d=2pie(r)\nh=T\nS=d/h\n=mph
202828
The average distance from the Earth to the Sun is 149,597,890 km. Therefore in one year the Earth travels a distance of 2*Pi*(149,597,890)km. This means that the velocity is about: \n\nvelocity=2*Pi*(149,597,890)km/ year\n\nand if we convert that to more meaningful units (knowing there is 365 days in a year, and 24 hours per day) we get: \n\nvelocity=107,300 km/h (or if you prefer 67,062 miles per hour) \n\nSo the Earth moves at about 100,000 km/h around the Sun (which is 1000 times faster than the speeds we go at on a highway!)
266539
let V1= 4 mph, V2= 5 mph\n\nV1*t + V2*t = 3\nt= 20 min\n\n19. a 20 min\n\n\n14.16/3 = 4.72\n\n4.72*14 = 66.08\n\n20 c 66.08
514565
1. d=t, any form of d=t+b would be parallel. b = any number. That is a line.\n\n2. d=t would also work here.\n\n3. Well if you go 60 MPH and you wnat to know how many miles a minute you can go convert using (60miles/hour)*(1hour/60minutes) and you get 1mile/minute. If you need feet, 1 mile is 5280 feet, so 5280 feet/minute
574541
r = d(pi)/180\nd = r(180)/pi\n\n\nr = -0.00923157  pi = 3.1415\n\nd = (-0.00923157)(180) / 3.1415\nd = -0.528929999
482103
v(t)= -gt - ve(ln((m - rt)/m))\n\nx(t) = ∫v(t) dt\n\n= ∫[-gt - ve(ln((m - rt)/m))] dt (Note ∫lnu du = ulnu - u + k)\n\nLet u = (m - rt)/m Thus du = -r/m dt ie dt = -m/r du\n\nSo x(t) = -½gt² - ve*-m/r (u(lnu - 1) + c\n= mve/r [ (m - rt)/m * (ln((m - rt)/m)  - 1] - ½gt²\n= ve(m - rt)/r * (ln((m - rt)/m) - 1) - ½gt²\n\nIf g = 9.8 m/s^2, m = 30000 kg, r = 155 kg/s, and ve = 3000 m/s and t = 1 minute = 60 s\n\nx(60) = 3000 * (30000 - 155 * 60)/155 * (ln (30000 - 155 * 60)/155 - 1) - ½ * 9.8 * 60²\n\n≈ 138390 m\n\n= 138.39km
403122
I solved the problem by stating that P would travel distance d in time t\nd=20*t+.5*.3*t*t\n\nand Q would travel d+20 in time t\nt is the time to collision\n\nd+20=30*t\nd=30*t-20\nSubtract the equations and solve for t:\n\n0=.15*t*t-10t+20\nt=2.064 seconds\n(the second root is superfluous)\nd=41.9m\n\nWhich proves collision\n\nj
469570
Let D = distance between home and school.\n\nUsing the formula : distance = speed * time, we have :\n\nFrom home to school : D = x * T1\nFrom school to home : D = kx * T2\n\nTherefore, x * T1 = kx * T2\nDividing through by x gives :\nT1 = k * T2\n\nWe know that T1 + T2 = 1 (total time travelled)\nTherefore, T2 = 1 - T1.\n\nSubstituting this into T1 = k * T2 gives :\nT1 = k (1 - T1)\n\nFrom this, we find that :\nT1 = k / (k + 1)\n\nSubstituting into 'home to school equation' gives :\nD = x * k / (k + 1)\n\nTherefore total distance covered\n= 2D\n= 2 * x * k / (k + 1) miles.
433168
Listening to music always helps!\nw=w’*t, w=500rpm=500/60 rps=2pi*25/3 rad/s, t=2.6s, then \nw’=w/t=2*pi*25/3/2.6=20.14 rad/s^2;\nu=w’*t^2/2=w*t/2=25/3*2.6/2=10.83=~10 full revs
516833
d = rt   distance equals rate times time.\nd = distance = 5216\nt = unknown\nr = 375 and 275\n\n5216 = 375t + 275t\n5216 = 650t\nt = 8 hours\n\nThe two jets traveling away from each other in opposite directions is like one standing still and the other moving at the rate of the combined velocities.
510144
use the formula\nA=2pir(h+r)\n=2pi*7*21\n=294pi
235659
Start by multiplying both sides by D this gives you\n(D*P) = (2*s*t)\n\nthe divide both sides by 2t which gives you\n(D*P)/(2*t) = s\n\nor\n\ns = (D*P)/(2*t)
504874
If initial speed is V0 and the speed after a time t second is V,\n\nAcceleration is change in speed per unit time.\n\nAcceleration a = (V - VO) / t.\n\nThe average speed during this time t is \n\nV (average) = (V + V0) / 2.\n\nNote; if any quantity is changing uniformly, the average is found by adding the initial and final value and dividing the sum by 2.\n\nThe average speed can also be determined by dividing the total distance by total time of travel.\n \nV (average) = S / t.\n\nTherefore S/t = (V + V0) / 2.\n\nS = (V + V0)* t / 2.\n\nFrom a = (V - V0) / t we have V = V0 +at.\n \nTherefore S = {V0 +at + VO} * t /2 \n--------------- = {2 V0 t + at t } /2\n\n-------------- S = V0t + 0.5 a tt
338948
D(istance) =  S(peed)   x   T(ime)\nor   D / T   =  S\n12 minutes =  .20 hr  \n11km /  .20h  =    55 km/h
365984
dx/dt=1/2\ndy/dt=2t\ndy/dx=dy/dt/dx/dt\n=2t/1/2=4t\nat t=-1 slope=-4\nat t=1 slope=4
512901
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
384322
P=P (momentum before = momentum after)\nmv=mv\n(55kg)(4.6m/s) = (230kg)v\nv= ((55kg)(4.6m/s))230kg\nv=1.1m/s
63705
1. Dy = 1/2 A*t^2 + Vy*t, Vy = 0 (speed is horizontal),\nso t = sqrt(66m/(1/2*9.8m/s^2)) = 3.67 s.\n   Dx = Vx*t = 6 m/s * 3.67 s = 22 m.\n\n2.(a)  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso t = sqrt(1.150m/(1/2*g)) = 0.484 s\n  (b)  Dx = Vx*t, so Vx = Dx/t = 0.385m/0.484s = 0.795 m/s\n\n3.  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso Dy = 1/2*9.8m/s*(2.9s)^2 = 41.2 m\n\n   Dx = Vx*t = 2.4 m/s * 2.9 s = 6.96 m\n\n4. Dy = 0, Dx = 97 m, a = 50 degrees\n   Dx = Vx*t = V*cos(a)*t, \nso t = Dx/(V*cos(a))\n   -1/2*g*t^2 + Vyt = Dy = 0 ( Vy = V*sin(a))\ndivide by t, so -1/2*g*t + V*sin(a) = 0, or\nt = V*sin(a)/(1/2*g)).  Equate the two expressions for t:\nV*sin(a)/(1/2*g)=Dx/(V*cos(a)), multiply both sides by V\nV^2*sin(a)/(1/2*g) = Dx/cos(a)\nV= sqrt(Dx/cos(a)/sin(a)*(1/2*g))\nV = 31.1 m/s