Physics questions (easy ones) Help?!?
970729
Looks like this is your homework. anyway, i'll only answer two of your questions.\n\n1. the density of water is almost 1 g/mL. and 1 liter = 1000 mL. hence the mass of 1 liter of water is 1000 g.\n\n2. since 1 liter of water is 1000 g or 1.00 kg, its weight is 1.00 * 9.8 = 9.80 N
730564
1. friction\n2. W=mg  , he has a mass of 100Kg. Weight is the force=newtons, g=pull of gravity N=100 x 9.8m/sec^\n98Newtons\n3. correct
898375
I know the answer is in my basic physics book.  I would have to look it up.  You should too.
630060
Speed of.. gravity? :S
986088
a) W = mg = 30*9.90662 = 294.1986 N\nb) Ft = 294.1986sinθ = 155.0897 N\nc) cosθ = 250/294.1986, θ  = 31.81376°\nd) a = 155.0897/30 = 5.19657 m/sec^2
833368
find the force of her weight. \nthat should be the answer
757859
haha, k  Draw a picture and draw a free-body diagram.\n\nI don't feel like busting out a paper/pencil  so I'll try it in my mind\n\nF(sub g) cos Theta  =  Mg cos(20 degrees)\n\nbecause the Normal force is equal to the y-component of gravity.\n\nTwo similar triangles, the incline and one which has the Force of gravity straight down, and the y compentn equal opposite of the normal force, and X component equal opposite to F\n\n\nlots of physics i'm skipping in my explanation tho
741402
a) F = ma \nYou're given F = -6070 N (negative because it's in the opposite direction) and m = 2600 so a = -2.33 ms^-2. Then over 1.76s, the change in velocity is (-2.33)(1.76) = -4.10 m/s. Car velocity is then 34.9 - 4.10 = 30.8 m/s.\n\nb) v^2 = u^2 + 2as\nv = final velocity = 30.8 m/s (from above)\nu = initial velocity = 34.9 m/s\na = acceleration = -2.33 ms^2 (from above)\nRearrange to get s = (v^2 - u^2)/2a = 155.61 m\n\nc) v = u + at\nYou want v = 0. So if the force is constant then acceleration is constant at 2.33ms^-2. \nu = initial velocity = 34.9 m/s\nSo rearrange to solve for time: (v-u)/a = t = 14.98 seconds.\n\nHope that helps. And learn your equations!
7127
here's a site with pics that might give you some ideas:\n\n\nhttp://www.physics.usyd.edu.au/competition/comp1998/comp98.html
179131
Try this:\nhttp://kr.cs.ait.ac.th/~radok/physics/f3.htm
160821
1. Dy = 1/2 A*t^2 + Vy*t, Vy = 0 (speed is horizontal),\nso t = sqrt(66m/(1/2*9.8m/s^2)) = 3.67 s.\n   Dx = Vx*t = 6 m/s * 3.67 s = 22 m.\n\n2.(a)  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso t = sqrt(1.150m/(1/2*g)) = 0.484 s\n  (b)  Dx = Vx*t, so Vx = Dx/t = 0.385m/0.484s = 0.795 m/s\n\n3.  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso Dy = 1/2*9.8m/s*(2.9s)^2 = 41.2 m\n\n   Dx = Vx*t = 2.4 m/s * 2.9 s = 6.96 m\n\n4. Dy = 0, Dx = 97 m, a = 50 degrees\n   Dx = Vx*t = V*cos(a)*t, \nso t = Dx/(V*cos(a))\n   -1/2*g*t^2 + Vyt = Dy = 0 ( Vy = V*sin(a))\ndivide by t, so -1/2*g*t + V*sin(a) = 0, or\nt = V*sin(a)/(1/2*g)).  Equate the two expressions for t:\nV*sin(a)/(1/2*g)=Dx/(V*cos(a)), multiply both sides by V\nV^2*sin(a)/(1/2*g) = Dx/cos(a)\nV= sqrt(Dx/cos(a)/sin(a)*(1/2*g))\nV = 31.1 m/s
383748
Facts to remeber while solving the physics problem\n1. A force that causes an object with a mass of 1 kg to accelerate at 1 m/s is equivalent to 1 Newton.\nWe will also be using the principles based on Newton's Laws of motion.\nEarth attracts a mass of one kilogram with a force of about 10 newtons.\nIn Physics weight and mass are different. The mass of a body does not change, the weight will depending on the gravitational pull acting on it.\n   \nF = m x a       \nWhere    m = mass (kg) \n      F = force (N) \n      a = acceleration (m/s/s) \nTo calculate the downward pressure on the bottom of the bucket .\nWe first calculate the force on a 1 Kg bucket (given) due to 30 cm(change to metres 1 m = 100cms) of water\nF =  1 x 0.3 x 0.3 = 0.09N\nTherefore total downward force on the bottom of the bucket is the addition of the two.\nTotal downward force = 10N + 0.09 N = 10.09N\n\nTo calculate weight of full bucket.\nWe know bucket weighs 1 Kg\n12 litres of water = 12 kg\ntherefore total mass = 13 kg\n1 Kg = 10N\nTherefore 13 Kg = 130N\nWight of full bucket = 130 N
1144961
c) R=v^2/g*2*sin(angle)cos(angle)=(2)*(5^2)*(sin30)*(cos30)/10\n       =2*25*0.5*0.8/10\n       =2 meters.\n\nb) Maximum height=(square of vertical componenet velocity)/(2*        accelaeration of gravity)\nvertical component of velocity= velocity*sine of angle=5*sin30=2.5 m/sec.\n\nH=(2.5^2)/(2*10)=2.5/20=0.3125 meters.\n\na) time=range/horizontal velocity\n   horizontal vel=v*cos(angle)=5*cos30=5*0.8=4 m/sec\n  time=2/4=0.5 sec(range is from first answer)
1132678
Directions how to do this can be found in the text book.\n\nDo your own homework!
762251
If 2 billiard balls of mass m (ma and mb) are perfectly elastic and ball A approaches from the left with velocity 2 m/s and hits ball B which has velocity 1 m/s, and is also going to the right, what's the status after the collision. \n\nA is going to the right with velocity 1 m/s and B is going to the right with velocity 2 m/s.\n\nma*2 m/s + mb*1 m/s = ma*1m/s + mb*2m/s\nmomentum is conserved
280308
u can get answer at www.howstuffworks.com\nok, if u get yr answer, then let me know,if don't get it then also let me know.
224079
try this site... it shows step by step how to go about sloving these equations, good luck: