Susie small finds that she weighs 300 newtons. Calculate her mass?
831091
w = mg\n(w is weight, m is mass, g is acceleration due to gravity (9.8m/s^2))\n :. m = w/g\n        = 300 / 9.8\n        = 30.6 kg
783790
F=-kx and F=ma\n\nfind mass from period     T = 2pi(m/k)^(1/2)\n\nhmmmm,  you seem to be missing k\n\nfor fixed ratio m/k the period of 0.51s and amplitude 4 cm is not enough to determine the mass (and therefore acceleration).
609994
Wouldn't be nM would it? Newtonmeters ?\n\none footpound equals  1.3558179Nm
310603
What is weight but a number...she looks great!!
55693
the weight of your mom
621321
If the density is 'd' kg/m^3 then the weight is '9.81d' Newton. [remember weight is force]
948
33.3 lbs\n\nThe moon's gravity is 1/6 that of earth.
367853
F = m*a \n\nm = 2 kg (two masses; two strings)\na = gravitational constant which happens to be 9.81 m/s^2 on the surface of the earth (or 32.2 ft/s^2 in EE units)\n\nsum the forces in the y-direction:\n\nT = F = 2 * m * a = 2 N\n\n(assuming massless strings)
508636
same mass, same weight
119498
no.. u couldnt\n\nfor some reason i worked it out and its impossible.\n\nfirst u take 6.30g * 1,000,000=63000000g. then u take that number and multiply it by 0.00220462262lbs which is equal to 1 gram. ur answer comes out to be approx. 13889.13lbs. in other words, someone would have to weigh 13889.13lbs to be equal to the weight of 1 million $1 coins
769928
T = 2pisqrt(l/g)\nl = g(T/(2pi))^2 = 9.80(1.0/(2pi))^2 = 0.25 m\n\n(The object's weight is irrelevant for small oscillations.)
150852
On Earth, at surface level, the acceleration due to gravity is approximately 9.81 m/s^2. Weight = mass*acceleration, and so mass of flea = 5.097*10^-7 kg
997326
Use Newton's universal gravitation equation, then sum the forces.\n\nF = G*m1*m2/r^2\n\nF = force (newtons)\nG = gravitational constant (6.6742 x 10^-11)\nm1 = mass of object 1 (kg)\nm2 = mass of object 2 (kg)\nr = distance between objects (meters) ==> don't forget to convert the units\n\nIn this case, you do it twice:\nF1 = G * 390 * 390 / 0.11^2 = 0.84mN to the left\nF2 = G * 390 * 390 / 0.39^2 = 0.067mN to the right \n\nNet Force = 0.84 - 0.067 = 0.772mN to the left
955176
Force - Newtons\nMass - Kilograms\nAcceleration - metres per second
737967
Draw a free body diagram as shown below: \n\n                [m1]------> T<---[m2]------->89N\n              <--------            <---------\n                f1 =19.6N        f2=29.4N\n\n\nFirst solve for f1 and f2:\n\nf=ukR is the general formula where f is the force of friction, uk is the coefficient of friction, and R is the force normal to the surface. In this problem R is the weight of the block=mg. Now substitute known values\n\nf1=0.2*10*9.8\n   =19.6N to the left\nf2=0.1*30*9.8\n   =29.4N to the left\n\nLooking at the free body diagram:\n\nfor m1:\n\nnet force, f1=T-19.6\n                      \nFrom the formula, f=ma  where f is the net force, m the mass, and a is the acceleration.:\n\nf1=10a\n\nThus,\n\nT-19.6=10a\n         a=(T-19.6)/10\n\nfor m2:\n\nNet force, f2=89-29.4-T\n                  =59.6-T\n\nSimilarly as for m1:\n\n59.6-T=30a\na=(59.6-T)/30\n\nNow a for m1 is also the same a as for m2, therefore:\n\n(T-19.6)10=(59.6-T)/30\nThe zero on each side cancels out. \n3T-58.8=59.6-T\n4T=59.6+58.8\n4T=118.4\n  T=29.6N in the direction shown in the above free body diagram.
264256
CALCULATING THE MASS OF THE EARTH\n\n1. F = GmM/r2 = ma, where F is the gravitational force, G is the gravitational constant, M is the mass of the Earth, r is the radius of the Earth, and m is the mass of another object (near the surface of the Earth). \n\n2. GM/r2= a (The m's canceled out.) Now solve for M, the mass of the Earth. \n\n3. M = ar2/G, where a=9.8m/sec2, r=6.4 x 10 6m, and G=6.67 x 10-11m3/(kg sec2). \n\n4. M = 9.8 x (6.7 x 106)2/6.7 x 10-11 = 6.0 x 1024 kg
192230
approx 300lbs is what I read.