how do you find acceleration?
824387
The ball goes from 0 to 9/s in .15 secs\n\nSo:    9meters           1\n        ------------  x    ------                  =9 /.15 m/sec2\n           second          .15secs\n\n                                                      =60 m/sec 2
861425
The unit of acceleration is metres per square second\n(or) m/(s)^2\n\nThe formula for force is mass*acceleration\nThe unit is kgm/(s)^2 (or) Newton(N)\n1kgm/(s)^2 = 1 N
872212
Use the certical component to find the time in the air and since the horizontal velocity does not change; Dx = vx * time\n\nDx = v^2 / g\n\nheight = v^2 / (2g)\n\nHope this helps without actually solving it for you.
987378
150m.\nThe equation is:-\nd=ut + (1/2).a.t^2\nWhere u= velocity=40m/s\nt= time =6sec.\na=acceleration/retardation= -5m/s^2 \n\nSource(s):
912569
I think F is the answer. Taking acceleration due to gravity is constant at 9.81ms^-2, but this assumes that air resistance is neglected, and g is a constant.\n\nIf air resistance is NOT neglected, and acceleration due to gravity is not taken to be constant, H is the correct answer. The rest of the options are wrong.
466570
Mass.\n\nF=ma\n\nF is the force\nm is the mass\na is acceleration
750537
acceleration = v^2/r or v*v/r (this is m/s^2 so divide by 9.81 to get g's)\n\nwhere v is the velocity\n\nr is the radius\n\nBe careful, think about the one g the pilot feels from the pull of the earth. ;)
844762
There are jthree  formulae cncerning velocity.These are v=u+at , V^2=u^2+2as ,and s=ut +(1/2) at^2 ,where u= initial velocity,v=final velocity ,a=accelaration.,s= distance travelled and t= time.Now how to decide which formula is to be used?in the problem,just see what they want you to find out and note down what is are already given in the problem, if they give you u,a and t and ask you to find out v,apply formula no.1.But if they gdonot give you time but give the distance or displacement and asks you to find v,donot hesitate to apply formula no,2 and so on.if you have any difficulty to understand.please do not hesitate to write to me through yahoo answersand i shall be glad to assist you
861574
first find the centripital acceleration\na_c = v^2 / r\na_c = 140^2/1000\na_c = 19.6\n\nthe total acceleration at the bottom of the loop is:\na = a_c + g\na = 19.6 + 9.8\na = 29.4\n\nF = m*a\nF = 70*29.4\nF = 2059 Newtons\n\n29.4/9.8 = 3 so the pilot is pulling 3 Gs at the bottom of the loop
382082
use d=V1*t + [(a*t*t)/2]\nd=distance=490m\nV1=initial velocity = 0m/s\nt=time = ?\na=acceleration due to gravity = 9.81m/s*s\n\nusing equation above, t will come up to be:\nt=(square root)(2*d/a)\nt=(square root)(2*490m / 9.81m/s*s)\nt=9.99 sec or 10 sec.
123977
Are you asking for the time it takes the car to accelerate 13.89 m/s....well then the answer is in the question.\n\nPerhaps you are asking for the distance  traveled while accelerating?\nchange in distance = 1/2at^2 + v_i*t\nwhere a is the constant acceleration, t is the time, and v_i is the initial velocity.\nthe car's acceleration = change in velocity / change in time
1082685
Simply equate gravity function with centripetal acceleration.\n\nGMe/(r+h)^2 = V^2/(r+h)....where r is earth radius and h is height\n\nsolve for v
853525
Both the above answers are correct, provided the acceleration is horizontal. If it's not then it's a vectors and trig. question.
979740
During the first 10 m the force increases in a linear fashion. Because it's linear, you can average it for the first 10m: 5 N. This easy way will give the same velocity at the end of the 10 m as if you did it by area under the curve. According to Newton's 2nd law, this gives the object acceleration a1:\nF = m*a1 where F = 5 N and m = 6.90 kg.\n\nUse a1 in the equation V^2 = Vo^2 + 2*a1*10 m where Vo = 0. Solve for V.\n\nFor the second 10 m section, the force is steady at 10 N. Again use Newton's 2nd law to get a2:\nF = m*a2 where F = 10 N and m = 6.90 kg.\n\nUse a2 in the equation V^2 = Vo^2 + 2*a2*10 m where Vo = V from the last part.  Solve for V.
797092
Once an object reaches terminal velocity, acceleration = 0. Thus, you'll be able to determine the terminal velocity by finding an area of the graph where it's a straight line, then taking the slope.
441442
you have your units wrong, do you mean 5m/s or 5m/s^2 and you mean 20m from the ground?\nIf acc=5 when height=20 speed= 14.14 in upwards direction (using 2ax=v^2-u^2)\nIf initial speed is therefore -14.14m/s then at t=1 speed=-4.34m/s so it is still going upwards\nt=2 speed= 5.45m/s it is now going down. Both last results worked out using v=u + at\n(Sorry I assumed you meant 5m/s^2 straight up, you need the angle at which it accelerates aswell)
268064
acceleration = change in speed / time taken\n\nThe units are metres per second per second\n\nFor example: \nIf a car starts at 0 m/s and over a period of 10 seconds increases its speed to 25 m/s then its acceleration is 25/10 = 2.5 m/s^2.\n\n-------------------------------------------\n\nAcceleration is the rate of change of velocity, hence the units.  If you were to plot a graph of velocity against time the gradient would be the acceleration.\n\nJust as velocity can be written v=ds/dt acceleration can be written a=dv/dt.