Sunday

April 20, 2014

April 20, 2014

Number of results: 1,322

**physics**

PE = mgh Therefore m = PE/(g x h) m = 24/(9.8 x 3) m = 0.82kg Now the object is moved to 1m above the ground. PE = mgh PE = 0.82 x 9.8 x 1 PE = 8.036 Round off to 8 (A) Alternatively, you could divide 24J by 3, as this is the change in height difference. But you may be marked ...
*Thursday, October 22, 2009 at 12:54pm by James*

**physics**

(a) PE = mgh (b) KE = PE (c) mv²/2 = PE, v =sqrt(2PE/m)
*Thursday, May 10, 2012 at 11:51am by Elena*

**physic**

PE = mgh Therefore H = PE/(mxg) H = 7/(1.5 x 9.8) H = 0.48m PE (object 2) = 4.5 x 9.8 x 0.48 PE (object 2) = 21.17J
*Thursday, October 22, 2009 at 12:51pm by James*

**PHYSICS**

No. The PE relative to the surface below the ice will be PE=mgh=52.5*g*(-2.8m) notice the PE is negative, it takes energy to get the object back to the surface. PE is always relative to some arbitary reference point.
*Sunday, January 31, 2010 at 10:48am by bobpursley*

**Physics**

PE ->KE KE ->PE(spring) => PE= PE(spring) mgh=kx²/2 x=sqrt(2mgh/k)=
PE(spring) -> KE->PE => the same h (due to the absence of friction)
*Wednesday, December 25, 2013 at 4:47am by Elena*

**Physics**

OK, i was told by my teacher that mg would equal together 2 kg. If your including g then: PE = mgh PE = (2)(9.8)(5) PE = 98 J
*Monday, January 4, 2010 at 4:38pm by Priscilla*

**physics**

It doesn't matter where the zero of PE is when you are talking about differences in PE, as you are here. difference in PE = M g (H2 - H1) = 52* 9.81 * 17.7 = __ J
*Wednesday, December 15, 2010 at 3:14am by drwls*

**physics**

PE at surface= GMm/r PE mars= GMmars*m/rm PE earth= GMe*m/re so on mars, you have to overcome the PE at the surface, or 1/2 m v^2=GMmars*m/rmars velocity= sqrt(2G*MassMars/radiusMars)
*Saturday, June 2, 2012 at 6:50pm by bobpursley*

**physics**

KE + PE = 480 Joules 220 + PE = 480 PE = PE = 260 J. mg*h = 260 34*9.8*h = 260 333.2h = 260 h = 0.780 m.
*Monday, November 4, 2013 at 5:24pm by Henry *

**physics**

PE = mgh PE = 2 x 3 x 9.8 PE = 58.8J
*Thursday, October 22, 2009 at 12:47pm by James*

**Economics**

Hi My three criterion are now, if the share is blue-chip or not, the companies PE and its comparison to the S&P/ASX200. Regarding the companies I have compared it to a rival companies PE, I said that it is good that its PE is higher than the other company but it is still a bad...
*Saturday, May 28, 2011 at 3:24pm by Em*

**Physics**

It has a potential energy of PE = mgh due to its position PE = .2 x 9.8 x 10 If all of the PE is translated to KE, what is the KE?
*Thursday, March 7, 2013 at 2:07pm by Dr. Jane*

**physical science**

Potential energy, as the boy is not in motion and is suspended above the ground. If you wanted to find the energy, the equation is; PE = mgh PE = 30 x 9.8 x 3 PE = 882J
*Saturday, September 10, 2011 at 12:05am by James*

**Physics (Help!)**

KE+PE=a constant first find the constant. AT the beginning, PE is zero, and KE is K KE+PE=K when at half height, PE=1/2 K KE+1/2 K=K KE=1/2K
*Sunday, January 9, 2011 at 5:26pm by bobpursley*

**science**

(a) recall that potential energy is stored energy, and is given by the formula: PE = mgh (units in Joules) where m = mass (in kg) g = acceleration due to gravity = 9.8 m/s^2 h = height (in m) if the reference position is at the floor (that is, h = 0), the PE is equal to PE = 4...
*Saturday, October 22, 2011 at 12:30am by Jai*

**Physics**

1. PE = W(fr)+KE W(fr) =PE-KE= =mgh - mv²/2 = =2.5(9.8350 - 60²/2)= = 4075 J. PE =mgh = 2.59.8350=8575 J. W(fr)/PE =4075/8575 =0.48. 2. PE =mgh = 359.812 =4116 J. k = PE/W =4116=5500 =0.75
*Sunday, June 17, 2012 at 3:34am by Elena*

**Physics**

A 2.0-kg rock falls from a height of 25 m to a point 20 m above the ground. How much potential energy does it lose? GIVEN: mg = 2.0 kg h/d = 5 m PE = mgh PE = (2)(5) PE = 10 J
*Monday, January 4, 2010 at 4:38pm by Priscilla*

**physics**

Since KE + PE = constant = PEmax , when KE = PE/3, (4/3)PE = PEmax PE = (3/4)PEmax The height must be 3/4 of 433 mm, or 325 mm
*Saturday, December 3, 2011 at 11:11pm by drwls*

**Physics**

Total energy (kinetic plus potential energy) at any point in the oscillator is a constant. KE = (1/2)m v^2 PE = (1/2)kx^2 TE = KE + PE KE1 + PE1 = KE2 + PE2 0.2435J + PE(3.60cm) = 0.1293J + PE(5.10cm) 0.1142J = PE(5.10cm) - PE(3.60cm) 0.1142J = (1/2) k(5.10cm)^2 - (1/2) k (3....
*Tuesday, January 8, 2013 at 2:47pm by Amy*

**College Physics**

You know the energy in the spring: PE=massprojetile*g*heightitgoes so now in the spring. PE=1/2 k x^2 solve for k speed thru equilibrium? 1/2 mv^2=PE above solve for v
*Sunday, September 30, 2012 at 8:44pm by bobpursley*

**Physics 11th grade 2**

1) PE = mgh PE = 20 kg * 9.8 m/sec^2 *100m KE = PE KE = 1/2 * m*v^2 2) W=F x d F = mg Force = 55 x 9.8 m/sec^2
*Monday, March 11, 2013 at 5:13pm by Dr. Jane*

**physics**

PE = m x g x h PE = 2.48 x 9.8 x 28.1 PE = 682.94J at 28.1m KE at 28.1m = 0, because the object is still at rest. The kinetic energy at 0m transfers from the previously stated gravitational potential energy. KE = 682.94J at 0m Therefore PE = 0J at 0m Mechanical energy is the ...
*Monday, February 28, 2011 at 12:41pm by James*

**Physics - Re-post**

A .20 kg ball is attached to a vertical spring. The spring constant is 28 N/m. The ball is supported initially so that the spring is neither stretched or compressed, is released from rest. How far does the ball fall before brought to a momentary stop by the spring? Our teacher...
*Tuesday, January 18, 2011 at 3:57pm by Jennifer*

**Physics-Help, confused!**

A .20 kg ball is attached to a vertical spring. The spring constant is 28 N/m. The ball is supported initially so that the spring is neither stretched or compressed, is released from rest. How far does the ball fall before brought to a momentary stop by the spring? Our teacher...
*Monday, January 10, 2011 at 8:36pm by Jennifer*

**Physics ~Conservation of Energy~ part 2**

1) a) W=mgh b) PE=mgh c) E(toral) = PE = mgh d) KE= PE=mgh e) mv²/2=mgh v=sqrt(2gh) 2) pendelum ... car ???
*Wednesday, January 30, 2013 at 6:12pm by Elena*

**Physics**

work this as energy. Initial PE=Final PE+friction work Initial PE=mgh final PE=1/2 k x^2 where k=2250, x-.030 frictionwork= mg*cosTheta*mu*6 So, put this into the equation, solve for mu. However, the angle of the ramp needs to be known to solve it, I don't see it given, or ...
*Wednesday, April 25, 2012 at 6:32am by bobpursley*

**Physics**

(a) PE= mgx=1.99.8(1.13-3)= -
(b) PE =mgh=1.99.8(3-1.13)= +
(c) ΔPE=0
*Wednesday, November 21, 2012 at 12:49am by Elena*

**physics**

PE=mgh h↓ => PE↓ at the ground PE=0
*Thursday, February 7, 2013 at 2:13am by Elena*

**physics**

A) PE of gravity= m*g*h m=mass of object g=acceleration due to gravity h= distance mass= weight/gravity=4.383 PE=(4.383kg)(9.81 m/s^2)(1.7m) PE=73.095 J
*Wednesday, February 8, 2012 at 9:13am by alexis*

**Physics**

Potential energy of the charges is PE=kq^2/r=9 10^9(910^-6)^2/0.052=14 Joules. According to the Law of conservation of energy PE=2KE KE=mv^2/2. 2KE= mv^2 =PE v=sqroot(PE/m)=sqroot(14/10^-3)=118.3 m/s.
*Saturday, February 25, 2012 at 5:07pm by Elena*

**Economics URGENT**

Hi My historical EPS is the 2004/05 one which is 24.9. The cost was 4.30. So the PE is 4.30/24.9? Because my calculator is telling me the PE it is 0.17... I'm thinking it should be a PE of 17, but I don't know where I went wrong
*Saturday, May 28, 2011 at 11:30pm by Em*

**physics**

Find the real height, rather than the slant height. Use Sin Cos or Tan. Plug that value into PE=mgh solve for PE. PE=KE KE=1/2mv^2 Solve for v
*Wednesday, December 19, 2012 at 9:22pm by PhysicsPro*

**Physics**

What potential energy does a 60.0-kg gymnast acquire in climbing 5.00 m up a vertical pole? GIVEN: mg = 60 h/d = 5.00 m PE = mgh PE = (60)(5.00) PE = 300 J
*Monday, January 4, 2010 at 4:34pm by Priscilla*

**science(physics)**

how long does it take to fall 30m? 30 = 5t^2 t = √6 what's its speed then? v = 10t = 10√6 what's its KE then? KE = 1/2 mv^2 = (1/2)(8)(600) = 2400 J Or, using PE lost = KE gained: starting PE = mgh = (10)(8)(100) = 8000 PE at 70m = (10)(8)(70) = 5600 PE lost = 2400...
*Thursday, June 27, 2013 at 11:16am by Steve*

**8th grade physical science**

Did I do these one correctly? A baby carriage is sitting on top of a hill that is 21 m high. The carriage with the baby wieghs 12N. Calculate the energy? PE= mgh PE= 12N(21m) PE= 252 Nm or do i multiply by 9.8m/s^2?
*Thursday, January 19, 2012 at 11:33am by Misty*

**physics Bob Pursley plz help**

ok so I calculate PE when R is (1/1.019)*3.38e6 and then calculate Pe when R is just 3.38e6? Or are you saying multiple (Pe)(1/1.019-1/1)?
*Monday, November 30, 2009 at 5:00pm by Andy *

**physics**

a. it has mg(1.50) PE less that at the ceiling b. it has mg(3.36-1.50) PE more than at floorlevel. c. at the same mpoint it has the same PE
*Wednesday, January 22, 2014 at 10:22pm by bobpursley*

**Physics**

letting the zero PE system at A, then initial PE=mga final PE =-mga cosAlpha-MG*2a*CosAlpha solve for cosAlpha, setting final=initialPE CosAlpha= 1/(-1-2)=-1/3 where alpha is measured from the vertical downward, clockwise
*Saturday, May 26, 2012 at 6:14am by bobpursley*

**Physics**

Assuming PE and h are measured from ground level, solve for h in PE=mgh
*Tuesday, November 30, 2010 at 1:05am by MathMate*

**physics**

PE = mgh m=PE/gh= 800/69.8= 13.61 kg
*Thursday, May 2, 2013 at 10:03am by Elena*

**physics**

What is the PE of the electron at 2cm? That PE must convert to KE.
*Wednesday, March 5, 2008 at 2:37am by bobpursley*

**Physics**

What would the gravitation PE and Elastic PE be like on the moon?
*Friday, June 24, 2011 at 1:20am by Janice*

**Physics**

1. KE=original PE mgh 2. PE at bottom is zero
*Saturday, September 22, 2012 at 6:48pm by bobpursley*

**physics**

PE=mgh=102=20 J. PE->KE
*Friday, April 5, 2013 at 2:00pm by Elena*

**AP PHYSICS**

The PE at infinity is zero. THe PE at r is Vq which is equal to q^2/r Watch units.
*Wednesday, March 26, 2008 at 6:02pm by bobpursley*

**physics**

Initial PE= Final PE mgh=mg2+ 1/2 k(h-12)^2 check my thinking.
*Wednesday, April 23, 2008 at 1:58pm by bobpursley*

**physics**

The change is PE is kQ1*Q2(1/Ri - 1/Rf) What you computed is potential, not PE
*Wednesday, March 4, 2009 at 10:36pm by drwls*

** pe**

missed school cause I had flu. now haf to write 5 papers for pe.
*Wednesday, March 2, 2011 at 1:42am by pam*

**Economics**

Where did you find a PE of 0.09? Which company? I've never seen such a low PE.
*Saturday, May 28, 2011 at 3:24pm by Ms. Sue*

**Economics**

That came out badly Retype PE = Market value per share/EPS EPS = net income-dividends/no of shares available EPS = 417000000-0/19000000 EPS = 21.94 PE = 2.08/21.94 PE = 0.09
*Saturday, May 28, 2011 at 3:24pm by Em*

**Physics**

a. put a coordinate system where one is fixed. The potential energy of the other is GMM (1/.24-1/.29) that is PE of the starting position minus the final position (24cm cneter to center). That PE turns to KE (1/2 M V2),but in reality, both are moving, so the KE is divided ...
*Thursday, November 4, 2010 at 4:15pm by bobpursley*

**11th grade Physical Science**

1/2 m v^2=Ke v= sqrt 2KE/m PE=mgh h= PE/mg
*Monday, November 2, 2009 at 6:07pm by bobpursley*

**mechanics_ physics**

PE -> PE(spring) mgh =kx²/2 x=sqrt(2mgh/k)
*Sunday, January 6, 2013 at 1:31am by Elena*

**Physics**

yes. KE+work= increase in PE+ original PE
*Sunday, November 14, 2010 at 9:20pm by bobpursley*

**Physics**

A 2000-kg truck is raised 15 m in an elevator. What is the increase in its potential energy? GIVEN: mg = 2000 kg h = 15 m PE = mgh PE = (2000)(15) PE = 30,000 J
*Monday, January 4, 2010 at 4:36pm by Priscilla*

**Accounting**

a. UR b. PE c. AE d. AR e. PE
*Sunday, July 26, 2009 at 10:46pm by Jesse*

**PHYSICS**

Vq= PE kq1*q2/r=PE
*Wednesday, February 24, 2010 at 8:24pm by bobpursley*

**PE**

any common activities between PE and science
*Thursday, September 27, 2012 at 3:11am by Salah*

**Physics**

PE=CU²/2 U=sqrt{2PE/C}
*Monday, February 18, 2013 at 3:18pm by Elena*

**Physics Mehanics**

your initial PE turns to final PE and final rotational KE, right? initial PE= mg(h/2) final PE= mg(0) rotational KE= 1/2 I w^2=1/2 I v^2/r^2
*Thursday, February 24, 2011 at 2:13pm by bobpursley*

**Physics**

change in PE=change in KE PE=1/2 mass(1130^2-(3*1130/4)^2 ) Now PE change is Mass*GMm/(altitude+radiusmoon) so you have to look up the mass of the moon Notice the mass divides out.
*Sunday, November 15, 2009 at 2:28pm by bobpursley*

**physics**

the gravitational constant G =6.6710⁻¹¹ Nm²/kg², PE₁=Gm₁m₂/R₁ PE₂=Gm₁m₂/R₂ E=ΔPE= PE₁-PE₂= =Gm₁m₂/R₁-Gm₁m₂/R₂= =Gm₁m₂(1/ R&#...
*Sunday, November 25, 2012 at 10:35pm by Elena*

**Physics**

a) mechanical energy= PE+KE PE= m(mearth*g*(Re/(Re+altitude)) *^2* altitude KE= 1/2 m v^2 where v is found by mv^2/(Re+h)= PE/h solve for v. b) mechanical energy at end? zero
*Thursday, November 17, 2011 at 8:08am by bobpursley*

**Physics Urgent**

A 3.2 kg block is hanging stationary from the end of a vertical spring that is attached to the ceiling. The elastic potential energy of this spring/mass sytem is 2.2 J. What is the elastic potential energy of the system when the 3.2 kg block is replaced by a 5.6 kg block? For ...
*Tuesday, April 24, 2007 at 5:26pm by Papito*

**physics**

a squirt gun is tied to a string and released when it is 1 meter above the ground. the mass of the gun is .1kg. what ist its initial PE? 1Joule what is its PE when it has fallen to .75meters? .75 J What is its KE at this point? .25 J What is its velocity? 0 m/s Above is the ...
*Monday, January 28, 2013 at 11:27pm by Anonymous*

**math alegebra**

The PE it gave up is changed to KE. PE gave up= GMe*m/re-GMe*m/(h-re) KE end= v change to miles to meters. then ke=PE chnange solve for h.
*Sunday, December 8, 2013 at 10:32pm by bobpursley*

**dynamics**

Sum work done and energies. Frictional force = F N Stopping Distance = D = 5 m Speed uphill = v = 6 m/s before braking: KE=(1/2)mv^2 PE=0 W=0 after braking: KE=0 PE=mg*Dsin(θ) W=D*F Equate KE+PE+W before and after braking, and solve for F.
*Monday, July 22, 2013 at 9:56am by MathMate*

**physics**

PE at top=PE at launch+Initial KE I recommend you change feet to meters... mgh=mg(distancehigh)+1/2 m vi^2 solve for h.
*Saturday, January 5, 2013 at 3:29pm by bobpursley*

**PE**

does gravity affect PE activities?
*Thursday, September 27, 2012 at 3:11am by bobpursley*

**science**

It is pretty easy... consider energy. The ball has an initial KE, and initial PE. The ball has a final PE you know, so calculate the final KE. Intialtotalenergy= final PE+finalKE
*Friday, January 22, 2010 at 9:02pm by bobpursley*

**Physics**

PEhightest: mgh PE at catch: mgh For the last two, you have to know more, like the angle it was hit at. Lets assume then it had no horizontal velocity, so all KE was vertical velocity. KE caught then is the difference between PE hightest and PE caught mg(21.8-2.5) Velocity ...
*Wednesday, October 24, 2012 at 7:38pm by bobpursley*

**physics**

Halfway? starting energy= 1/2 k .358^2 that is PE PE at half way= 1/2 k (.358/2)^2=1/4 starting PE so KE must be 3/4 starting energy. 1/2 m v^2= 3/4*1/2 k (.358)^2 solve for v.
*Saturday, November 20, 2010 at 5:45pm by bobpursley*

**biology**

I have a question about reduced and oxidized chemicals and potenital enery. Reduced have weaker PE because they hold on to their electrons and oxidized have strong PE because they can let go electrons right? If I am right then why is that CO2 has a low PE because Oxygen is ...
*Tuesday, October 20, 2009 at 2:39pm by Anonymous*

**quantum physics**

a) follows from conservation of energy: Photon energy + m = photon energy after collision plus gamma m (in c = 1 units) In b) you have to take into account that both the photon after the collision and the electron after the collison are not moving in the same direction the ...
*Wednesday, May 7, 2008 at 6:08pm by Count Iblis*

**Physics**

All of the gravitational PE the idiot has must translated to elastic PE mgh=1/2 k x^2 mg*32=1/2 k x^2 so my question is x. How long is the stretched cord? It is not clear to me where it is attached.
*Monday, October 12, 2009 at 2:11am by bobpursley*

**PHY**

PE at the surface: GMm/r but you are given g=Force/mass=GM/r^2 =.77 PE at Surface then=GMm/r*r/r=mgr escape velocity for KE=PE 1/2 mV^2=mgr Vescape= sqrt (2gr) checking book answer Vesc=sqrt(2*.77*1.2E6)=text answer.
*Thursday, July 5, 2012 at 3:35pm by bobpursley*

**physics**

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*

**physics**

(PE+KE)initial = (PE+KE)final mgh + mv0^2/2 = 0 + mv^2/2 => h = (v^2-v0^2)/2g
*Thursday, August 23, 2012 at 6:51am by Ajayb*

**Physics**

PE of ball when released is mgh KE of ball at bottom of swing is the same, and is 1/2 mv^2. get PE, solve for v
*Monday, July 22, 2013 at 9:02pm by Steve*

**Physics**

The PE is maximum when the spring is compressed or stretched. KE develops when the spring is moving. KE+ PE= constant. So, at the neutral point, where PE is zero, that is where KE is maximum, ie, it is moving fastest at that point.
*Thursday, June 23, 2011 at 5:51pm by bobpursley*

**college physics**

well, if amplitude is 1/2, and PE is related to the square of amplitude, then PE must be 1/4, so KE is 3/4 max, so ratio is 3/1 check my thinking.
*Wednesday, June 29, 2011 at 7:21pm by bobpursley*

**physics**

Energy = PE = PE₁₂+PE₁₃+PE₁₄+PE₂₃+PE₂₄+PE₃₄= =kq₁q₂/a +kq₁q₃/a+kq₁q₄/a+ kq₂q₃/a+kq₂q₄/a +kq₃q₄/a . Since q₁=q₂=q₃=q...
*Monday, November 11, 2013 at 12:44pm by Elena*

**AP PHYSICS**

That is not the correct answer. I did (Initial KE+ Initial PE)- (FinalKE+Final PE)= losses
*Saturday, November 13, 2010 at 9:42am by Indina*

**Physics**

work= PE1-PE= q1V1-q2V2=.25 (-30)=-7.5J PE change = - work done
*Monday, April 18, 2011 at 3:42pm by bobpursley*

**physics**

KE of the objects transforms into PE, and then PE→KE => v(start) =v(land)
*Wednesday, November 7, 2012 at 11:14am by Elena*

**Physics**

a. work=avgforce*distance b.PE=above work. KEwhenreleased=InitialPE d. zero, no PE, it has KE. e. Initial energy=1/2 m v^2 solve for v.
*Monday, May 10, 2010 at 2:59pm by bobpursley*

**physics**

You're quite right, the deceleration as calculated above accounts for gravity, but the force does not. From v=14, v1=0, we calculate a=(14²-0²)/2/2m=49 F=ma=65*49=3185 N On this we have to add the weight of the diver, 65*9.8 to give 3822 N. As suggested by Drwls, it ...
*Thursday, December 24, 2009 at 11:15am by MathMate*

**physics**

PE= mgh for the top boulder, M is larger, and h is larger than the other bolder. So which has the greater PE?
*Monday, December 7, 2009 at 9:50am by bobpursley*

**pe**

pe and science objectives
*Friday, November 2, 2012 at 3:15am by salah*

**Physics E&M**

We need to find the work done by the proton and the other two electrons on one electron in moving it in from infinity to its final location.. Then triple it. I will call a side of the triangle s You can put in the nanometer later. I will call proton +q, electron -q Now in ...
*Monday, February 18, 2008 at 4:41pm by Damon*

**physics Bob Pursley plz help**

Find the total PE available PE=GMm/Re (1/2.019 -1/1) Now, that has to be the work done heat+finalKE=work done
*Monday, November 30, 2009 at 5:00pm by bobpursley*

**Physics URGENT!!!!**

KE=1/2 m v^2 force*distance=workaborbed=KEinitial 2. change in PE= 1400*g*6m change in PE=finailKE-initialKE = 1/2 m (15^2-7^2)
*Saturday, April 9, 2011 at 8:48pm by bobpursley*

**physics**

Assume the fixed electron is at zero PE, zero KE. Then the PE of the original electron is -ke^2/1E-15 at the new position, energy is conserved, so new PE+KE=original PE -ke^2/1E-15=-ke^2/1E-3 + KE find KE, then KE= 1/2 m v^2
*Tuesday, July 3, 2012 at 5:38pm by bobpursley*

**Physics**

PE -> KE, PE=KE, KE=mv²/2, PE=mgh, mgh =mv²/2, v=sqrt(2 gh).
*Monday, June 11, 2012 at 3:14pm by Elena*

**physics(help)**

Gravity exerts a normal force on the dome of forcenormal=mg*CosTheta where theta is measured from the vertical to the point of contact. c. when mv^2/r>normal force, it takes off. now v is the result of gravity PE converting to KE. let Theta=90 deg=zero PE point, or referece...
*Friday, November 1, 2013 at 7:04pm by bobpursley*

**physics**

h = ssinα =100.5 = 5 m PE = mgh =509.85=2450 J. W =ΔPE =2450 J.
*Monday, April 30, 2012 at 8:15am by Elena*

**Physics**

when the KE is = PE... PE=k(Q)q/d where Q is the +charge on the gold nucleus, q is the charge on the alpha particle
*Sunday, February 10, 2013 at 8:43pm by bobpursley*

**Physics**

At top of hill: KE + PE = mg*h 0 + PE = mg*h PE = mg*h At bottom of hill: KE + PE = mg*h KE + 0 = mg*h KE = mg*h = 0.5m*V^2 0.5m*V^2 = mg*h V^2 = 2g*h V = Sqrt(2g*h). V = Sqrt(19.6*h). Sqrt means Square root.
*Tuesday, December 3, 2013 at 10:36pm by Henry *

**Physics**

PE = 0.0061 kg * g * 0.017 m = 0.010 J KE = PE = 0.010 J
*Monday, October 26, 2009 at 9:28pm by MathMate*

**PE**

list the common objectives between science curriculum and PE curriculum ?
*Wednesday, September 26, 2012 at 4:52am by Salah*

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