physic question I have the answer . I need the way of solution.?
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
856026
Use the gas law\nPV=nRT
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.
899235
A) I think I got it...Use energy, except for g, use   Fg cos(19 degrees), found from a free-body diagram.\n\nso 1/2 mv^2 = mgh\n\n1/2 (4.7)^2 =  9.8(cos (19))  h    solve for h\n\nB)  now that you have h...maybe you can use   \n\nd y = V(sub 0 sub y) t + 1/2 (a x cos (19)) t^2\n\nyea, tell me how it comes out.
898375
I know the answer is in my basic physics book.  I would have to look it up.  You should too.
966112
u can consult ur physics book
1186593
Mix with a 0% solution in the relation 33:2\n\nSo use 2 units of your solution (like 2 liters) and 33 units of the 0% solution (like 33 liters).\n\nThat's all.
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
918781
Where is your freebody diagram? There are other forces acting on the block than just gravity...\n\nGood Luck
510279
Jimbo, you gave up too soon......\n\nV = L*W*H = L*W*x in this problem\n\nV = (6-2x)*(4-2x)*x\n\nV  oila!
855447
if it is equations of motion then \nv=u+ft\ns=ut+1/2ft^2\nv^2=u^2+2fs\nhope this is what you wanted
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
687306
Check the Michaelis-Menten equation.It's too long to write it all down for you ;)
905049
First assume that we neglecting any losses from air resistance or internal friction.\n\nWith that simplification, imagine two states:\n1) you are on the bridge 100m above the water and at rest\n2) you are suspended by the spring at its point of maximum extension, and therefore with zero velocity\n\nSince we assumed that there were no losses, the energy of both of these states is equal. Where is that energy located?\n -->1: your gravitational potential energy (no kinetic energy, no stored elastic energy)\n--> 2: your remaining gravitational potential energy, and the stored elastic energy in the spring (by definition, no kinetic energy)\n\nSo what has happened is that the spring has stored elastic potential energy that is equal to your loss of gravitiational potential energy. That's the only way the energies of the two states are the same.\n\nNow you can set up some equations. Let's say that you start at height H0 and end up at height H1.\n\nchange in PEg= m * g * (H1-H0)\nchange in PEk = 1/2 * k * [(H1-H0)-30m]^2\nwhere we have accounted for the 30m long unstretched length of the spring.\n\nNow by observing that -PEg = PEk, and plugging in the given values for m and k and the known value for g, you can readily solve for the change in height (H1-H0) (it should be < 0) It's a quadratic, so you expect to find one answer that is not physically realizable - reject that one, and select the other one.\n\nDoes that help?
660156
http://www.physlink.com/education/askexperts/ae25.cfm
857152
I'll agree that you need help.  Density is mass/vol.  You've given the density, you need another value to calculate the remaining value.
1008978
well this is really a chemistry problem but the formula is q=mC(delta) T, meaning heat= mass times the specific heat capacity times the change in temperature, so to set this up we use, 1.68^10 j = m (4.18 j/g) (80 degrees) becuase 4.18 is C and 80 degrees is the amount u wish to raise it, so then u get 1.68^10 j = 334.4m, divide both sides by 334.4 and you get .53558=m so your mass is .536g of water