7. [Glencoe99 11.P.57.]\nAn archer puts a 0.25 kg arrow to the bowstring. An average force of 203 N is exerted to draw the string back 1.3 m. Unless otherwise directed, assume that air resistance is negligible. Draw energy bar graphs to solve the problem.\n(a) Assuming that all the energy goes into the arrow, with what speed does the arrow leave the bow?\nm/s\n(b) If the arrow is shot straight up, how high does it rise?\nm\n\n8. [Glencoe99 11.P.62.]\nFrom what height would a compact car have to be dropped to have the same kinetic energy that it has when being driven at 95 km/h? Unless otherwise directed, assume that air resistance is negligible. Draw energy bar graphs to solve the problem. \nm \n\n9. [Glencoe99 11.P.64.]\nKelli weighs 405 N, and she is sitting on a playground swing seat that hangs 0.40 m above the ground. Tom pulls the swing back and releases it when the seat is 1.00 m above the ground. \n(a) How fast is Kelli moving when the swing passes through its lowest posit
201130
Yep, someone can.\n\nDO YOUR OWN HOMEWORK.
67570
1) V2^2=V1^2+2d*g ----> 0=47^2+2d*-9.8 ----> d=112.7 m\n2)remeber when mesuring the time we take the velocity on the Y axes and when mesuring range or distance we take velocity on the X axes got it?\nd=V1+0.5*g*t^2 and since the rock is thrawn horizontaly the velocity on the Y axes once it's thrawn is zero, by the way the decplacement would be minus cuz the rock is projecting downwards \n-3.5 m = 0 + 0.5 * -9.8 * t^2 -----> t=0.845 seconds\nd=V*t=47*0.845= 40 m\n3)this is a tough one you'd better draw it :\na.V2^2=V1^2+2d*g ----> 0= (47sin37)^2 + 2d*-9.8 ----> d=40.8m\nb.V2=V1+g*t ---> 0=47sin37+-9.8*t ---> t=2.886 seconds\nrange=V*cos37*t=47cos37=108 m\nc.V2=V1+g*t=47sin37+(-9.8*2)=8... m/sec (velocity on the Y axes)\nvelocity on the X axes is constant and equals=47cos37=37.5m/sec\nthe net velocity^2=37.5^2+8.685^2 ---->velocity of rock after 2 seconds of launch=38.5m/sec
442007
The choice of the violin was a bit unfortunate since violins have a special frequency response that is not a parabolic wave when properly bowed:  \nhttp://plus.maths.org/issue31/features/woodhouse/\n\nWell, that may not be the point here since the author of the book probably assumed that the string follows the harmonic frequency behavior as described by:\n\nf=sqrt(T/u)/(2*L)\nwhere T is the tension\nu is the linear mass (Mass per unit length)\nL is the length\nAnd f is the frequency harmonic.\n\nThe act of "fingering" the string is to depress it and thereby change the length.  You can likely assume that this has negligible effect on the tension of the string.  So if when fingered the new l = 2L/3\nyou just plug that into the equation and solve using a ratio:\n\nf=K/L\nf'=3K/2L\nf"/f=3/2\n\nj
454743
approximately 1107 feet i think\nof course this is neglecting air resistance and mass of the bullet along with loss of energy heating\n\nmy work:\nfind the time that bullet is in air:\nwe know that height is 5 feet and the initial velocity of the downward component is 0 because the gun was fired horizontally.\nusing second kinematics equation :\n intial velocity(time) + 1/2(acceleration)(time)^2 = distance\nwe fill in:\n0(t)+1/2(32.67ft/sec^2)(t^2) = 5 and solve for t which = .553283\nNow that we know the time bullet spends in the air we can find the distance by using time and the same equation but using all horizontal component.\nwe will in:\n2000(.553283)+1/2(0*)9.5532830^2 which equals =1106.57 or about 1107\n\n* = there is no acceleration since there is no gravitaional effect on horizontal component\n\ni made have made a mistake so please check
553884
For 4 a&b)\nThere would be three arrows.  The first would be a force in the upward direction (i.e. normal force) second would be force due to gravity(downward direction, with equal magnitude o that of the normal force).  The third arrow is an arrow that is in the same direction as the force applied to the large box.  Its magnitude is proportional to the normal force.  It s static friction that is applying that force.   \n\nAs far as part c goes It's been a while for me.  You've got the textbook, look up some formulas.  Besides, I think you'd need to include some masses in order for me to solve those.\n\nThe force diagram for the Large box is very similar to that of the small box, except that it has forces applied to it from the small box. (this is because the large box is supplying the normal force for the small box.)
213342
At the top of the halfpipe, the ball is just sitting there with potential energy stored up (PE = mgh, where m = mass, g = acceleartion due to gravity, and h = height).  As the ball rolls downward, its PE lessens and its kinetic energy increases (KE = 1/2(mv^2), where v = velocity of the ball).  At the bottom of the halfpipe, all PE is gone and KE is at maximum.  As the ball goes up the other side, the KE decreases and the PE increases.  In a perfectly frictionless halfpipe, the ball would return to the same height as it was before it started rolling, and the KE would go to 0 and the PE would be at its maximum.\n\nIf you just dropped the ball (no halfpipe - just a freefall), the KE just before impact would be the same as the PE before dropping.  The two are inversely proportional - as one goes down, the other goes up.\n\nNow if you knew the exact shape of the halfpipe, you could start plugging in numbers to find the PE and KE of the ball at various points.
414085
Ei - A= Ef (A = work done by air resistance, "i" means initial and "f" means final)\n\nPEi + KEi - A = PEf + KEf\n\nA = PEi + KEi - PEf - KEf\n\nA = mghi + (1/2)mv²i - mghf - (1/2)mv²f\n\nA = m[ghi + (1/2)v²i - ghf - (1/2)v²f]\n\nA = 0.664[(9.80)(2.27) + (1/2)(7.00²) - (9.80)(3.31) - (1/2)(5.23²)]\n\nA = 0.664[22.2 + 24.5 - 32.4 - 13.7]\n\nA = (0.664)(0.6)\n\nA = - 0.4 J (since the work done decreases the total energy, it's negative)\n\n*** 0.4 J of work are done BY the nonconservative force air resistance ON the ball.\n\n\n*** Believe it or not, the engineer above me did it completely wrong, even though he understood the concept.  First, work and energy are NEVER expressed in Netwons (N), but in other units such as joules (J).  Also, when calculating the potential energy, the formula is PE = mgh.  The engineer forgot the "g," which is equal to 9.80 m/s^2.  I am fresh out of AP Physics, and I got a 5 on the AP test.  All throughout physics and AP physics, I was drilled with all kinds of problems.  Energy/work problems are my specialty.  You can even verify my answer with your friends and your teacher, and you'll see that it is correct, and the engineer's answer is very wrong.\n\n*** To the engineer: Sometimes, even professionals can be wrong.  I'm sure you understand all the concepts of physics.  You obviously have done a LOT of math and physics in the past, since you're an engineer.  However, it's probably been a while, and you may have forgotten a few formulas (and I don't think the potential energy formula can be easily derived).
478364
this seems an easy question but most of the student get it wrong since it is so easy to mess up your mind. it was a part of an exercise for my University exams. it makes you realise what is going on with spring ...just consider it as a opposing force. the body will drop and will expirience g --> it will accelerate. at the same time it will also expirience the sping;s opposing force.\n\nit will accelerate till the force become equal. this is the importnat factor. weight will keep on accelerating the body till it is completely cancelled by the spring force.\n\nso when \nFs = Weight then Velocity is maximum. \nso kx =mg =>\n\nx = mg / k.  (at this displacement the velocity  will be maximum)\n\nat this specigic point due to the theorem of conservation of energy we get that.\n\nkinetic energy + sping stored energy = potential energy at the start.\n\nso \n1/2mu^2 = (m*g)^2 / k => (be carefull mass = 76.7/1000kg)\nU = \n\nU = sqrt (( m*g^2)/k)) => u = 0.361m/s 9\n\nIF THE ALGBRA IS CORRECT\n\nNOW AFTER THAT POINT the body will still travel due to inertia and the springs force  larger now than gravity wll slow it down and bring it at rest \nat that point all the potential energy (due to height difference) will be transfornmed to springs stored energy therefore,\n\n1/2 k xmax^2 = mgXmax =>\n\nXmax = 2*m*g / k =>\n\nXmax = 0.02658m \n\n\n( i am at work and i believe the algebra is correct. the main point is to realise where certain things happen.)
339467
Consider a force diagram at the point that the 499 N load is applied.  You have 499 N in the negative "y" direction, and two equal Tension forces (call them T), each at an angle to the horizontal.  Thus, the tension forces have each an "x" and "y" component.  One tension force goes to the tree, the other to the car.  You've got the angle from the horizontal from trig.\n\nNow do a force balance on those three forces, in the "y" direction. There is no acceleration, so sum of forces = ma = 0\n\nT*sin(4.76) + T*sin(4.76) - 499 = 0\n\nOnly one unknown here, solving.......\n\nT = 3006.67 N\n\nThat's a lot of force!\n\nCheers,
645938
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!
383339
Equation A is correct since momentum is a vector quantity it has direction.\n\nkinetic energy in equation B is scalar, so you should not subtract.\n\nI got the quadratic equation:\n\n75.83*V1^2-198.3*V1-904.167=0\n\nWhich gives two roots\nV1=5, or -2.384\nthe negative is the one I chose since the collision must be elastic and there is only one direction of motion (either positive or negative right or left).\n\nGiven this, I computed\nV1=-2.384\nV2=5.6\n\nj
40752
Energy is conserved so the gravitational potential energy (ge) of the projectile is equal to the spring potential energy (se) expended by the spring.\n\nge=mgh and se=1/2 k(x2-x1)^2, where x1 is the spring position when compressed and x2 is the equilibrium spring position. (x1-x2) is spring displacement.\n\nGiven: (x1-x1)=0.120m, m=0.035kg, h=20.0m.\n\nA) Find spring constant, k.\nee=se\nmgh=1/2 k(x2-x1)^2\nk=2mgh/(x2-x1)^2\n2*0.035kg*9.8m/s^2*20m/.12m)^2\nk=953N/m\n\nB) Find projective speed as it moves through spring equilibrium position.\n\nIf you set your 0 on the x-axis to the position where the spring is compressed, then x1 would be 0 and x2 would be 0.12m. So the spring equilibrium position is at x2=0.12m.\n\nNow use the kinematic equations of motion to find the projectile speed when spring displacement is 0.12m. You can assume constant acceleration to be 9.8m/s^2.\n\n(vf)^2=(vi)^2+2ad where vf and vi are final and initial velocity, respectively, vi=0, and d=(x2-x1)=x2.\n\nvf=sqrt(2*a*x2)\nvf=sqrt[2(9.8m/s^2)(.12m)]\nvf=2.325m/s
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a) As the movement is circular, the total of the forces is in this plan and to the centre and the value is mw²R.\n\nSo P + S = mw²R where P is the weight and S the tension of the string and R the radius and a the angle between the string and the vertical\nIn projection, we have :\nmw²R = S sin a\nmg = S cos a\n\nDividing the 2 equations :\ntg a = w²R/g\nand R = L sin a\n\nso w² = g/(Lcos a)\nso cos a = g/Lw² = g T²/(L 4 pi²) \n\nso w²= g/(L cosa)  \n\nso v = r sqrt(g/(Lcos a)) = 1.4 sin 30  sqrt( 9.8/1.4 cos30) = 1.99 m/s\n\nb) 4 = 1.4 sin a sqrt (9.8 /1.4 cos a)\n\nc) As mw²R = S sin a\n    and mg = S cos a\n\nw = sqrt (S²- m²g²) /mR
353952
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!
368542
we know m=0.0043kg, Vfinal=304m/s, d=0.86m and Vinitial=0\n\nwhat we need is the value of acceleration ( remember F=ma )\n\nso the formula we need is\n\nvfina^2=viinitial^2 + 2ad\n304^2=0 + 2*0.86*a\na=(304^2)/(2*0.86)\na=53730.23256\n\nso F=m*a=53730.23256*.0043=231.04 N\n\nForce is 231.04 N\n\n*****************************************************\n\nHere's an alterative if you want to double check  \n\nthe kinetic energy (KE)=1/2 *m*(velocity)^2\n\nthe KE of the bullet must equal the Work needed to fire it\n\nKE=0.5*0.0043*304^2 =198.6944 j\n\nWork=F*d=198.6944\n\nso F=(198.6944)/0.86=231.04 N\n\nAgain we get  force = 231.04 N\n\nSo we're pretty sure this is the correct answer
366261
SRY!! Matt is going to a basketball game out of town. If he could travel 75mph, He would arrive one hour early. If he could travel 45mph he arrives one hour too Late. How far is Matt's trip\n\ndis=75*(time-1)\ndis=45*(time+1)\n\nthen\ndis= 225mile,    time=4hour
520160
Your best bet is to go to an archery shop with it. They can measure the draw weight and check to see if the draw length is set for you. They can also cut your arrows to the correct length, if needed, all of which can determine the proper weight of the arrows you will need!