April 18, 2014

Search: Science (Physics)...........F=ma and tension

Number of results: 162,344

The force on the rope is tension, which is equal to mg-ma But this same tension is spinning the wheel: angularacceleration=Tension/I a/R=Tension/I aI=R(mg-ma) solve for a, then you have it. check my thinking.
Sunday, May 11, 2008 at 11:26am by bobpursley

Physics help!
F= ma= Tension-Weight ma+Weight=Tension (4125kg)(0.0300 * 9.8)+ (4125kg)(9.81)=Tension Calculate and this will give you the answer for maximum force. For part (2) You have to subtract them (ma-weight)
Monday, September 26, 2011 at 6:59pm by J

Tension= friction+ma-gravity down the ramp tension= mu*mg*cos30+ma-mgsinTheta
Thursday, September 30, 2010 at 8:18pm by bobpursley

a = [(mB-mA)/(mB+mA)]*g = (3/4)g You can derive it by writing free-body motion equations for each mass, and eliminating the cord tension force, T. mA*a = T - mA*g mB*a = mB*g -T (mB + mA)*a = (mB - mA)*g
Friday, November 26, 2010 at 1:43am by drwls

well, look at the net accelerating force. Net accelerating force=totalmass*a 1kg*9.8-.2*2.5*9.8=(3.5kg)a solve for acceleration. Tension? Look at the tension pulling the cart: Tension= ma+friction=2.5*a+.2*2.5*9.8 check all that against your force diagram. If you wish, compare...
Friday, January 21, 2011 at 7:55pm by bobpursley

find the moment of inertia for the two disks (each has half the mass). tension*radius= momentinertia*angular acceleartion and tension= mg-ma=mg-m*angacceleration*r so you have two equations, two unknowns, solve for tension, and angular acceleartion. on the way up, tension*...
Saturday, March 17, 2012 at 5:22pm by bobpursley

THe monkey is putting a tension on the rope of mg+ma. THe opposite monkey is putting a tension of -mg. The resultant force on the rope is ma. F=ma ma= (2m)A A= 1/2 a Therefore, both monkeys accelerate upwards at the same rate, which is half the rate as the acceleration (...
Wednesday, April 11, 2012 at 8:23am by bobpursley

Science (Physics)...........F=ma and tension.
Oh okay Thanks Bob
Saturday, September 29, 2007 at 9:14pm by ~christina~

Science (Physics)...........F=ma and tension.
the woman has a force pulling her back (ie, the boxes are pulling back with 50N, and the shoes on the floor are pulling forward 50N.) When doing F=ma, M is in kg a=50/10= 5m/s^2 The acceleration of the smaller box is the same, 5m/s^2, so T= ma=4kg*5m/s^2=20N
Saturday, September 29, 2007 at 9:14pm by bobpursley

tension= force repulsion= kQ1*Q2/distance^2 when cut, then T=ma or a= tension/mass
Wednesday, February 24, 2010 at 9:06pm by bobpursley

T1 is the tension in the coupling between the locomotive and the first car, T2 the tension in the coupling between the first and the second car. Newton’s 2 law for the first car is ma = T1-T2, .......(1) for the second car: ma =T2,............(2) Plug (2) in (1) ma = T1 – ma, ...
Monday, June 11, 2012 at 6:35pm by Elena

Three boxes are connected by cords, one of which (mA) wraps over a pulley having negligible friction on its axle and negligible mass and two boxes (mB & mC) are hanging. The masses are mA = 22.0 kg, mB = 40.0 kg, mC = 16.0 kg. -When the assembly is released from rest, what is ...
Thursday, February 8, 2007 at 8:16pm by **chocolatekisses

Net force= Tension + normal force + weigh I used net force= ma. I got Normal force= ma + mgCos(angle) Normal force to be 505N. Tension= ma + mgSin(angle) = 396 Is this correct? or is the Net force = o.
Saturday, December 8, 2012 at 5:53pm by Antonio

An elevator with a wieght of 27.8 kN is given an upward acceleration of 1.22 m/s^2 by a cable. a) what is the tension in the cable. b) what is the tension when the elevator is declerating at the rate of 1.22m/s^2 but is still moving upwards? what exactly is a kN? to find the ...
Monday, November 13, 2006 at 9:50pm by Chris

mass on light side = M tension = T T-Mg = Ma T = Mg + Ma mass on heavy side = M+m (M+m)a = (M+m)g - T so (M+m)a = (M+m)g - Mg - Ma Ma + Ma + ma = m g (2 M+m)a = m g a = m g/(2M+m) which we could have guessed in the beginning :) Forces on m Reaction force up = F force down = m ...
Saturday, February 8, 2014 at 11:05am by Damon

ma = tension - weight so ma + weight = tension so 4700(.06)(9.8) + 4700(9.8)
Thursday, February 3, 2011 at 7:42pm by nine

AP lost
Tension= mg +- ma where a is the acceleration. + when the elevator is going up (it adds to tension), and - going down, it reduces tension when falling. a= change velocity/time (rest means zero velocity)
Monday, October 15, 2007 at 11:57pm by bobpursley

Then for FTQ (force of tension of string Q): F=ma =(8)(25) =200 m/s^2 FTR F=ma =(4)(25) =100 m/s^2 ??
Monday, March 9, 2009 at 9:18pm by Anonymous

The tension is mg+ma The work done by tension is tension*13 work done by weigh? mg*13 worktension=changePE+KEchange solve for KE change, then velocity from that.
Wednesday, October 29, 2008 at 1:41pm by bobpursley

physics help- tension
Is the time measured from the time the weights are released? If so, you know that the acceleration rate is .35/0.7 = 0.5 m/s^2. For which rope do you want to know the tension? They will be different. Writing the free body F = ma equation for both weights will let go solve for ...
Sunday, December 6, 2009 at 10:01pm by drwls

Sorry, Anonymous, we do not have latex installed, at least yet. I presume the figure looks like this, but a mirror image of it: o-----T-----mA | | T | | mB The letter o stands for a pulley. The answer to the question lies in the formula F=ma Force exerted on the system is ...
Saturday, October 23, 2010 at 7:53pm by MathMate

You need a tutor. Tension=ma+mg is in SI units. Tension was given as 29,400 Newtons. You have to consider units.
Wednesday, February 8, 2012 at 2:38pm by bobpursley

a) steady? no acceleration. Tension= mg b) tension= mg-ma
Wednesday, June 22, 2011 at 5:40pm by bobpursley

physics help plz
(a) When the buckets are at rest, the tension in the cords just equal the weight that is suspended beneath them. Each bucket weighs m•g = 3.5• 9.8 = 34.3 N. The lower bucket has gravity down of 34.3 N and the tension (T) in the lower string up, the acceleration is 0, so the ...
Sunday, May 6, 2012 at 4:26pm by Elena

well, if the blocks are moving, then a= F/Masstotal=1/eachmass tension=ma= 107*eachmass*1/eachmass=107N b.tension= massbehind*a= 54N
Sunday, October 9, 2011 at 6:04pm by bobpursley

I wondered if someone would check my work concerning an Atwood's Machine. (I am led to believe the answers for these two questions should be the same but I'm not coming out that way.)Thank you! 8.) Caluclate the tension in the Atwood's Machine string for the case when m2 is ...
Saturday, September 22, 2012 at 1:37pm by Miaow

The tension in the string will be equal to T, which connects both masses A and B. For mass A, the force down the incline is Ma*g*sin(θ) which equals T. For mass B, the force due to gravity is Mb*g which also equals T. Thus, equating the two values of T, we get: Ma*g*sin...
Wednesday, June 20, 2012 at 11:52am by MathMate

To solve this problem, you'll need the tension formula which is T=Ma. T is tension, mass M is weight Mg and tension or in other words the mass that is on the surface, & a is acceleration. M=3kg a=2.5m/s2 T=? Just plug it into the formula... T=(3kg)(2.5m/s2) T=7.5N
Saturday, November 30, 2013 at 5:32pm by Lily

tension= mg+ma= m(g+a)
Friday, February 5, 2010 at 12:30pm by bobpursley

tension=mg+ma a=3.5/5 m/s^2
Wednesday, February 13, 2013 at 8:21pm by bobpursley

Tension of rope = T N Mass of girl, m = 45 kg Acceleration = 3.0 m/s² T = ma = 135 N Mass of boy, M = 65 kg T = MA, solve for acceleration A.
Wednesday, October 28, 2009 at 10:08pm by MathMate

25 gram mass is given an upward acceleration of 30 m/s^2 by a rope. need help finding tension in the rope? not sure how to set up the problem?? Tension is the reaction force. Clearly you have two forces in opposite directions. Let T be the tension in the rope. The net force ...
Monday, September 25, 2006 at 12:11am by greg

Tension=mg+ma solve for a
Monday, June 1, 2009 at 8:47pm by bobpursley

tension=mg+ma solve for a
Wednesday, August 31, 2011 at 8:06pm by bobpursley

f=ma kx=ma x= ma/k horizontally If at 30 deg above. x= ma/k (1/cos30)
Thursday, June 11, 2009 at 1:23pm by bobpursley

Tension=mg+ ma= m(g+a) where a is + going up.
Tuesday, October 4, 2011 at 8:47pm by bobpursley

Science (Physics)...........F=ma and tension.
2 crates shown in the figure below are on a horizontal frictionless surface. The woman is wearing golf shoes for traction as she applies a horizontal force of F= 50.0 N to the 6.00kg crate. The ropes are of negligible mass. Smaller box weighs 4.00kg a) draw a free body diagram...
Saturday, September 29, 2007 at 9:14pm by ~christina~

TRIAL 1 A. ) Bag A + 6 stones Mass (kg) - 0.037 kg Mass of weight hanger + slotted weights - 210 g/ 0.21kg Tension 1 (N) - ?? Acceleration (m/s2) - ?? B. ) Bag B + 4 stones Mass (kg) - 0.08 Mass of weight hanger + slotted weights - 210 g/ 0.21kg Tension 1 (N) - ?? Acceleration...
Wednesday, May 4, 2011 at 4:07pm by nik

revise that to tension=Ma, it is the mass on the table * acceleration
Saturday, October 10, 2009 at 1:29pm by bobpursley

physics grade 11
tension in rope= mg+ma
Sunday, October 17, 2010 at 2:10pm by bobpursley

tension=mg+ma work done= force*distance
Sunday, October 17, 2010 at 7:11pm by bobpursley

tension= mg+ ma constant speed means a=0
Tuesday, August 7, 2012 at 9:40am by bobpursley

friction on the ramp=mg*mu*cosTheta downward force due to weight=mg sinTheta Net force=mass*acceleration Tension-friction-gravityforcedown=ma solve for tension check my thinking.
Thursday, September 27, 2012 at 7:41pm by bobpursley

F=totalmass*a F=(2M)a a= F/2M but this is the a for each block, so the tension pulling the second block is tension= ma= m(F/2m)=1/2 F
Thursday, November 18, 2010 at 4:06am by bobpursley

Consider the following equation in the y-direction. Fy=T+Fg=ma Or in this case... T+Fg=(mv^2)/R Tension equals to 0 because when the pail of water is at the highest point, tension is nonexistent. Therefore... mg=(mv^2)/R Cancel out mass g=(v^2)/R (Rg)^(1/2)=v Plug in values.
Monday, October 19, 2009 at 10:46am by Young J. Hong

Tension= mg+ma where a is the accelearation upward (downward will be negative)
Thursday, October 14, 2010 at 1:56pm by bobpursley

AP Physics
First, find the Force in the y-direction. Fy=T+Fg=ma ..or in this case T+Fg=(mv^2)/R Tension equals to 0 because when the pail of water is at its highest peak, there, tension does not exist. so now we have mg=(mv^2)/R Cancel out mass to get... g=(v^2)/R (gR)^(1/2)=v Plug in ...
Thursday, September 24, 2009 at 1:52am by Young J. Hong

Tension=mg-ma solve for a. Then, speed equals Vf^2=2a*8
Tuesday, September 30, 2008 at 4:58pm by bobpursley

12th grade Physics
tension=mg+ma=m(9.8+.8)= to two decimal places.
Saturday, October 24, 2009 at 1:09pm by bobpursley

the pulling force is m2*g the masses being pulled is m1+m2 F=ma a= F/m= m2*g/(m1+m2) tension in the cable: Tension=m1*a
Monday, September 26, 2011 at 6:22pm by bobpursley

A light rope passing over a frictionless pulley connects two objects. One object has a mass of 10kg. The tension in the rope is 90N. Calculate the acceleration and the mass of the second object. Object 1 Fg-Ft = ma 100 - 90 - 10*a 10/10 = a a = 1m/s^2 Object 2 Ft-Fg = ma 90-...
Tuesday, April 1, 2008 at 10:20pm by Alex

A light rope passing over a frictionless pulley connects two objects. One object has a mass of 10kg. The tension in the rope is 90N. Calculate the acceleration and the mass of the second object. Object 1 Fg-Ft = ma 100 - 90 - 10*a 10/10 = a a = 1m/s^2 Object 2 Ft-Fg = ma 90-...
Tuesday, April 1, 2008 at 11:40pm by Alex

sum of forces-> F=ma F(friction)= ufn* you don't need this F(applied)-F(friction)=ma f(a)-f(f)=ma f(a)/ma=f(f) add numbers
Monday, November 22, 2010 at 4:35pm by matt

dmA/dt = cmB^2/mA where c is a constant mB=m-mA so dmA/dt = c (m-mA)^2/mA = (c/mA)(mA^2-2mmA+m^2) = cmA-2cm+cm^2/mA integrate with respect to t
Tuesday, March 9, 2010 at 8:14am by FredR

the tension T=ma, since the only acceleration is that due to gravity.. a=g (9.8m/s^2) so T=0.442*9.8
Monday, February 25, 2008 at 2:10pm by matt

centripetal acceleration is v^2/r everywhere on the circle. Tension=ma+-mv^2/r at the bottom, +, at the top, -
Tuesday, October 1, 2013 at 2:18am by bobpursley

Tension, T = 100N mass = m kg acceleration = 0.816 m/s² T = mg + ma Solve for m
Thursday, November 11, 2010 at 2:18pm by MathMate

tension= weight held up= weight chain below+ weight bolder tension= (790+600)g That is the tension at the top, and of course, the tension at the bottom is 790g, so tension varies on the length. If it breaks, it is at the top. Here, tension cannot be greater than 3*600*g. Is it?
Wednesday, February 8, 2012 at 2:06pm by bobpursley

college physics
Consider, T1 - T2 - mg = ma Also, T2 - mg = ma Then, T1 = 2ma + 2mg & T2= mg + ma so, T1 = 93.48 N & T2 = 46.74 N Now to find max acc using given Tension Force 96. Set equation 1 using T1 as 96 and now find a. your answer should equal a = 1.90 m/s2 Enjoyyy...
Tuesday, September 21, 2010 at 8:16pm by M@ro

As I already stated, it is the ratio of ma to mg that gives you the tangent of the angle. The mass cancels out. So does the tension, T T sin theta = ma T cos theta = mg (sin theta)/(cos theta) = tan theta = a/g
Sunday, April 5, 2009 at 12:48am by drwls

a)What is the tension T in the string? in this case, tension = mass2 X gravity so, tension = 1.10 kg X 9.8 m/s^2 tension = 10.78 For letter b, I am afraid I have no clue :( c)What is the speed of the puck? F = (m(v^2))/r the tension is the force in this case, so: tension = (m1...
Tuesday, November 30, 2010 at 2:13pm by Brianna

you have mg downward, and a force at the ceiling pulling upward. Now, when going upward, there is a force = ma that accelerates the mass upward (if it is accelerating). Total tension = ma + mg if falling (accelerating downward, a is negative. If it is at constant speed, then a=0
Wednesday, December 17, 2008 at 5:17pm by bobpursley

F=ma so the tension in a coupling depends on the mass behind the coupling In the case of the just last car, tension=masscar*a now between the 30and 31 car, there are 20 cars behind tension=20(masscar)a
Saturday, September 22, 2012 at 8:08pm by bobpursley

In a vertical circle, max tension is at the bottom. Tension=mg+mv^2/r solve for v In a hoizontal circle, tension is more complicated. Tension=m * sqrt((v^2/r)^2+g^2)
Wednesday, May 30, 2012 at 1:32pm by bobpursley

AP physics
You have distance, time, intial velocity, compute acceleration. d=1/2 at^2 2j) Tension= mg-acceleration*m 3) Tension= mg+ma in the case of each of those, m is the mass of the block being investigated. one knows tangential acceleration, a. angular acceleartion= tangential ...
Sunday, January 27, 2008 at 9:34am by bobpursley

A loaded elevator with very worn cables has a total mass of 1800 kg and the cavles can withstand a maximusm tension of 28000N a) What is the maximum upward acceleration for the elevator if the cables are not to brake? b) What is the answer to part a) if the elevator is taken ...
Thursday, November 16, 2006 at 3:28pm by Chris

Are you sure the 1.65 +/- 5 mA is correct? That could be any value between 6.65 mA and -3.35 mA. Did you mean 1.65 A +/- 5 mA ? The rheostat in the circuit allows the current in the resistor to vary with a fixed-voltage power source (like a battery).
Tuesday, February 10, 2009 at 1:08am by drwls

isnt v=sqrt(tension) but f*lambda= v= sqrt(tension). So leaving the lambda constant, then f is proportional to sqrt tension 230/220= sqrt (tension/orgtension) square both sides, solve for tension, then find the percent increase.
Monday, November 15, 2010 at 6:26pm by bobpursley

Use the fact that it goes 1 m in 1.9 s to get the acceleration, a. (1/20 a t^2 = (1/2)a*(1s)^2 = 1.7 m a = 3.4 m/s^2 The tension T in the cord minus the weight equals ma. T = m (g + a) m = 5.3 kg and g = 9.8 m/s^2 Do the numbers.
Sunday, September 19, 2010 at 11:28pm by drwls

pulling force mg acceleration= mg/(M+m) tension= ma
Saturday, October 10, 2009 at 1:29pm by bobpursley

Let T be the string tension. That and the angular acceleration alpha are the unknowns. Separate equations for the acceleration of the block and the angular acceleration of the cylinder will allow you to solve for both variables. The moment of inertia of the cylinder in (1/2) M...
Friday, January 22, 2010 at 3:17am by drwls

tension= mg-ma 400N=100(9.8-a) solve for a. Then, 200m=at solve for t.
Wednesday, March 14, 2012 at 8:29pm by bobpursley

Force applied =5kgx9.8= 49N Force Against (friction)=4kgx9.8x.4=15.68 first to find the Acceleration... 49-15.69=Total Force= 33.32N F=MA F/M=A 33.32/9(both blocks)= 3.702m/ss Now for Tension.. 9.8-3.702=6.098 6.098X5=30.49N=Tension
Tuesday, February 2, 2010 at 7:36pm by Anonymous

A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limv and back down to a 15kg package on the ground. a) what is teh magnitude of the least acceleration the monkey muct have if it is to ligt teh package off the ground? If after the packafe has been ...
Tuesday, November 21, 2006 at 5:52pm by Bri

physics (newton's laws)
A 5.00 kg object placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fasted to a hanging 9.00 kg object. Find the acceleration of the two objects and the tension in the string. First, there is just one force acting on the ...
Tuesday, February 20, 2007 at 1:05am by Eric (please help me)

tension=mg+ma=mg+mv^2/r=you do it. at hightest point, gain of PE= KE at bottom, gain of PE=1/2 .380*2.03^2/.831 mgh=gain of PE h=gainofPE/(.380*9.8) cos Theta= (r-h) /r tension at highest point. mg*cosTheta
Saturday, March 8, 2014 at 8:16pm by bobpursley

frequency*wavelength=speed=sqrt(tension/mass/length) 200*1=sqrt(tension/.0003) square both sides solve for tension 40000*.0003=tension=hangingmass*g
Tuesday, April 16, 2013 at 6:07pm by bobpursley

Physics( Explain and choose right answer)
The tension is smallest at the top, if the speed is in fact enough to make the top without the tension going to zero. At top: Tension + m g = m v^2/R tension = m (v^2/R - g) if the centripetal acceleration is not greater than g, the rope goes slack.
Monday, March 24, 2014 at 6:40pm by Damon

Physics Classical Mechanics
A crane is configured as below, with the beam suspended at two points l1 and l2 by each end of a cable passing over a frictionless pulley. The two ends of the cable each make an angle θ with the beam. A counterbalance object C with mass mC is fixed at one end of the beam...
Tuesday, November 26, 2013 at 10:24am by A

Physics- Mechanics
Find the tension in the system(F=Ma)the divide this by the number of pulleys that have shortened the y distance. the angle is just there to throw you off. This is a bad assumption.
Friday, September 24, 2010 at 4:55pm by maximus Corbetti

A lamp hangs vertically from a cord in a descending elevator that declerates at 2.4m/s^2 a) if the tension in the cord is 89N what is the lamp's mass? b) what is the cord's tension when the elevator is ascends with an upward acceleration of 2.4m/s^2? I know T=w where w=mg. ...
Monday, November 13, 2006 at 5:24pm by Chris

Start with the tension. T= mg-ma Then the moment of inertia... Torque=I *angular acceleration Force*radius= I * linear acceleration/rad solve for I.
Monday, March 24, 2008 at 4:18pm by bobpursley

a rock suspended from a string moves downwards at constant a=g. Which of the following is true concerning the tension in the string? a) the tension is zero b)the tension is equal to the weight of the rock c)the tension is less to the weight of the rock d)the tension is greater...
Wednesday, November 14, 2007 at 2:54pm by Andrea

College Physics
You did not say what the mass M is. As turns out, you don't need to know M to get the answer. First convert the mass to weight, W. W = Mg The vertical component of string tension is T cos30 = Mg The horizontal component of string tension is the centripetal force T sin30 = Ma, ...
Wednesday, November 10, 2010 at 1:47pm by drwls

The figure below shows a collection of wires. The currents at the numbered points are: I1 = 2.4 mA right; I2 = 3.6 mA down; I3 = 0.60 mA right; I4 = 4.0 mA up; and I5 = 1.6 mA up. i. imgur. com/SCyljHG. png a. What is the current at point A (size and direction)? b. What is the...
Wednesday, March 20, 2013 at 4:35pm by k

tension= totalmass*(acceleration+ g) solve for acceleration now, given acceleration, scale reads mg+ma where m is the mass of the woman.
Wednesday, November 10, 2010 at 9:29pm by bobpursley

Look at energy. Intial gravational energy=mass*g*h rotaltional energy=1/2 I w^2=1/2 I(v/r)^2 (look up the moment of inertia for a solid disk I) rotational energy+1/2 mv^2=Initial GPE solve for v when it hits the floor (c) this will verify the following. a) net force=torque=...
Thursday, January 3, 2013 at 8:41am by bobpursley

MA=ma so the bigger acceleration goes to the smaller child, by a factor of 2. a= M/m * A
Monday, October 17, 2011 at 7:15pm by bobpursley

1. due to symettry, assume each half is supporting 25N So you are looking for tension. SinTheta=1m/5m based on lengths SinTheta=25N/tension based of forces 25/Tension=1/5 tension= 125N check that.
Wednesday, October 6, 2010 at 7:37pm by bobpursley

Both blocks accelerate at the same rate. Draw free body diagrams and write Newton's second law (F = ma) for each block. Solve for the two unknowns: accleration and string tension.
Sunday, October 19, 2008 at 4:41pm by drwls

You can use the law of cosines here, but it is much more complicated. Two sides a,b are the vectors (head to tail), and c is the unknown side. The angle between the vectors is 180-60=120 draw a triangle to prove that. c^2=a^2+b^2-2ab*cos120 = 2*tension^2( 1-cos120) =2 tension^...
Sunday, April 3, 2011 at 1:36pm by bobpursley

Physics Block System
Here is the diagram: h t t p :// *delete spaces in "h t t p :" or it won't load What is the mass of block M that will allow it to accelerate from rest to the right with a a = 2 m/s^2? Surfaces are frictionless and pulleys are ...
Thursday, May 1, 2008 at 2:36pm by Will

Physics URGENT!!!
The rope is doing two things: lifting up, reducing the normal weight (and friction), and pulling the sled horizontally. consider horizontal: tension*CosTheta*distance=work=friction*distance tension*CosTheta=friction=mu*(mg-tension*sinTheta) solve for tension. work=tension*...
Friday, October 5, 2012 at 10:00pm by bobpursley

science pls help
Looks like you need to brush up on one of the most basic and useful identities in mechanics: F = ma F(N) = m(kg) * a(m/s^2) 1 N = 1 kg-m/s^2 So, with that in mind, #1: F = ma = 40kg * 10m/s^2 = 400 kg-m/s^2 = 400N #2: F = ma = 60kg * 2m/s^2 = 120N Note that you can eliminate ...
Thursday, March 13, 2014 at 5:21pm by Steve

Tension on the elevator cable= mg+ma vf^2=vo^2+2*Acceleration*distance. find distance when acceleration is -9m/s/3s= -3m/s^3
Wednesday, January 27, 2010 at 10:14pm by bobpursley

tension= (mg+ma) solve for a. b. Vnext= vfirst+ a t solve for vnext
Wednesday, February 8, 2012 at 2:38pm by bobpursley

assuming each side holds half the weight.. I see a triangle where sinTheta=(1/2weight)/Tension sinTheta=(4.2N)/Tension solve for Tension.
Monday, February 14, 2011 at 1:53pm by bobpursley

AP Physics
You need to break the forces into components along the direction of travel. First, the hanging box, its weight is mg, downward, in the direction of travel. Second, the box on the plane. The weigh mg has a component along the plane, mgsinTheta, and a component normal to the ...
Sunday, November 4, 2007 at 7:57pm by bobpursley

F=ma=kx k=ma/x so what is acceleration a? vaverage=3.13m/.511s vfinal then must be twice that, or 6.26/.511 but we know Vfinal^2=Vi^2+2ad so solve for a Then, solve for k from k=ma/x
Friday, July 19, 2013 at 10:01pm by bobpursley

Physics( Explain and choose right answer)
A ball on the end of a rope is moving in a vertical circle near the surface of the earth. Point A F T is at the top of the circle; C is at the bottom. Points B and D are exactly halfway between A and C. Which one of the following statements concerning the tension in the rope ...
Monday, March 24, 2014 at 6:40pm by Dan

Pages: 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | Next>>