How many revolutions does the wheel turn through during this time interval?
970150
We calculate in radians for convenience.\n\nFixed part of rotation from initial v is\n48 * 7 = 336\n\nVariable part from additional acceleration is\nd = 1/2 a t^2\nwhere acceleration a is\n(120 - 48) / 7 = 10.3 r/sec^2\ntherefore d is\nd = 1/2 * 10.3 * 7 * 7 \nd = 252\n\nTotal d is then\n252 + 336 = 588 radians\nand total revolutions is 588 / (2*pi) = 93.6
791900
Ok.  From the hint start by figuring out the ratio of wheel speeds\nCircumference of A =2*pi*10\nCircumference of C = 2*pi*25\nso C/A=2.5  ie A must go around 2.5 times to make C go around 1 time.\nThere are 2*pi radians to 1 revolution so 100 rev/min =200*pi rad/min.  A must go 2.5 times as fast as this or 2.5*200*pi=500 rad/min\nAccelleration is the derivative of velocity which means that we must integrate the angular acceleration to obtain the speed at any given point of time.\nIntegrating 1.6rad/s^2 with respect to time and setting velocity equal to zero at time equals zero we obtain angular velocity= 1.6*time elapsed (rad/sec).  Note that this speed is in "per seconds" and we need it to be in "per minutes" so we must multiply by 60 for the units conversion.\nThis gives us the angular speed in rad/min for any chosen passage of time.  We shall set this equal to the goal speed of 500*pi rad/min as mentioned above. So\n1.6*time elapsed*60 =500*pi\nSolving for time elapsed we obtain a time of 1.63 minutes.
893172
Since the propeller has three blades it will look the same every 1/3 of a revolution.\n\nThe propeller does 13.4 rev/s, or 0.0746 s/rev (Just take 1/13.4).\n\nMultiply how many seconds it takes to do one revolution by 1/3, because this is how often the propeller will look like it did in the beginning:\n\n0.0746*(1/3) = 0.0249s\n\nSo, the shortest time between flashes to make the propeller appear stationary is 0.0249s.\n\nThe next shortest time will be twice this (0.0498s) because in 2/3 of a revolution the propeller will again look as it did in the beginning.
174325
Yes.  One ROTATION is one day (which varies for whatever planet you're on...for Earth it is 24 hours)\n\nOne REVOLUTION is one year for the planet (again, varies for each planet...earth is 364 1/4 days...and the reason for LEAP YEAR).
725445
You can't without the radius of the wheel...m/s is a tangential speed.
727926
88 revolutions per 5 seconds = 88/5 rps = 17.6 rps (revolutions per second) or in rpm...\n=  88*60/5 rpm = 1056 rpm (revolutions per minute).
2556
You're really close! The answer is actually 1:03:11. You were off one minute because you forgot to borrow a minute from the 34 so you could subtract the seconds.
626709
Four hours.  The earth rotates through a full 360 degree circle in 24 hours, this means it rotates at the rate of 15 degrees and hour (360 divided by 24).  So it will take 4 hours to rotate through 60 degrees.
92660
jack rear of vehicle up. put a small mark on the driveshaft one you can see and spin wheel real slow and count how many times ur wheel turns b4 driveshaft makes a full revolution.
745809
1 revolution  =2*3.1416 (pi) radian.\n\nSo, a revolution = pi radias i.e; 3.1416 radian
1099800
24,901.5 miles. Spinning one revolution in 23 hours, 56 minutes and 4.1 seconds. Sit on a hillside in Equador with some great local product and your flying at like over 1000 mph. Dizzy yet?
426870
I picture an axle with one wheel bigger than the other for this problem. Since they're not spinning, wheel A has to cover the same amount of distance wheel B covers in the same amount of time.\n\nWheel A has radius 12cm and rotates at 35 radians/second:\n12 cm * 35 rad/sec = 420 cm / sec.\n\nWheel B must also cover 420 cm each second. So we work backwards:\n\n(420 cm / sec) / 17 cm = 24.7 rad / sec\n\nSo you're right, B is the correct answer.
746380
Torque = change in angular momentum over change in time\n\nT=Iw/t      T=mrrw/t     T=1.1(.09)(93.6)/1.7  T=5.45 N/m
959353
All of the power from the motor goes to friction because they just let the wheel coast and burn up all the energy.  But, I'm sure that wasn't the intended answer for the problem\n\nYou need to make a few preliminary calculations:\nFirst you need the Inertia of the flywheel which you can find given the mass and radius of the wheel.\n\nThen you need to calculate the angular acceleration required to bring an object at rest up to 5.88 rad/s in 0.531 sec.  Using this, you can get the torque of the motor - friction.\n\nFinally, calculate the angular acceleration required to bring an object spinning at 5.88 rad/s down to rest in 26.8 revolutions.  Using this, you can get the torque of the friction by itself.\n\nAdd the second result from the first to get the torque of the motor.  Take the friction torque and divide by the motor torque times 100 to get percentage of friction.
819223
If I understand correctly, one revolution of the disk is equal to one period.  You need to know the time (T) taken for one period.\n\nFrequency (f):\n f = 1/ T\n T = 1/ f\n T = 1/ 2200\n T = 0∙000 454 545....\n T ≈ 454∙454 µs\n\nAngular speed (w):\n w = Θ/ t\n w = 2π/ 454∙454 x 10^-6\n w = 6∙283 185 307.../ 454∙454 x 10^-6\n w = 13823∙007 68...\n w ≈ 13,823 rad/s\n\nIf the period is 20 for one cycle then:\n w ≈ 13,823 / 20 rad/s\n w ≈ 691∙15 rad/s
858881
12.25cycles/20sec = 0.6125 cycles/sec, = 0.6125Hz, which is the frequency.  The period is the reciprocal of the frequency, which equals approx. 1.6327 seconds.\n\nI didn't count that first half cycle because it looks longer than the rest.  I started counting at the first positive zero-crossing at time 2.6, and counted 12.25 cycles from there to time 22.6\n\nThis is coming from someone who has made his living for the last few decades looking at and interpreting thousands of such displays.
763203
12 times\nin 12 hours the hour hand only completes one circle when each hour the minute hand completes one circle.\n\nits 11 since at 12 they dont cross just over lap