What is the acceleration of a car moving with a constant acceleration?
Q. What is the acceleration of a car moving with a constant acceleration if in 5s it covers two pints at a distance of 50m? The speed of the car after passing the second point is 20m/s? What is the acceleration of a car moving with a constant acceleration if in 5s it covers two points at a distance of 50m? The speed of the car after passing the second point is 20m/s? Can you solve it for me
Asked by chelzee - Mon Feb 16 03:42:19 2009 - - 5 Answers - 0 Comments

A. I came in 4hours after the question was posted, sorry. . ...___ .../___\ [[[MMM]]]..==> OO...O...I.t=5s ...I<=50m==>I ...Vi=?...==>Vf = 20m/s Here' the simplest way: Displacement = S ...S = (ave. V) t ...Vi +Vf .S = (--- ) t...where S = 50m., Vi = ? , Vf = 20m/s, t = 5s ...2 ...Vi + 20 50 = (---) 5...divide by 5 ...2 ... (Vi + 20) .10..= --- ...2 20 = Vi + 20 .Vi = 0 (It started from rest, the figure shows) Next, from the formula===> a = (Vf - Vi) / t a = ( 20 - 0 ) / 5 ...= 20 / 5 a = 4m/s per sec
Answered by Questor - Mon Feb 16 08:49:56 2009

What is the angular acceleration of the turntable?
Q. The turntable of a record player rotates initially at the rate 33 rev/min and takes 14.9 s to come to rest. What is the angular acceleration of the turntable, assuming the acceleration is uniform? Answer in units of rad/s^2. How many rotations does the turntable make before coming to rest? Answer in units of rev. If the radius of the turntable is 18.6 cm, what is the magnitude of the tangential acceleration of a point on the rim at t = 0? Answer in units of cm/s^2.
Asked by i.p - Sun Apr 15 10:45:22 2007 - - 2 Answers - 0 Comments

A. a) Angular acceleration is the rate of change of angular velocity divided by time. = dw/dt if acceleration is uniform then = (w1 - w0)/(t1-t0) first convert 33 rpm in radians 33 rpm = 33 * 2Pi Rad / 60 s = 3.456 rad/s Then divide by time = 3.456 rad/s / 14.9s = 0.232 rad/s^2 b) angle a is given by equation a = 1/2 ^2t + wt (notice similarity with distance equation) Also is negative since it is a decelaration a = 1/2 (-0.232) *14.9^2 + 3.456*14.9 a = 25.74 rad one rotation is 2Pi so 25.74 rad = 35.74rad/2pi = 4.10 rotations c) At t=0, = 0.232 rad/s^2 and a = *r = 0.232 rad/s^2 * 18.6cm = 4.315 cm/s^2
Answered by catarthur - Sun Apr 15 12:10:52 2007

How do I find acceleration once I know the speed and acceleration?
Q. A plane must reach a speed of 33 m/s for takeoff. How long a runway is needed if the constant acceleration is 3.0 m/s/s?
Asked by Melanie - Wed Aug 5 20:12:17 2009 - - 2 Answers - 0 Comments

A. use the equation v=V(initial) + at to get the time to get to that speed. 11 seconds if t (v initial is zero) then use x=x(initial which is zero) + v(initial) x t + 1/2 a t^2 v initial here is also zero so the equation becomes x = 1/2 a t^2 x=181.5 meters.
Answered by Kaartheek G - Wed Aug 5 20:29:11 2009

What kind of acceleration rate and top speed does the 2008 harley sportster XL 883 and 1200 have?
Q. I am 16 and buying my first harley. I have narrowed it down to either the 2008 sportster XL 883 and the 2008 sportster XL1200. Does anybody know what the acceleration rate and top speed of these bikes? Thanks.
Asked by Condor - Sat Feb 16 22:33:29 2008 - - 12 Answers - 0 Comments

A. Hahahahha a 16 year old on a harley nice... if u looking for acceleration go with a sport bike. I recommend a 2008 kawaski ninja 250r i ride the 07 models and they are a blast
Answered by XLR8 - Sat Feb 16 23:35:09 2008

What is the acceleration of the boat just after the rain starts?
Q. A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 3.00 m/s when it starts to rain. The rain is falling vertically, and it accumulates in the boat at the rate of 10.0 kg/hr. The drag is proportional to the square of the speed of the boat, in the form F_d= 0.5 v^2. What is the acceleration of the boat just after the rain starts? Take the positive x axis along the direction of motion.
Asked by RelientKayers - Fri Apr 13 00:17:14 2007 - - 2 Answers - 0 Comments

A. Make a momentum balance in the direction of motion (the x-direction). The change of the momentum equals the forces acting on the boat. There is only the drag force, which acts against the direction of motion. : d(m v)/dt = -F_d = - k v where k = 0.5 Ns /m Expand the time derivative on the LHS, by applying product rule of differentiation: v dm/dt + m dv/dt = - k v where dm/dt = 10kg/hr is the change of the mass of the boat The acceleration is the time derivative of time velocity: a = dv/dt = - (k v + v dm/dt) / m For the moment,w hen the rain starts all the value on the right hand side are known, thus: a = - (k v + v dm/dt) / m = - (0.5Ns /m (3m/s) + 3m/s 10kg/3600s ) / 250kg = 18.03 10 m/s
Answered by schmiso - Fri Apr 13 03:00:06 2007

What is the acceleration at the coins highest point?
Q. A coin is thrown straight up and reaches its highest point then falls back down again. The up direction is positive. When the coin is at its highest point what is the direction of the acceleration? On the way down? Any help appreciated. Thanks.
Asked by JoJo - Sat Aug 23 09:37:01 2008 - - 4 Answers - 0 Comments

A. The acceleration will always be directed downward, that is because it is the acceleration due to gravity, which is equal to 9.8 m/s^2. The acceleration will stay the same at any point of a free-fall motion. Only the velocities will change.
Answered by rigelali - Sat Aug 23 09:46:43 2008

How to find acceleration from a graph given distance and time?
Q. If given a table of values of distance and time of a cart that began at rest, how do I create a graph in order to receive acceleration? Would it come from the slope? What kind of equation would I use? Thanks!
Asked by ash13007 - Wed Apr 22 21:11:25 2009 - - 4 Answers - 0 Comments

A. graph the equation then to find acceleration take the second derivative.
Answered by Paulie Walnuts - Wed Apr 22 21:29:52 2009

What is the highest acceleration ever reached with a centrifuge?
Q. I'm curious about the possibility of using accelerative forces in fuel refinement, but I need to know how severe of an acceleration people have achieved. I can convert just about any unit you give me, but I'd really like something objective and numerical.
Asked by kevinthenerd - Wed Dec 26 12:59:22 2007 - - 2 Answers - 0 Comments

A. Centrifuges for petroleum fractionation were looked at about 20 years ago. At least one demonstration unit was built but judging from their total absence today you can presume that the technology had big problems relative to distillation. I can think of many issues, safety being at the top of the list followed by capital cost, reliability and maintenance expense.
Answered by goblin - Wed Dec 26 18:01:23 2007

What is the angular acceleration as the blades slow down?
Q. The blades in a blender rotate at a rate of 7340 rpm. When the motor is turned off during operation, the blades slow to rest in 2.15 s. What is the angular acceleration as the blades slow down?
Asked by Sarah - Tue Nov 3 21:19:18 2009 - - 1 Answers - 0 Comments

A. 3413.9535 rad/sec squared 543.347575010877 rev/sec squared
Answered by ITSMEYO - Fri Nov 6 04:11:07 2009

How do you find acceleration based on distance and velocity?
Q. A gun is fired and the bullet is accelerated in the gun barrel, which is 1.00 meters long. The bullet leaves the barrel with a velocity of 600.0 m/s. Calculate the acceleration of the bullet while it is in the barrel.
Asked by skiman322 - Wed Sep 9 22:39:00 2009 - - 2 Answers - 0 Comments

What is the centripetal acceleration and the linear speed?
Q. A drive pulley 8.0 cm in diameter is set to rotate at 9.0 rev/s. A)What is the centripetal acceleration at the edge of the pulley? B)What is the linear speed of a belt attached to the pulley?
Asked by brett h - Sun Sep 27 11:00:21 2009 - - 1 Answers - 0 Comments

A. A) a = r, calculate in rad/sec and plug in. B) v = r
Answered by Engineer-Poet - Sun Sep 27 11:35:11 2009

What is the angular acceleration of the flywheel?
Q. A flywheel 0.600 m in diameter pivots on a horizontal axis. A rope is wrapped around the outside of the flywheel, and a steady pull of 40.0 N is exerted on the rope. The flywheel starts from rest, and 5.00 m of rope are unwound in 2.00 s. A) What is the angular acceleration of the flywheel?In rad/s^2 B) What is its final angular speed? C) What is its final kinetic energy? D) What is its moment of inertia about the rotation axis?
Asked by Natiphy2007 - Tue Nov 20 15:08:06 2007 - - 1 Answers - 0 Comments

A. ___- Radius of flywheel=r=0.600/2=0.300 m The flywheel starts from rest, and 5.00 m of rope are unwound in 2.00 s. initial angular velocity=wi=zero angular displacement=theta(O)=dis tance/radius angular displacement=theta(O)=5/0 .3=16.666 radian angular acceleration= alpha=a! theta(O)=[1/2 ]a!*t^2 angular acceleration= a!= 2*O / t^2 angular acceleration= a!=2*16.66/4=8.333 rad /s^2 A) The angular acceleration of the flywheel is 8.333 rad/s^2 ___ final angular speed=wf=? wf=wi+a!t but wi = 0 wf =a1T=8.333*2=16.666 rad/s B) final angular speed is16.666 rad/s ___ Final kinetic energy =KE=work=force*distance=4 0*5=200 J C) Final kinetic energy is 200 J ___ Kinetic energy=KE=(1/2)I*wf^2 moment of inertia about the… [cont.]
Answered by ukmudgal - Tue Nov 20 16:52:21 2007

How to calculate radial acceleration of the earth around the sun?
Q. How do you calculate the radial acceleration of the earth in its orbit around the sun? Use f=ma to find radial forcethat holds earth in orbit. Check by calculating gravitational attraction between earth and sun using newton's law of gravity. can anyone help me answering this question please? thanks!
Asked by sytycdc403 - Tue Jun 24 15:25:49 2008 - - 5 Answers - 0 Comments

A. The orbit is close enough to a circle that we can model it as such for a simple back-of-the-envelope calculation. Centripetal acceleration = omega^2 r = (2 pi / T)^2 r So you know the period, T, is one year. Convert that to seconds. You can look up the orbital radius, r. Plugnchug. By setting that equal to the gravitational acceleration from newton's law: a = GM/r^2 we determined the mass of the sun.
Answered by ( )Mistress Bekki - Tue Jun 24 15:42:32 2008

What is the acceleration in meters per second^2 and g's during impact?
Q. Falls resulting in hip fractures are a major cause of injury and even death to the elderly. Typically, the hip's speed at impact is about 2.0 m/s. I this can be reduced to 1.3 m/s or less, the hip will not fracture. One way to do the is by wearing elastic hip pads. (a) if a typical pad is 5.0 cm thick and compresses by 2.0 cm during the impact of a fall, what acceleration in m/s^2 and g's) does the hip undergo to reduce its speed to 1.3 m/s? (B) The acceleration you found in part (a) may seem like a rather large acceleration, but to fully assess its effects on the hip, calculate how long it lasts.
Asked by John Q - Wed Sep 5 10:57:19 2007 - - 1 Answers - 0 Comments

A. One bit of data you are missing is the distance of the fall. I will assume 1 meter from the ground to the hip and that there is no resistance in the fall. Using conservation of energy, the speed at impact is v=sqrt(2*10*1) v=4.472 m/s In order to reduce the speed to 1.3 m/s in 2.0 cm, the following equation must be satisfied: 1.3=4.472-(a+9.81)*t and 2.0/100=4.472*t-.5*(a+9.8 1)*t^... the t is the time of compression and a is the deceleration. The easiest way to solve is to isolate a*t from the first equation and substitute into the second 4.472-1.3+9.81*t=a*t 1/25=2*4.472*t-(4.472-1.3 -+9.8... solve the quadratic for t and then solve for a j
Answered by odu83 - Wed Sep 5 12:34:03 2007

What is the magnitude of the angular acceleration of the grindstone?
Q. A grindstone is initially spinning with an angular speed of 8.0 rad/s. The angular speed is then increased to 16rad/s over the next 4.0 seconds. Assume that the angular acceleration is constant. What is the magnitude of the angular acceleration of the grindstone?
Asked by xxtruflipxx - Wed Dec 19 14:07:35 2007 - - 1 Answers - 0 Comments

A. You can use the modified kinematics equations for rotational motion. vf - final velocity. vi - initial velocity. a - acceleration. t - time. vf = vi + at Just plug in your variables. 16 = 8 + a(4) 8 = 4a a = 2 Your angular acceleration is 2 rad/s^2
Answered by Sowmya - Wed Dec 19 14:12:35 2007

Why do we think acceleration of space bodies are due to dark energy?
Q. After the big bang, the stars and other bodies are moving away from each other. As they are moving away from each other, gravitational force by each body on others reduce due to the increase in distance. When the pulling force reduces obviously there should be acceleration. Am I wrong?
Asked by Ananth Murthy - Thu Sep 20 14:54:05 2007 - - 6 Answers - 0 Comments

A. Mastermind is confusing dark energy with dark matter. Dark matter is assumed to exist in order to make observation of galaxy rotation conform with gravitational theory. Dark energy, on the other hand, has been suggested as a possible reason for the apparent acceleration of the expansion of the known universe. In my own Fractal Foam Model of Universes, the expansion of our universe is a result of reverse-time expansion of a sub-universe, whose cosmic foam is our ether foam. In the sub-universe, great walls of galaxies are stretched to the breaking point; like a slowly fizzing foam, the bubbles pop; this causes pressure waves to radiate thru the rest of the sub-universe's cosmic foam, which is our ether foam. Due to time reversal from… [cont.]
Answered by Philip J - Thu Sep 20 15:50:14 2007

How do you graph instantaneous acceleration given velocity graph?
Q. I am confused! I know instantaneous acceleration is tangent to acceleration graph. But then does that mean instantaneous velocity equal to acceleration since acceleration is tangent of velocity?
Asked by Rushi P - Sat Sep 12 23:33:40 2009 - - 1 Answers - 0 Comments

A. Instantaneous acceleration is the tangent of the **velocity** graph, not the tangent of the acceleration graph. That's probably why you're confused. If you have a graph of acceleration versus time, and you want to know the instantaneous acceleration at any particular time, just read the graph at that time - you don't need to measure the tangent or calculate the slope. You only need to use the tangent or slope if you need to figure out the acceleration from the graph of velocity versus time.
Answered by unknown - Sat Sep 12 23:44:15 2009

What is the maximum acceleration of the top of the building?
Q. During an earthquake, the top of a building oscillates with an amplitude of 32 cm at 1.4 Hz. What is the maximum acceleration of the top of the building?
Asked by 1st Baby Due 04-22-10 - Wed Apr 15 17:56:50 2009 - - 1 Answers - 0 Comments

A. In SHM, max velocity = w*amplitude and max acceleration a = w^2*amplitude. w = 2pi*1.4 rad/s, so a = (2pi*1.4)^2*0.32 m/s^2.
Answered by kirchwey - Wed Apr 15 20:40:58 2009

How can you have different size forces but the same acceleration?
Q. The forces on each mass are different but, in the same direction as the common acceleration. How can you have different size forces but the same acceleration? ~ I don;t really get it don't you have to have them be equal to get the same acceleration?
Asked by Kaye Y - Thu Oct 8 01:57:38 2009 - - 6 Answers - 0 Comments

A. F=ma m1a=m2a, m1~=m2 implies forces different but acceleration, a, is the same
Answered by unknown - Thu Oct 8 02:05:53 2009

What acceleration and speed would the proton attain if the field accelerated a distance of 1 cm?
Q. Beams of high speed protons can be produced in "guns" using electric fields to accelerate the protons. (A) what acceleration would a proton experience if the guns electric field were 2.00x10^4 N/C? (B) What speed would the proton attain if the field accelerated the proton through a distance of 1.00 cm?
Asked by Aggie Love :) - Mon Jul 13 22:45:31 2009 - - 1 Answers - 0 Comments

A. A) acceleration a = F / m = E e / m = 2 x 10^12 m/sec^2 (nearly) after using the values and simplifying. B)using V^2 - 0 = 2 a S so that final speed V = 2 x10^5 m/sec (nearly) after using the values and simplifying.
Answered by knr - Mon Jul 13 23:00:21 2009