration and you just got acceleration and her mass is 65kg so you just multiply by 65kg to get F = 3089.36N\n\n\n4. well F = ma and you have the force. when they say it weighs this much newtons, then it means  use gravitational acceleration which is 9.8. so 2350N = m x 9.8m/s^2\nm = 2350/9.8\nm = 239.79kg\n\ni hope you actually understand this stuff. try to do your homework by yourself   dont take advantage of this site. youll just end up screwing yourself over. and if u dont get anythin, let me know and ill try to help
728641
acceleration can be found using\na=(m1-m2)g/(m1+m2) = 3*9.8/5 = 5.88 m/s^2
787381
m = (7-4)/(6-8) = -3/2
939750
W=mg ... where g is the gravitational acceleration
210676
Use free body diagram to balance the forces and u can derive the equation of motion \nIf u want i can give u full derivation.\n"::