Another Physics Question.?
483972
In what position is the bolt and what are the bolt's physical dimensions? If the bolt is laying in a horizontal position, and it's 1.60 cm thick (16 mm or about 5/8"), I'd say that the oil's refraction index is 1.00.\n\nOTBPOC
533021
there are some solved physics problems on \nhttp://www.dctech.com/physics/help/\nbut I don't know if it will help you out.
747360
Well gosh....  \n\nsince the only linear unit mentioned in the question is 18 ft/s  let's assume that the 60 is going to give us acceleration in ft/sec^2 \n\nSo, the position function for this is the second integral of the acceleration function, right? \n\nso...  the first anti-derivative of 60t is  30t^2   which gives the velocity, so at the end of the boost phase, our rocket is moving 30 * 3 * 3 = 270 ft/s \n\nand the second antiderivative is 10t^3  right?  \n\nSo, the position at time T is 10 * 3 * 3 * 3 = 270 feet.  (odd coincidence, but there you are...) \n\nNow, the rocket begins a SECOND acceleration by gravity of -32 ft/s negative 'cause we're accelerating DOWN now... ....   but this time the acceleration is constant,  so \n\nV = gt   and s = 1/2  a  t^2 so, we're slowing from 270 feet per second to zero at the top...  and  t = v/g = 270 / 32 = 8.44 seconds and the height we coast to is the 270 we had at the end of boost plus   1/2  *  32  * 8.44 * 8.44 = 1140 feet.  \n\nThe parachute is going to open after 14 seconds...   so, 14-8.44 is 5.56 seconds of free-fall (and we'll ignore air resistance again) so, we go from zero to something in 5.56 seconds....   and position is 1140 - ( 1/2  * g  * t^2 ) again...   so, the parachute opens at \n\n1140 - (1/2  * 32  * 5.56 * 5.56 )  = 645 feet. \n\nad the velocity is  gt = -32 * 5.56 = -178 feet per second...  \n\nNow, for the NEXT five seconds, we slow under the effect of the parachute   to  -18 ft/sec -- so,  we slow by 160 feet per second in five seconds, so the acceleration of the parachute is (oddly) 32 feet/second squared (huh, how odd is that? )  \n\nSo, the distance traveled during that phase is Vo t + 1/2 at^2  \n\nSo, we have (-160 * 5) + (1/2*32*5*5) =  -800 + 400 = 400 feet.  \n\nand from 400 feet up, we continue down at 18 ft/s for an additional  22 seconds...  \n\nand the rocket lands at t = 22 + 5 +  14 + 3  = 44 seconds \n\nAnd yes, I dropped a bunch of decimal points.  Your original data was only to two significant digits....
463641
see\nhttp://hyperphysics.phy-astr.gsu.edu/hbase/surten2.html#c1\n\n here you can find the answer for your question with MATHEMATICAL EXPLANATIONS.
1006189
Explain Newtons laws of motion.
50475
A)come on, the weight of the man is just his mass times gravity.\nB)this sounds like a torque problem, you need to know where on the beam the man is standing... since the beam is in equilibrium, you know that the sum of all torque is 0. 9im going to use T for torque because i cant get the greek letter i usually use) T=Fr (force times the distance from the pivit point) so all torques directed up minus all torques directed down equal 0. so that means all torques pointing up equal all torques pointing down. ok, so that means that:\n Tbeam + Tman = Tcable.\n and the cable is at an angle so it would actually be FrsinX instead of just Fr.  so now since the beam is uniform, we say that it's weight is in the center, so r is half of the length of the beam. so\n Tbeam=Fr=(288)(2.6) and now we convert the mass of the man to newtons, just multiplying by gravity(which im rounding to 10 because i dont feel like getting a calculator) and getting 750 Newtons. so that means\n Tman=Fr=750r. this has to stay as r because you didnt tell me where on the beam the man is standing. it probably indicate's where he's standing on your diagram. r would be the distance between the wall and the man because we're making the wall our pivot point, because if the end with the cable was our pivot point, everything there(including the cable) would be 0 because it would be 0m away from the pivot point, so we couldnt solve for Ft... so you'll have to plug in the "r" for the man yourself.  and Tcable=FrsinX=F(5.2)sin67 where F is the force of tension which is what you are solving for overall, r is 5.2 because the cable is 5.2m (the length of the beam) away from the pivot point. so now, we plug into what i told you before(Tbeam + Tman = Tcable.) which is true because the beam is in equilibrium. so \n\nTbeam + Tman = Tcable\nFr    + Fr   = FrsinX\n(288)(2.6) + (750)(r) = F(5.20)sin67\n\nplugging in the r which should be on your diagram, you can solve for F whichi is the tension in the cable.
48084
F = ma    \nNewton's second law.
121693
cmon...dont u know basic physics...just calculate the resultant of these 2 forces\nanswer= [(132)^2 + (181)^2]^1/2\nif this makes sense!!
756774
P=P (momentum before = momentum after)\nmv=mv\n(55kg)(4.6m/s) = (230kg)v\nv= ((55kg)(4.6m/s))230kg\nv=1.1m/s
954443
Hi Yo-Yo!\nEnergy conservation law: m*g*h=E1+E2, m=0.12kg, h=1.2m, g=9.8m/s^2, E1 is translation kinetic energy, E2 is rotational kinetic energy.\nE1=m*v^2/2, v is translation speed at full length h, E2=I*w^2/2, w=angular speed at full length h.\nNow v=w*r, r=3.2*10^(-3) m, thus m*g*h=m*(w*r)^2/2+I*w^2/2, hence w=sqrt(2*m*g*h/(m*r^2+I)). Plug numbers in and find w & v (be a good boy!).\nMeanwhile v=a*t & h=a*t^2/2, ‘a’ is linear acceleration.\nThus t=v/a & h=a*(v/a)^2/2, so a=v^2/(2*h). Plug numbers in and find a & t. Angular acceleration being w’=a/r. My regards!
806029
1st link \n\nb) F = 40 N\n\nc) F = 60 N\n\n2nd link\n\nb) linear velocity = 0.59 m/s\n\nc) Taking g = 9.8 m/s^2\n\nv = 0.83 m/s\n\nd) It would not effect the answer as the expression for v which is sqrt(mu * r * g) is independent of the mass. \n\n3rd link\n\na) Rotate the setup, note the reading on the balance. Find the time required for 1 circuit.\n\nb) No work is done\n\nc) T = 5.48 N\n\nd) 5.9 %\n\ne) If the cord is exactly horizontal, it won't touch the rod and hence cannot be rotated.
34477
It is 'B'. But I don't understand why you put up this question under 'Physics'!
933787
plz ask in physics category
987931
you are making a mistake with the units\n\ndid you mean kg/s^2? if it is so the answer is 10300 kg/s^2\nif did you mean kgf/m it is 1050 kgf/m \nif not your units are wrong. \nremember that 1 N (newton) is 1 kgxm/s^2\nand 1 kgf = 9.8065 kgxm/s
714752
This problem is a conservation of momentum problem:\n\n(a) In an elastic collision, the formulas for calculating the final velocities are derived from both the conservation of momentum formula and the conservation of kinetic energy formula. Check out my source for more information, but for now just trust me that they are valid:\n\nv1 = (v1(m1 - m2) + 2m2v2)/(m1 + m2)\nv1 = (1*(4 - 7.2) + 0)/(4 + 7.2) = -0.3 m/s\n\nv2 = (v2(m2 - m1) + 2m1v1)/(m1 + m2)\nv2 = (0 + 2*4*1)/(4 + 7.2) = 0.7 m/s\n\n(b) In an inelastic collision, just use the conservation of momentum formula:\n\nm1v1' + m2v2' = (m1+ m2)v\n\n4*1 = (4+7.2)*v\nv = 4/(4+7.2) = 0.36 m/s\n\nI hope this helps!
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
863271
mass doesn't matter here, just the acceleration\nv=u+at\na=(v-u)/t\na=(9-0)/0.15=60 m/s^2