try solving this question of physics.?
921196
If you run x1 meters against wind, the work you produce is A1=F*x1+f*x1; F is the force of your muscles to overcome the resistance of air with no wind, f is the force of the wind in your face. Let x1 be covered in time t1, then P1=A1/t1 is your power against the wind. \nIf you run x2 meters with the wind, the work you produce is A2=F*x2-f*x2; F & f being the same.\nLet x2 be covered in time t2, then P2=A2/t2 is your power with the wind. They say that dP=P1-P2=50W or dP=(F+f)*(x1/t1)-(F-f)*(x2/t2). As x1/t1=x2/t2=v, then dP=2*f*v. Thus f=dP/(2*v)=50/7=7.14N  is the force of the wind, m=80kg is excessive datum!
643953
depends on the direction of the force
574012
I've alread answered two of your questiions in my last answer....\n\nIn this one, the guys above are correct - however, I'd try working it out for myself if I was you. Your tutor will soon realise something is wrong if you are giving answers to a subject you don't understand. Read the chapter, ask you teacher if you need more help. Learn Newtonian mechanics for yourself.
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...
124298
Answer to #2\nt=1.565 seg\nV= 2.556 m/seg\nt=1.565 seg, the same.
1102516
by any chance is this a question from your maths homework? lol :D\n\nI cant work it out myself because to put it simple, im crap at maths.
933787
plz ask in physics category
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.
415527
First of all, the sum is incompleted.  Is it for maximum height, or for range?  Secondly, let us assume you are asking for maximum height.  Then differentiating the given equation of motion, d = -5(t+1)^2 + 258 with respect to 't', we get dt/dt = -10(t+1).  At maximum height, the velocity is zero. \nTherefore, -10(t+1)=0, or t = -1.  This is absurd time can not become negative once the motion is initiated.  Check up there is something wrong in the problem.
706663
The idea is that all forces at the nodes (connecting points) are balanced. Draw the two free body diagrams.\n\nLeft point:\nFx\n0=-T1cos53+T2cos11\nFy\n0=T1sin53-T2sin11-57\nRight point:\nFx\n0=-T2cos11+T3cos42\nFy\n0=T2sin11+T3sin42-W2\n\nBy inspection we see only the first 2 eqns are required. Solve the system:\n0=-T1cos53+T2cos11\n0=T1sin53-T2sin11-57\n\nT1=83.62 N\n{=}8.5 kg
35590
I concur with previous answerer. The acceleration is 1.5 m/s^2. Rearrange Newton's Second Law, to solve for the acceleration. Then, substitute what you know.
826674
use conservation of angular momentum in these kind of problems\nNOT energy conservation\nif you dont know how to use the conservation equation, better read up on examples
976575
Hi Mark, I will answer your question if you answer mine first.\nWhy are you asking this question, it seems like a lot of people in school are trying to take the easy way out of a physics or math problem.\nWill answer if you are not doing said above!\nCheers, Susie
1014316
would you like to supply me with some formulas? im out of grade 12 physics and i forgot most of the formulas. trust me from experience, just plug numbers into a formula and things will work\n\nF=mg = ma\nFk=μkN\na= vf^2 - vi^2 / 2d
833368
find the force of her weight. \nthat should be the answer
769928
T = 2pisqrt(l/g)\nl = g(T/(2pi))^2 = 9.80(1.0/(2pi))^2 = 0.25 m\n\n(The object's weight is irrelevant for small oscillations.)
666681
1/2 * mass * velocity^2.