How much lift is exerted on an object that has a total surface area of 100 m squared, with 4% atmo. pressure?
146130
Do you mean 100 m squared, or 100 m^2, i.e. 100 square meters?  I ask, because 100 meters squared could mean an area 100 m x 100 m in size, or 10,000 m^2.  Since that is the size of several football fields, I assume you mean the later.\n\nStandard atmospheric pressure is 1.013 x 10^5 N/m^2.  Hence, if you take 4% of that pressure, (i.e. multiply by .04) and multiply by the total area over which the pressure is exerted, 100 m^2, you get the force produced by that pressure, which is the lift you wanted.  I will leave the arithmetic for you.
938540
assume shot putter moved the shotput with constant acceleration\nhence we need to find acceration\n\nv^2= u^2 + 2as\n(13)^2= 0 + 2a (2.8)\n169 = 5.6a\na= 30.17857142m/s^2\n\naverage force exerted \n= ma\n= 7 x 30.17857142\n= 211.2N\n\nyey
952952
one third of the elastic limit is 8x10^7 Pa and the total area is 3x10^(-4) m² so the total force is 2.4x10^4 N.  It takes 12,740 N to hold the elevator stationary agains the acceleration of gravity which leaves 11,260 N available to accelerate the elevator upwards, so the maximum acceleration would be about 8.66 m/s².\n\n\nDoug
821783
Bernoulli's equation can be used to estimate the pressure difference on the upper surface versus the lower surface, using the velocities you are given. Then recall that pressure is force per unit area, so take the pressure difference times the total area of the wings (remember you have two wings), and you have force!
612367
Volume of cylinder = Area of the base X height = πr2h.
473588
Momentum is mass times the velocity.\n\nMass is 100g = 0.1 kg\n\nvelocity is 2.8 m/s.\n\nMomentum is 0.1 kg x  2.8 m/s. = 0.28 kg.m/s
338219
the resistive force that air will exert upon an object moving through it at a given velocity.\n\nthe equation is Drag= 1/2 cpav2 (squared)\n\nc is a constant.\np is the density of the fluid (air in this case)\na is the cross sectional area of the object\nv is the velocity of the object.
1120781
do your own homework\n\nits probably 0.26 kg M/s ^2 since its a velocity\n\nyou also probably need to convert it to rad/s whatever that is
188896
K = mv^2 = (10kg)(4.5 m/s)^2 = 202.5 J\nI would suggest you try doing your homework, asking for all the answers will only hurt you in the future.
915365
initial surface area=6a^2\nnew surface area=2[2a^2+4(a^2/2)]\n=2*4a^2=8a^2\nit will be 4/3 times the original surface area
880761
People above neglected the vertical component of 460 N.  This must be added to the weight of the box.  Also, not sure why they took 350 N and multiplied it by acceleration.\n\nso F_vert=460sin35.6=267.78 N (downward)\n     F_hor=460cos35.6=374.0 N\n\ncalculating F_frict=mu x (F_normal)\n\nF_normal=weight of box + F_vert = 267.78+350=617.78\n\nso F_frict=0.57x617.78=352.13 N\n\nnow the net force on the box = 374.0 - 352.13 = 21.87 N\n\nand we need mass of the box, m=350/9.81=35.7 kg\n\nso a=21.87/35.7=0.6126 m/s2\n\nfinally,  d=V_initial x t + 0.5 x a x t^2, V_initial=0\n\nso t=((4.2)/(0.5x0.6126))^0.5=3.7 sec\n\n**********************************************************\n\nHowever, should you wish to ignore the vert component\n\nF_frict=mu x (F_normal)=0.57 x 350 = 199.5 N\n\nso net horizontal force is:\n\nF= 374.0 - 199.5 = 174.5 N, a=174.5/35.7= 4.89 m/s2\n\nplug a into:\n\nd=V_initial x t + 0.5 x a x t^2, and solve for t\n\nt=((4.2)/(0.5x4.89))^0.5 = 1.31 sec
861817
10 kg load weight 98 Newtons (10 x 9.8).\nA force of 98 newtons acting over a distance of 4 meters does 392 joules (= 4 x 98 newton-meters) of work.\nIf the engine does 500 joules of work, its efficiency (output/input) is 392 / 500 = .784, or 78.4%.
909202
a= r * w^2 = .12 * 2^2 = .12 *4 = .48 m/s^2
847764
Surface Area = 4*PI*r^2\n\nr = SQRT(SA/4/PI)
864950
Gravitational Potential Energy = m*g*h joules\n        where   m - mass of the object in kg\n                     g - accelaration due to gravity in m/s^2\n                     h - height of the object from the reference in m\nHere the height from reference is 500m\ng is taken to be a constant and its value is 9.81m/s^2 though it may vary slightly with latitude, altitude and other such factors.\nMass is given as 0.400kg\nThus Potential energy = 0.400 * 9.81 * 500\n          Potential Energy = 1962 joules\n\nHope so u got wat u needed.
680534
Use the equation from the link below.  You will find that \nv/c = 0.14365, or\nv = 0.14365 c\nor\nv = 43,100 km/s\n\nThe object is moving away from the observer at this rate.\n\nBy the way, you probably won't get credit unless you can work out the math steps I left out.
1135961
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!