physics qts. plz help?
824682
The energy absorbed is that which is not outputted.  In this case, 75% of the energy is absorbed, so the energy absorbed per cycle is .75*8500 J.  One watt = 1J/sec.  So seconds = J/watt.  The power is constant, so the power in each cycle is 6,00 kW; this times the duration of the cycle will be the energy in each cycle.
966112
u can consult ur physics book
163823
Q=mc(Tf- Ti)\nWhere Q is the heat added\nm is the mass\nc is the specific heat capacity\nTf is the final temperature\nTi is the initial temperature
865948
KE = 1/2 mv^2\nfind the KE before the collision using the formula.\nfind the KE after the collision using the formula.\nsubtract the two values to find the change in kinetic energy
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.
1034911
R'(q) = 5(4q)/(2q^2 + 5)\nR'(4) = 5(16)/(32+5)\nR'(4) = 80/37
759307
here we go  kinetic energy at point of shot equals potencial energy at top of the squirt\n\n1/2 m v^2 = mgh\n\nm's cancel  (Rained in seattle)\n\nv = Square root (2gh)\n\ncheck my algebra... Note, you must change cm to meters before you can answer
899593
Hello,\n\nIn this link you will have what you need:\n\nhttp://www.bsu.edu/web/fseec/pie/Lessons2/Physics/Potato%20Battery.doc\n\nI believe now will be OK.
773114
F=F1-F2\nF=qm(M1/R1^2-M2/R2^2)\nWhere \nq- universal gravitational constant (6.672 x 10^-11 N m^2/kg^2) \nM1- mass of  407kg\nM2 – mass of  102kg\nM - 21.7 kg\nR1-distance between M1 and the object\nR1-distance between M2 and the object\n\nIn the first case R1=R2=Rso we have \n\nF=qm(M1 -M2)/R^2\nF=6.672 x 10^-11 x 21.7(407 + 102kg) / (0.475)^2=3.266e-6 N\n\n\n\nFor the second question \n\nF=F1-F2=0\n\nSo \n\nM1/R1^2=M2/R2^2\n\nR1/R2 = (M2/M1)^.5\n\nAnd we know that R1+R2=0.475 m\nSo R2=0.475 –R1\n\nThen we have \n\nR1 = (M2/M1)^.5 (0.475 –R1)\nR1=((M2/M1)^.5  x 0.475)/(1+(M2/M1)^.5)\nR1=(1.997 x.475)/(1 + 1.997)= 0.317 m\nAnd R2 must me then 0.475-.317=0.158  m
516392
What? xplain ur qstn well somebody may able 2 help u
36697
We know Q = mass x specific heat x increase in temp.\n          = m x s x (dt)\n\nQ(A)/Q(B) = (m(A) x s(A) x (dt)(A))/(m(B) x s(B) x (dt)(B))\nGiven, Q(A) = Q(B) as they absorb same amount of energy\n       m(A) = m(B), dt(A) = 8, dt(B) = 16\n\nPutting values we get s(A) = 2 x s(B)\n\nSo 1. F 2. T 3. T 4. F 5. F
737967
Draw a free body diagram as shown below: \n\n                [m1]------> T<---[m2]------->89N\n              <--------            <---------\n                f1 =19.6N        f2=29.4N\n\n\nFirst solve for f1 and f2:\n\nf=ukR is the general formula where f is the force of friction, uk is the coefficient of friction, and R is the force normal to the surface. In this problem R is the weight of the block=mg. Now substitute known values\n\nf1=0.2*10*9.8\n   =19.6N to the left\nf2=0.1*30*9.8\n   =29.4N to the left\n\nLooking at the free body diagram:\n\nfor m1:\n\nnet force, f1=T-19.6\n                      \nFrom the formula, f=ma  where f is the net force, m the mass, and a is the acceleration.:\n\nf1=10a\n\nThus,\n\nT-19.6=10a\n         a=(T-19.6)/10\n\nfor m2:\n\nNet force, f2=89-29.4-T\n                  =59.6-T\n\nSimilarly as for m1:\n\n59.6-T=30a\na=(59.6-T)/30\n\nNow a for m1 is also the same a as for m2, therefore:\n\n(T-19.6)10=(59.6-T)/30\nThe zero on each side cancels out. \n3T-58.8=59.6-T\n4T=59.6+58.8\n4T=118.4\n  T=29.6N in the direction shown in the above free body diagram.
1017997
The equations are wrongor insufficient. Post the correct data if you want me to help/ Alternately, you could also message me.
510692
Try: allaboutphysics.com I think that might help.
757859
haha, k  Draw a picture and draw a free-body diagram.\n\nI don't feel like busting out a paper/pencil  so I'll try it in my mind\n\nF(sub g) cos Theta  =  Mg cos(20 degrees)\n\nbecause the Normal force is equal to the y-component of gravity.\n\nTwo similar triangles, the incline and one which has the Force of gravity straight down, and the y compentn equal opposite of the normal force, and X component equal opposite to F\n\n\nlots of physics i'm skipping in my explanation tho
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
685032
plz tell me wat is meant by x.wat angle is it.as i understood ur question QT is a line outside the figure & perpendicular to PQ.hence the angle should be 180 degrees