Two physics questions?
919061
(a)\n(b)\n(b)\nfor your answer if i had four books and a piece of wood in my head that depends how long is the nail to get seriously hurt or have a big migraine with to much hammering in my head
458976
The answers for these questions:\n\nAnswer 1: (c) 27 meters\nAnswer 2: (d) .5 km\n\nAlthough the answer to 2nd question should be 0.4 kms, but I reckon it's just approximate. \n\nYou can email me for further help by clicking Contact Nitro in my answers profile.
1093339
No one on here is going to actually DO your homework for you.. but here is a place for you the get help -\n\nhttp://www.themathpage.com/aPreCalc/logarithms.htm\n\nhttp://en.wikipedia.org/wiki/Logarithm\n\nhttp://physics.mtsu.edu/~wmr/log_2.htm
74435
Try reading this, it should answer your question.\nhttp://phyun5.ucr.edu/~wudka/Physics7/Notes_www/node58.html
935341
1 sum the forces and set equal to zero. 2 forces, gravity and electric.\n\n2. conservation of energy. initially all kinetic, finally all potential. KE1=U2. U2 is electric potential. solve.\n\ngood luck
398633
Part 1: Mike\nWork=Force x Distance(moved) x cos (theta) (angle between work and distance)\nF=215\nD=64.7\nTheta=35 degrees\n\nW=215*64.7*cos35\nW=11,394.815 Joules\n\nPart 2: Chris\nF=8N\nD=3.9\nTheta=90 degrees\nW=8*3.9*cos90\nW=0\n\nThats all you get...good luck
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
901635
1\na.  distance, slower\nb. weaker gravitational pull farther away\n2\na. 750.353 km\nb. 4.67 cm\nc. Not enough info\n\nPS please award best answer
12692
you left out the number of seconds.  I'll give you the formulas you need though:\n\n(a)  a=v/t  so 6 divided by the time you forgot to put.\n(b)  F=ma  so .2 times the acceleration you just solved for in (a)\n(c)  same as (b), the puck exerts an equal force on the stick, the puck just goes flying farther because it's smaller.
446591
1)X=V1t+1/2gt^2\nat the interval between getting to the highest point and being dropped back to the first place is half of the full time = 2 seconds\nV1 at the highest point = 0\nX=displacement\nX=0+1/2*9.8*4=19.6 m \n2)V2=V1+gt\nV2 = velocity of landing\nV1 = velocity at highest point\nV2 = 0 + 9.8 * 2 = 19.6 m/sec
894578
Don't add the two. Only the vertical distance counts in determining work because only in lifting the block did you move it against the force of gravity.
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...
515562
Friction, Heat generated to name 2 reasons.
553994
1st - An object in motion/rest will stay in/at motion/rest unless acted upon by another force.\n\nExamples - a ball will keep rolling until its friction with the ground stops it. A rock will stay put unless someone transfers energy to it to move it.\n\n2nd - Force = Mass x Acceleration\n\nExamples - Some car has 5000 kg mass and 10 m/s^2 acceleration. Find its force.\nForce = 5000 x 10\nForce = 50,000 newtons\n\nSome truck has 250,000 newtons of force, and an acceleration of 5 m/s^2. Find its mass.\n250,000 = Mass x 5\nMass = 250,000/5 = 50,000 kg   \n\n3rd - For every action, there is an equal and opposite reaction. \n\nExample - you bounce a ball on the ground. The ball hits the ground, and compresses inward. The ground pushes back up on the ball causing it to land in your hands. You hit a volleyball in PE. The ball pushes against your hands and goes over the net.
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
802764
1) KE = ½mv²\n1.1 x 10^9 = ½ x 8 x 10^4 v²\nv² = 2.75 x 10^4\nv ≈ 165.8 m/s\n\n2) another physics question: what is the speed of a 0.145kg baseball if its kinetic energy is 109J?\n\nDo as above\n\nKE = ½mv²\n109 = ½ x 0.145 x v²\nv² ≈ 1503.4\nv ≈ 38.8 m/s\n\n3) ANOTHER QUESTION: TWO 3.0 bullets are fired with speeds of 40.0 m/s and 80.0 m/s, respectively. what are their kinetic energies? which bullet has more kinetic energy? what is the ratio of their kinetic energies?\n\nAssuming 3.0 g bullets as they would NOT be 3.0 kg bullets\n\nSo 3 g = 0.003 kg\n\na. v = 40 m/s\nKE = ½mv²\n= ½ x 0.003 x 40²\n= 2.4 J\n\nb. v = 80 m/s\nKE = ½mv²\n= ½ x 0.003 x 80²\n= 9.6 J\n\nratio of energies (slower to faster) = 2.4:9.6 = 1:4 = (ratio of velocities)²\n\n4) another physics question: a car has a kinetic energy of 4.32 x 10^5J when traveling at a speed of 23 m/s. What is it mass?\n\nKE = ½mv²\n4.32 x 10^5 = ½m x 23²\nm ≈ 1633 kg
768228
a. V ice = pi(0.12^2-0.10^2)*3 = 0.041 m^3\nb. V ice = pi[(0.10 + x)^2 - 0.10^2]*3 = pi(0.10^2 + 0.20x + x^2 - 0.10^2)*3 = 3pi(0.20x + x^2)\ndV/dt = pi[0.60 + 6(0.02)]dx/dt = 0.000002pi\n= [0.60 + 6(0.02)]dx/dt = 0.000002\ndx/dt = 0.000002/[.60 + 6(.02)]