I need help with this physics problem?
83134
you really should have given numbers \n\nkinetic energy will be 1/2mv^2\n\nIts linear speed can be determined by distance over time  (d/t)\n\nthe next 3 answers need me to know what data you have in order calculate their values
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!!
824936
a) You have to draw gravity as a straight line down. Then you have to realize that gravity does no work if the block was on a horizontal surface. Work done by gravity is alway M*G*H.  The mass is 4.8 kg and the gravity is 9.81 m/s^2 .  The height will be sin (30) = H/2.  the sin (30 degrees) is 0.5 so .5=H/2.  the height will be .5/2 = .25 m.\n\nsoo you will have 4.8*9.81*0.25\n\nb) the work done by the frictional force will always be the frictional force times the distance.  The frictional force is going to be the normal force times Uk.  The normal force has components to it, so its going to be sin(30)*g*m.  The you need to multiply that by the friction coefficient and then by the distance traveled.  You will have 2*0.436*sin(30)*4.8*9.8\n\nc.  The work done by the normal force is just going to be cos(30) (i think) times 2 ( the distance traveled) times 9.81 (gravity) times 4.8 (the mass)
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
898375
I know the answer is in my basic physics book.  I would have to look it up.  You should too.
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!
487526
Sounds like a Math/Science type question to me. Why are you posting it here?\n\nBesides, I have the the slightest idea.
380004
I don't understand.  Can you explain the problem a little better?
760426
Figure out the centripetal force.  After that, this becomes a sliding problem turned on its side.
891370
Get the unbalanced force from the first 2 #s in your ? :  2.8-2.6 = .2\n\nb)  F = ma---> m = F/a = 0.2/.11 = 1.818 kg
247689
Use analytical trig (hint: sin function).  That should give you the answer you are looking for.
926868
Solve for momentum.\n\nBefore collision total momentum  = mv\n\nDefine your coordinate system so that the particle was traveling along the x axis in the positive direction, and the collision happens at the origin. Then all of the momentum before the collision is in the x direction. the initial y component of momentum is zero.\n\nparticle 1 goes flying off at v1 at angle a1\n\nparticle 2 goes flying off at v2 at angle a2\n\nThe total momentum afterwards must be equal to the momentum before. So the y component of p1 plus the y component of p2 must equal the y component of the momentum before collision (zero).\n\nSo m*v1*sin(a1) = m*v2*sin(a2)\n\nSimilarly the x components must add up so \n\nm*v1*cos(a1) + m*v2*cos(a2) = mv\n\nNow just solve the first equation for V1 and substitute in the second equation. \n\nThen solve the second equation with substitutions to find the value of v2 in terms of v, and the sin and cos of the two angles. \n\nFinally substitute the value of v2 back into your equation for v1 to find the value of v1.\n\nI leave the simple algebra of solving the equations and the calculation of plugging the numbers into the equations to the student.
933787
plz ask in physics category
742316
Torqu equals force applied times radius, so the equation would be T=Fr    or    T=mar    so, T=(4)(9.8)(6), or 235.2 n/m\n\nMoment of inertia is not needed to solve this problem unless it later asks for another different answer.
771534
The problem is about conservation of angular momentum.  The tricky part is (b). To solve this use this definition of angular momentum and draw a picture to find out the value of r sin θ\n\nThe angular momentum of a particle of mass m with respect to a chosen origin is given by vector product L = r x p= mvr sin θ
1149825
I'll give you a head start, and we'll see if you can figure it out from there:\n\nFirst, the givens:\n\nd = height from the ground to the point at which the two balls are released = 122.5 meters\n\nv1 = initial velocity of the first ball = 0\nv2 = initial velocity of the second ball\nt1 = time ball 1 is dropping (seconds)\nt2 = time ball 2 is dropping (seconds)\nt2 = (t1)-2 seconds\n\na = gravitational acceleration = 9.8 m/sec^2\n\nFor both balls:\n\nd = vt + at^2\n\nFor the first ball, v1 = 0, so d = at1^2\n\nFor the second ball, d = v2t2 + at2^2\n\nThe distance in both case is the same, so:\n\nat1^2 = d = v2t2 + at2^2\n\nOkay, so here you go:  You have the basic equations,so solving is trivial arithmetic:\n\nd = 122.5 meters = (9.8 m/sec^2) (t1)^2\n\nand solve for t1.  Once you know t1, then you know that t2 is two seconds less, so with that value you can solve:\n\nd = 122.5 meters = v2t2 + (9.8 m/sec^2)(t2)^2\n\nI leave the rest to you!  (And I have calculated the answer if you want to check it)
105143
Draw a force diagram to help you figure this out.\n\nRemember, perpendicular vectors are independent.\n\nF=ma