if s= 0.5(u+v)t  find t if the distance travelled was 10.3m and the body reached 8.2m/s from rest.?
957344
t=2.512 seconds\n\n10.3=0.5(0+8.2)t\nTherefore \n\n1 = 4.1t/10.3\n\n1/t = 4.1/10.3\n\nt = 10.3/4.1 = 2.512 seconds
1009103
We have a formula:\nSpeed = Distance / Time\nhere Time= Distance /Speed.==D/S\nD=100km in both cases\nS1= 125km/hr; S2= 80km/hr\n\nT1=D/S1==100/125 hrs === 4/5hrs===4/5*60mins= 48mins\nT2=D/S2==100/80 hrs ===5/4 hrs ===5/4*60 mins= 75mins\n\nTime saved = T2-T1==75-48=27 mins.\n\nIs this clear buddy
751049
First find Vo:  Vo^2 = Rg/sin(28)  Where R = range & 8 = theta\n\nUsing your numbers, Vo = 41.523 m/s\n\nImpulse = change in momentum ----> Ft = mVo---> F = mVo/t\n\nSo, F = (.0431kg*41.523m/s)/.00438sec = 408.59 N\n\nVoila!
877088
E = hc/λ \nλ = 1.2 x 10^-10 m\nc = 3 x 10^8 m/s\nh = 6.626068 × 10^-34 m2 kg / s\n\nE = 6.626068 × 10-34 * 3 x 10^8 / 1.2 x 10^-10\n\nE = 1.7 x 10^-15 J.
29068
pier_haringsma is correct.  You do not need any additional information.  Let \nD = total distance of the trip  \nD/2 = the distance the driver has gone so far\nT1 = amount of time he took to travel the first half of the trip\nT2 = the complete length of time of the whole trip (both halves).\n\n   Now, he wants the average ( D / T2 ) to equal 30.  His average so far is ( (D/2) / T1 ) = 15.\n\nD/T2 = 30, so D = 30*T2\n(D/2) / T1 = 15, so D = 30*T1\n\nD = 30*T1 = 30*T2, so T1 = T2.\n\n  So the driver wants the total time for the trip (BOTH halves), T2, to equal the time for the first half of the trip, T1.  This means he would have to travel the second half of the trip in no time, which is impossible.
929521
V=(4/3)pir^3\ndV/dt=4pir^2*dr/dt\n10=4pir^2*dr/dt\nA=4pir^2\ndA/dt=8pir*dr/dt\n=4pir^2*dr/dt*/r\n=10/r\nwhen r=8 dA/dt\n=10/8cm^2/sec
83305
I suppose the problem was already due so I'll finish it for you now.\n\nF=m*a\na=F/m= 2*10^6N/1.2*10^6kg = 1.67m/s^2 * g/9.81m/s^2 = 0.17g
1149306
I think you can use a simpler equation than the one for distance.\nWe don't need to find distance, just time.\n\nv = u + a.t\nv= 0, u= 64 f/s, a= -32f/s^2\nsolve for t. \nt = 2 s\nThats the time it takes to go up just before it starts to fall again.\nThe time taken to fall is going to be the same: another 2 s\n\ntotal time: 4 s
124298
Answer to #2\nt=1.565 seg\nV= 2.556 m/seg\nt=1.565 seg, the same.
569320
v=vo + at\nv=0 + (9.8)(8)\nv=78.4\n\ny=1/2(vo + v)t\ny=1/2(78.4)(8)\ny=1/2(627.2)\ny=313.6m\n\nB
752231
Velocity = distance / time\n\nV = 800km/12 hours\nV= 66.667 km/hour\n\nOr, if you're looking for the velocity in meters/second, it's \n\nV=800000 meters/ 43200 seconds\nV=18.5 meters per second
816469
well F=mv/t given: m=78kg, v=24m/s, t=60s\n\nF=78(24)/60=31.2N
1120781
do your own homework\n\nits probably 0.26 kg M/s ^2 since its a velocity\n\nyou also probably need to convert it to rad/s whatever that is
125809
The struggle here is that you provide too little information to recognize this problem as coherent.\n\nWe may assume that the acceleration of gravity is 9.8 m/s/s or 32 ft/sec/sec.\n\nThe velocity(V) of an object is equal to its initial velocity(v) plus its acceleration(a) times time(t). \nV = v + at or V = v - 32t (in feet per second)\n\nYour object will reach maximum height when the velocity is zero or when the initial velocity is equal to 32 ft/sec/sec * time. With your list of times, you'd just multiply by 32 to solve for the initial velocity, if that is what the problem is actually looking for. There are many other permutations of this data, but hopefully that is a start for you.\n\nGood luck.
918281
Let's assume your range of vision was 0.2 miles infront of you, and 0.1 miles behind with the mirror.  (I'm pulling these numbers out of my behind.)\n\nAt t=0, the car was at position x = -0.1 relative to you.\nAt t=5 seconds, it was at x = 0.2.\n\nSo the speed (relative to you) is 0.3 miles / 5 seconds = 0.06 miles per second = 216 miles per hour.  Since you were doing 85 mph, it real speed was 301 mph.\n\nAre you sure this took only 5 seconds?  I picked some pretty small numbers for the range of vision and this still results in some wicked speed.
107382
v^2-u^2 = 2gh\nHere, v=1, g=-10m/s/s, h=0.6m\nHence, u^2 = 1+12=13\n=> u = 3.606 m/s
127000
This problem is solved with the application of work and energy principle which states that total work done on a moving particle is equal to the change of its kinematics energy:\nW=T2-T1\nHere the only work done on the particle is dot to gravitation force and is equal to -m*g*h:\n\n-m*10*0.45 = (1/2)*m*(1)^2-(1/2)*m*(v)^2 --->\nm*v^2=(4.5+0.5)*m*2=10*m --->\nv=sqrt(10) m/s\n\nv is equal to sqare root of 10.(approximately 3.162 m/s)