Number of results: 2,359
Physics (Classical Mechanics)
No friction? then the inertia on m1 tending to move it back (m1*a) is pulling on the cord, equalling in tension m2*g first, what is a: F=(M+m1+m2)a solve for a. then m1*a=m2*g m1(F/(M+m1+m2))=m2g solve for F, check my math.
Thursday, November 18, 2010 at 9:39pm by bobpursley
mechanics
a = [(m1 - m2)/(m1 + m2) ]*g m1*g -T = m1*a T = m1*(g-a) = 2*m1*m2*g/(m1+m2)
Thursday, November 22, 2012 at 11:19am by drwls
physics
m1 =0.0112 kg, v =? m2 = 1.01kg, u = 4.4 m/s The law of conservation of linear momentum m1•v + 0 = (m1+m2) •u, (a) v =(m1+m2) •u/m1. (b) ΔKE = m1•v^2/2 – (m1+m2) •^2/2.
Monday, April 30, 2012 at 9:12pm by Elena
Mechanics M1
Vb=a*t=.8*3 m/s
Wednesday, March 7, 2012 at 7:42am by bobpursley
Mechanics M1
thanks a lot Elena, your help is much appreciated
Sunday, April 29, 2012 at 5:58am by Jat
mechanics
two bodies of mass m1 and m2 are attached to two ends of a string which passes over a massless and frictionless pulley. find the acceleration of the bodies and the tension of the string. m1 is bigger than m2
Thursday, November 22, 2012 at 11:19am by andrew
Physics. PLEASE HELP!
two blocks of masses m1 and m2 (m1>m2) are placed on a frictionless table in contact with each other. A horizontal force of magnitude F is applied to the block of mass m1. if P is the magnitude of the contact force between the blocks, what are the net forces acting on ...
Thursday, October 11, 2012 at 10:48pm by Jack
Physics. Really stuck!
two blocks of masses m1 and m2 (m1>m2) are placed on a frictionless table in contact with each other. A horizontal force of magnitude F is applied to the block of mass m1. if P is the magnitude of the contact force between the blocks, what are the net forces acting on ...
Monday, October 8, 2012 at 1:38am by Lindsay
physic
I assume the m1 is hanging vertically. This is called an Atwood machine the force of gravity acts on both, on m1,it is vertical, m1*g On m2, it is downward,of course, but the component down the plane is m2*g*sinTheta Net force=totalmass*acceleration m1*g-m2*gsinTheta=(m1+m2)*a...
Friday, October 9, 2009 at 3:22am by bobpursley
physics-mechanics
a ball, m1, is held in place, 30 degrees from the roof it is connected to, from m1 hangs a 2nd ball m2, 1 pendulum connected to another pendulum. what is the acceleration of the 2nd ball, m2, the moment m1 is released?? my 1st thought was, simply -g, since m1 isnt held anymore...
Monday, December 1, 2008 at 11:14am by devan levin
physics
m1=1.7 kg, m2 =2.4 kg, L=1.29 m. PE=KE m1•g•L=m1•v²/2 v=sqrt(2•g•L) The velocities of the bodies after elastic collision are u1=(m1-m2)v/(m1+m2), u2=2m1v1/(m1+m2) “-“ means the motion in opposite direction
Tuesday, October 23, 2012 at 5:15pm by Elena
physics
Solve these two simultaneous equations with unknowns a (acceleration) and T (tension). M1*g - T = M1*a T - M2*g = M2*a This leads to a = [(M1 -M2)/(M1 +M2)]*g and T = [2*M1*M2/(M1+M2)]*g
Thursday, February 14, 2013 at 10:05pm by drwls
Physics
The law of conservation of linear momementum m1•v +0=(m1+m2) •u v=(m1+m2) •u/m1. The law of conservation of energy (m1+m2) •u²/2=(m1+m2) •g•h, u=sqrt(2•g•h). v=(m1+m2) •u/m1= (m1+m2) •sqrt(2•g•h)/ m1.
Monday, August 6, 2012 at 8:34pm by Elena
Physics (Classical Mechanics)
Hey.. I dont know exactly how to get the solution for this one... doubleudoubleudouble dot physikerboard.de/files/fahrzeug_223.jpg (its a sketch i made with paint^^) We assume there are no frictions at all and we only look at the masses M, m1 and m2. I want to know how much ...
Thursday, November 18, 2010 at 9:39pm by Toby
Physics
Work is definitely not zero for launching a satellite. The kinetic energy must increase to reach orbital velocity. The potential energy also increases because you are raising it farther above the Earth's surface. The formula Epi = -G*m1*m2/R can be used for the potential ...
Thursday, February 28, 2013 at 10:17pm by drwls
physics
Use the universal law of gravity. F = G*M1*M2/R^2 If M1 is the heavier mass, F = 2G*M1^2/R^2 = 2*10^-12 N R = 2.9 m Look up G and solve for M1
Saturday, December 11, 2010 at 10:53pm by drwls
physics (the algebra bit)
m1 a = m2 g - m2 a - m1 g sin theta - (Mu kinetic)m1 g cos theta add m2a to both sides: a(m1+m2)=g(m2-m1*sinTheta) divide both sides by m1+m2 a= g(m2-m1sinTheta)/(m1+m2)
Thursday, July 30, 2009 at 4:21pm by bobpursley
physics
m1•v =(m1+m2)u, u=m1•v/(m1+m2). (m1+m2)u²/2=kx²/2, x=u•sqrt[(m1+m2)/k}= = m1•v•sqrt[(m1+m2)/k}/(m1+m2)= =m1•v/sqrt[k•(m1+m2)]= =0.005•1000/sqrt[20(2+0.005)]=3.53 m.
Wednesday, July 18, 2012 at 2:36am by Elena
physics
The mass m2 is larger than m1 by a factor 3.30/1.50 = 2.2 The combined mass m1 + m2 is therefore m1 + 2.2m1 = 3.2 m1 When the same force F acts, acceleration will be ne 3.2 tikmes less than when it acted on m1, or a = 3.30 m/s^2/3.2 = 1.03 m/s^2
Thursday, November 20, 2008 at 1:50am by drwls
Physics
The acceleration rate is a = 4.1/7.6 = 0.5395 m/s^2 For an Atwood's machine (with no friction or pulley inertia) a = [(M1 - M2)/(M1 + M2]*g You don't need to use the 15.58 meter motion information. You know the accleration already. Total KE = (M1 +M2)Vfinal^2)/2 Use ...
Monday, October 31, 2011 at 9:36pm by drwls
Mechanics M1
(6,2) + (0,.5) = (6,2.5) |6,2.5| = 6.5 tanθ = 2.5/6 θ = 22.6°
Tuesday, March 6, 2012 at 11:10am by Steve
Physics
(a) 0.21*M1*g = 5.968 N (b) The maximum friction force that M2 can apply to M1 without M1 slipping is also 5.968 N. That will cause M1 AND M2 to accelerate at rate a = 5.968 N/M1 = 2.058 m/s^2 The force that would have to be applied to M2 to make this happen is (M1 + M2)*a = 8...
Friday, March 2, 2012 at 11:35pm by drwls
Physics
F=G (M1)(4-M1)/.21^2 solve for M1, then 4-m1
Tuesday, April 6, 2010 at 12:09am by bobpursley
physics
if you had written the separate equations of motion for m1 and m2, recognizing that the acceleration a is the same for both, you would find that the acceleration rate is: a = (m2 - m1)g/(m1 + m2) 1. S = (1/2)a*T^2 is the distance moved 2. T = m2(a + g) = 2(m1*m2)g/(m1 + m2) (...
Thursday, April 12, 2012 at 6:02am by drwls
Derivation of Physics Equation
How do we derive these laws: (most of the time, the numbers are subscript) v'2 = v1 * ( (2 * m) / (m1 + m2) ) + v2 * ( (m2-m1) / (m1+m2) ) & v'1 = v1 * ( (m1-m2) / (m1+m2) ) + v2 * ( (2 * m2) / (m1 + m2) )
Saturday, April 5, 2008 at 9:11pm by Anonymous
Physics URGENT!!!
m=2 kg , m1=70+2=72 kg, m2= 50 kg v =9 m/s, Δh=4 m, h=20 m (a) m1•v =(m1+m)v1 v1= m1/(m1+m) (b) KE1+ΔPE=KE2 (m1+m)•v1²/2+(m1+m)•g•Δh=(m1+m)•v2²/2 v2=sqrt(v1²+2g•Δh) (c) (m1+m)•v2...
Monday, October 15, 2012 at 5:37pm by Elena
Mechanics M1
velocity downstream is zero (she swims directly across) 1.25cosTheta=1 solve for theta 1.25sinTheta=V solve for V
Wednesday, March 7, 2012 at 7:48am by bobpursley
Physics
m1=2 kg, x1=2m, y1 6 m, v1x =2.4 m/s, v1y =4.1 m/s, m2 = 4.5 kg, x2=4 m, y2 = 1 m, v2x=4.5 m/s, v2y =4.6 m/s. x(c) =(m1•x1+m2•x2)/(m1+m2) =(2•2+4.5•4)/6.5 = 3.38 m, y(c) =(m1•y1+m2•y2)/(m1+m2) =(2•6+4.5•1)/6.5 = 2.54 m. v(cx) =(m1•...
Tuesday, June 12, 2012 at 12:14pm by Elena
Physics
Since they come to a dead stop, they had equal and opposite momenta beforer the collision Let M1 be the mass of the faster toy locomotive (speed V1) and M2 be mass of the slow the slow one (with speed V1/3). M2 = 4.48 - M1 M2*V1/3 = M1*V1 = (4.48 -M1)*V1/3 V1 cancels out. M1...
Wednesday, March 28, 2012 at 1:13am by drwls
physics (correction)
I mistakenly wrote the equation for the acceleration with two vertically hanging weights on a pulley. With one weight on a frictionless horizontal table, it accelerates faster. Let M1 = 5 kg and M2 = 9 kg and T = string tension T = M1*a M2*g -T = M2*a M2*g = (M1 + M2) a a = M2...
Wednesday, August 3, 2011 at 10:20pm by drwls
physics
The case of inelastic collision m1•v1+m2•0=(m1+m2)•v v= (m1•v1)/ (m1+m2). It is necessary to know the mass of a piece of wood m2. The change in kinetic energy is ΔK=K1-K2 = m1(v1)^2/2 - (m1+m2)v^2/2. For finding the percentage ((K1-K2)/K1) •100%
Thursday, February 9, 2012 at 9:48am by Elena
Physics
The law of conservation of linear momentum for inelastic collision m1•v = (m1+ m2)•u =(m1+ m2)•(v/2) Consenquently, m2=m1
Wednesday, March 21, 2012 at 3:33pm by Elena
physics
How do we derive these laws: (most of the time, the numbers are subscript) v'2 = v1 * ( (2 * m) / (m1 + m2) ) + v2 * ( (m2-m1) / (m1+m2) ) & v'1 = v1 * ( (m1-m2) / (m1+m2) ) + v2 * ( (2 * m2) / (m1 + m2) )
Sunday, April 6, 2008 at 3:47pm by Anonymous
Physics
In vector form m1•¬a = m1•¬g - ¬T1 m2•¬a = m2•¬g - ¬T2 I•¬ε =¬M Projections on the vertical axis: m1•a = m1•g - T1, m2•a = m2•g – T2, (mR^2/2)•(a/R) = (T1-T2)•R. Solving this ...
Wednesday, April 18, 2012 at 10:15am by Elena
Physics(Please respond)
The law of conservation of linear momentum m1•v = (m1+m2) •u v= (m1+m2) •u/m1 =(86.4+24.8) •2.99/86.4 =3.85 m/s
Sunday, June 3, 2012 at 1:34pm by Elena
physics
Apply Newton's Universal Law of Gravity and solve for M1*M2. F = 2.5*10^-10 = G M1*M2/(0.25)^2 Once you know M1*M2 and M1 + M2 (= 4 kg), you can solve for M1 and M2 separately.
Sunday, February 20, 2011 at 6:47pm by drwls
physics
Heat of fusion is the amount of heat energy required to change the state of a substance from solid to liquid. λ = 335000 J/kg is heat of fusion of water-ice c = 4185.5 J/(kg•K) is heat capacity of water Q1 = λ•m1 Q2 =c•m1 •(t-t1) =c•...
Monday, April 9, 2012 at 12:25am by Elena
physics
M1 Vi + 0 = (M1+M2)V if (1/2)M1 Vi^2 > (1/2)(M1+M2)V^2 then energy was lost as heat in the nasty crash.
Wednesday, February 15, 2012 at 8:49am by Damon
physics
Use Newton's universal law of gravity. F = G M1 M2/R^2 In this case, M2 = 2 M1. Therefore F = 2 M1^2/R^2. M1 is the mass of the smaller sphere. R is the separation. Look up the universal constand G; it must be in your notes, or textbook. Solve for the smaller mass, M1. ...
Monday, January 18, 2010 at 11:02am by drwls
Chemistry
M1V1 = M2V2 Where, M1 = x V1= 0.0250L M2 = 0.12M V2 = 0.0836L Substitute in the given values like so; 0.0250M1 = (0.12M)(0.0836L) 0.0250M1 = 0.010032mol Divide both sides of the equation by the coefficient in front of M1 to solve for it; M1 = 0.010032mol / 0.0250L M1 = 0....
Tuesday, December 8, 2009 at 7:17am by Alex
Physics- momentum please helppp!!!
s =5 m, m = 95 kg, m1= 1 kg, m2 = 15 kg, v=0.1 m/s, v1 = 10 m/s, v2= 2 m/s. 1. Hammer (m+m1+m2) •v = m1•v1 – (m+m2)•u1 u1 = {m1•v1 - (m+m1+m2) •v}/{m+m2} = ={1•10 –(95+1+15) •0.1}/{95+15} = - 0.01 m/s. Astronaut will move in ...
Wednesday, April 25, 2012 at 11:34pm by Elena
Physics URGENT!!!
x: m•v1=(m1+m2) •v(x), y: m•v2 =(m1+m2) •v(y), v(x)= m•v1/(m1+m2), v(y) =m•v2 /(m1+m2), v=sqrt[v(x)²+v(y)²], φ =arctan {v(y)/v(x)}. ∆K = Kf −Ki = =(m1•v²/2 +m2•v²/2) - (m1...
Wednesday, October 17, 2012 at 11:03am by Elena
physics
(a) 13 = F/m1 2.1 = F/m2 m2/m1 = 13/2.1 = 6.19 F/(m2 - m1) = F/[m2(1 -(m1/m2)] = (F/m2)/(1 -0.1615) = 2.1/(0.8385) = 2.5 m/s (b) F/(m1 + m2) = F/[m2(1 +(m1/m2)] You finish it.
Saturday, October 9, 2010 at 10:31am by drwls
Physics URGENT!!!
x: m1•v1=(m1+m2) •v(x) y: m2•v2 =(m1+m2) •v(y) v(x)= m1•v1/(m1+m2) =75•6/(75+82)=2.87 m/s v(y) =m2•v2 /(m1+m2)= 82•5/(75+82)=2.66 m/s v=sqrt[v(x)²+v(y)²] =3.91 m/s φ =arctan {v(y)/v(x)} =42.8⁰ &#...
Wednesday, October 17, 2012 at 11:03am by Elena
physics ..PLEASE HELP ASSIGN DUE SOON :(
m1•a=F-T-m1•g•sinα m2•a=T (m1+m2)a= F-T-m1•g•sinα+T a=(F-m1•g•sinα)/(m1+m2). T=m2•a=m2• (F-m1•g•sinα)/(m1+m2)=...
Wednesday, October 31, 2012 at 9:42am by Elena
physics
m1•v1+m1•v2=(m1+m2)•v
Monday, September 3, 2012 at 8:46pm by Elena
physics
Let M1 be the mass on the table and M2 be the mass hanging from the pulley. Let T be the tension force in the cable that connects them. Both masses accelerate at the same rate a, but in different directions. Solve this pair of equations. There are two unknowns, a and T. M1 g...
Sunday, September 13, 2009 at 10:11pm by drwls
Physics- PLEASE HELP
1. The acceleration of the system is a=F/(m1+m2) = 58/(15+17)=1.81 m/s². T1+T2 m1•a=T m2•a= F-T, T = m1•a = 15•1.81=27.5 N. 2. μ = 0.09, T1=T2=T, m1•a=T-F1(fr)=T- μ•N1=T- μ•m1•g m2•a=F-T-F2(fr)= F...
Wednesday, October 3, 2012 at 9:11pm by Elena
Physics
m1=75 kg, m2 =50 kg. L=h=3m Law of conservation of energy PE1=KE1 m1•g•h = m1•v1²/2, v1=sqrt(2•g•h) =sqrt(2•9.8•3)=7.67 m/s. Law of conservation of momentum m1•v1 =(m1+m2) •u u= m1•v1/(m1+m2) =75•7.67/125 =4.6 m/s...
Thursday, July 12, 2012 at 1:33pm by Elena
Physics- Elena please help!
1. The acceleration of the system is a=F/(m1+m2) = 58/(15+17)=1.81 m/s². T1+T2 m1•a=T m2•a= F-T, T = m1•a = 15•1.81=27.5 N. 2. μ = 0.09, T1=T2=T, m1•a=T-F1(fr)=T- μ•N1=T- μ•m1•g m2•a=F-T-F2(fr)= F...
Wednesday, October 3, 2012 at 8:07pm by Elena
PHysics
m1v1=(m1+m2)v but 1/2 (m1+m2)v^2=u(m1+m2)gd v=sqrt2ugd v1=(m1+m2)sqrt2ugd /m1 = (1.115)/.015 sqrt(2*.25*9.8*95) I dont get either of those answers. Your Vo is wrongly computed. vo=sqrt(2ad)=sqrt(2*2.45*95)=21.6m/s
Saturday, December 6, 2008 at 8:45pm by bobpursley
physics (the algebra bit)
ok my question I'm trying to solve this for a m1 a = m2 g - m2 a - m1 g sin theta - (Mu kinetic)m1 g cos theta my text book gives me this a = (m1 + m2)^-1 (m2 g - m1 g sin theta - (Mu kinetic) m1 g cos theta) ok I don't see how I get to this point tand dont know what ...
Thursday, July 30, 2009 at 4:21pm by physics (the algebra bit)
Physics
When an object of mass m1 is hung on a vertical spring and set into vertical simple harmonic motion, its frequency is 12 Hz. When another object of mass m2 is hung on the spring along with m1, the frequency of the motion is 4 Hz. Find the ratio m2/m1 of the masses. For ...
Monday, April 23, 2007 at 10:46pm by Papito
Mechanics M1
A rabbit runs in a horizontal straight line ABC across a field (a)the rabbit runs from rest at A with a constant acceleration of 0.8 m/s^2 and reaches B after 3 seconds, find its speed at B
Wednesday, March 7, 2012 at 7:42am by ahmad
physics
At the step "m2 a + m1 a = m2g," you can factor out an a using the distributive property: a(m2 + m1) = m2g Now divide both sides by (m2 + m1): a = m2g / (m2 + m1)
Tuesday, July 14, 2009 at 4:26pm by Marth
Physics. PLEASE HELP!!
m1 = 1.3 kg, m2 = 1.3 kg k = 0.89, g = 9.81 m/s2 M1•g•h = m1•v^2/2 v=sqrt(2•g•h) = sqrt(2•9.81•8) = 12.53 m/s. Law of conservation of linear momentum m1•v1 + 0 = m1•u1 +m2•v2. Since m1=m2, u2 =12.53 m/s. Given coefficient of ...
Saturday, April 7, 2012 at 8:40pm by Elena
chemistry
m1v1=m2v2 m1=m2v2/v1 m1=(0.20M NaCl)(.0036)/(0.01) m1=7.2x10^-4(0.01) m1=7.2x10^-6
Thursday, March 10, 2011 at 4:11pm by kris
Physics
m1 =4.3•10^-3 kg, m2 = 22.6•10^-3 kg, h =1.4 m, L= 3.1 m., v1 =? m1 •v1 = (m1+m2) •v, v= m1 •v1 /(m1 + m2) . h =g•t²/2 => t = sqrt(2•h/g) L= v•t = v• sqrt(2•h/g), v =L/sqrt(2•h/g), m1 •v1 /(m1 + m2) = ...
Wednesday, October 24, 2012 at 8:48pm by Elena
Physics - Elastic Collision
It is derived by requiring that both momentum and kinetic energy be conserved. The center of mass moves forward at velocity vcm = M1*V1/(M1 + M2). After collision, the CM retains that velocity but the velocities change directions in CM-fixed coordinates. In CM-fixed ...
Saturday, September 4, 2010 at 10:54am by drwls
Momentum
m1 =4.4•10^-3 kg, m2 = 21.3•10^-3 kg, h =1.3 m, L= 2.4 m., v1 =? m1 •v1 = (m1+m2) •v, v= m1 •v1 /(m1 + m2) . h =g•t²/2 => t = sqrt(2•h/g) L= v•t = v• sqrt(2•h/g), v =L/sqrt(2•h/g), m1 •v1 /(m1 + m2) = ...
Wednesday, May 16, 2012 at 6:43pm by Elena
Physics
Hey Ben, alright first you have to determine the acceleration of the system. Let m1=2.1kg and m2=4.9kg Use F=ma m2g-m1g=m2a+m1a (m2-m1)g=(m2+m1)a a= (m2-m1)(g)/(m2+m1) = (4.9-2.1)(9.8)/ (4.9+2.1) =3.92 ms^-2 Now we must determine the speed of m1 we use v^2=u^2+2as we know that...
Monday, April 1, 2013 at 5:14am by William Nguyen
Physics
Their initial velocities are in a ratio V1/V2 = m2/m1 Their kinetic friction forces are in a ratio F1/F2 = M1/M2 Gliding time t is proportional to V/a = V m/F Gliding distance D is proportional to V*t = V^2*m/F D1/D2 = 3 = [V1^2*m1/F1)][m2/(F2*V2^2)] = (V1/V2)^2*(m1/m2)*F2/F1...
Tuesday, April 1, 2008 at 3:51pm by drwls
physics
m1=10 kg, m2 = 4 kg, α = 40º. m1•a= m1•g -T, m2•a = T – m2•g•sinα, a•(m1 +m2) = m1•g –T+T- m2•g•sinα = = g•(m1 – m2•sinα), a= g•(m1 – m2•sin&#...
Tuesday, May 15, 2012 at 2:28pm by Elena
Physics
The law of conservation of linear momentum for elastic collision m1•v1 = m1•u1 + m2•v2. The velocities after collision are u1 =(m1-m2) •v1/(m1+m2) u2 =2•m1•v1/(m1+m2) Since m2=4m1, u1 =(m -4m) •82/(m+4m) = - 49.2 m/s u2 =2•m •82/(m...
Wednesday, March 21, 2012 at 3:53pm by Elena
science
Classical mechanics Relativity Thermodynamics Electromagnetism/Optics Quantum mechanics
Wednesday, August 29, 2007 at 8:17pm by ~christina~
physics
m1•v1+ 0 =(m1+m2) •V, V = m1•v1/(m1+m2) = 3•10/8=3.45 m/s
Friday, March 16, 2012 at 4:06pm by Elena
Physics
m1•v1+ 0 =(m1+m2) •V, V = m1•v1/(m1+m2) = 1.1•m/5.3m=0.21 m/s
Friday, March 16, 2012 at 9:26pm by Elena
math
In this case, the slope of the given line is m=-3. The slope of the perpendicular line is m1 such that m.m1=-1. Therefore m1=-1/(-3)=1/3 If it has to pass through a point (x1,y1), the equation of the line is then y=m1(x-x1)+y1 Can you continue from here?
Thursday, November 19, 2009 at 10:32pm by MathMate
Physics
m1 = 35 g = 0.035 kg, v1 =? m2 = 2.6 kg, v2 =0, u = 7.5 m/s m1•v1 =(m1+m2) •u, v1 =(m1+m2) •u/m1 = ......
Wednesday, May 9, 2012 at 9:09pm by Elena
Physics
m1= 399kg, m2=185 kg F(fr) = F =G•m1•m2/R² μ•m2•g= G•m1•m2/R² μ•g= G•m1/R² R =sqrt(G•m1/ μ•g)
Monday, December 12, 2011 at 10:35pm by Elena
Physics
Initial heights of the blocks: m1 -> H; m2 -> h Initial energy of the “two blocks” system is E1 = m1•g•H+m2•g•h When the blocks covered s=0.9 m the energy of ”two blocks+spring” system is E2= m1•g(H-s•sin&#...
Thursday, October 25, 2012 at 1:31pm by Elena
physics
1. h=at²/2 a=2h/t²= 2•1/1.2²=1.4 m/s² m2•a =m2•g –T = > T = m2•g-m2•a. m1•a=T-F(fr) = T-μ•N= T-μ•m1•g = m2•g-m2•a- μ•m1•g. μ=[m2...
Monday, October 8, 2012 at 1:07pm by Elena
Math Algebra 2
Assume all other variables are known. m1v1 + m2v2=(m1+m2)v3 expand each side: m1v1 + m2v2= m1v3 + m2v3 group terms containing m1 on the left, others on the right m1v1-m1v3 = m2v3 - m2v2 Factorize m1 and m2 m1(v1-v3) = m2(v3-v2) Divide by the coefficient of m1: m1 = m2(v3-v2)/(...
Tuesday, September 13, 2011 at 3:25pm by MathMate
Univ-Phys
M1 is the heavier car and the mass of the other car is M2 = M1/2 Orginally, (1/2) M1 V1^2 = (1/2)*(1/2)(M1/2)V2^2 which implies that V1^2 = (1/4)V2^2 V1 = V2/2 Also, (1/2)M1*(V1+5)^2 = (1/2)(M1/2)(V2+5)^2 = (1/2)(M1/2)(2V1 + 5)^2 (V1+5)^2 = (1/2)(2V1+5)^2 V1^2 + 10 V1 + 25 = ...
Wednesday, October 7, 2009 at 10:00am by drwls
AP physics
m1 = 0.0016 kg, v1 = 542 m/s, m2 =0.249, v2 = 0, u = ? (a) m1•v1 +m2•v2 = (m1+m2) •u u = m1•v1/(m1+m2). (b) KE1 =m1•v1²/2. (c) KE = ((m1+m2) •u²/2 (d) ΔKE = KE1 – KE. (e) (ΔKE/KE1) •100%
Thursday, May 10, 2012 at 8:45pm by Elena
Physics
Apply the law of conservation of linear momentum. Let the second (struck) car's mass be M2, and the first car's mass be M1. M1*19 = (M1 + M2)*10 Solve for M2 9 M1 = 10 M2 You finish it.
Wednesday, February 1, 2012 at 11:49pm by drwls
Physics- Elena please help!
Law of conservation of energy KE =PE (m1+m2) •u²/2 =kx²/2 u=sqrt{kx²/(m1+m2)}=..... Law of conservation og linear momentum m1v + 0= (m1+m2)u v=(m1+m2)u/m1 =....
Tuesday, October 16, 2012 at 4:16pm by Elena
Math
Let p and q be the roots of the polynomial mx^2 + x(2 - m) + 3. Let m1 and m2 be two values of m satisfying p/q + q/p = 2/3. Determine numerical value of m1/ (m2)^2 + m2/(m1)^2. Please work the complete solution.
Tuesday, April 23, 2013 at 8:33pm by Watson
Math
Let p and q be the roots of the polynomial mx^2 + x(2 - m) + 3. Let m1 and m2 be two values of m satisfying p/q + q/p = 2/3. Determine numerical value of m1/(m2)^2 + m2/(m1)^2. Please work the complete solution.
Tuesday, April 23, 2013 at 11:20am by Watson
Physics
Ok I got a question I asked before except there are other parts that I didn't ask so here we go Three blocks on a frictionless horizontal surface are in contact with each other A force F is applied to block 1 (mass m1). Draw a free-body diagram for each block ok I did this...
Sunday, July 12, 2009 at 12:06pm by Physics
Physics, still don't get it!
I'm sorry bobpursley I don't understand your response, please clarify. When an object of mass m1 is hung on a vertical spring and set into vertical simple harmonic motion, its frequency is 12 Hz. When another object of mass m2 is hung on the spring along with m1, the ...
Friday, April 6, 2007 at 2:56pm by Mary
Physics
The net force pulling the 5.7 kg block is Fnet = M1*a = 5.7*1.0 = 5.7 N That equals the string tension M2*g minus the friction force M1*g*U 5.7 = M2*g - M1*g*U M1 = 5.7 kg M2 = 4.0 kg Uk = kinetic friction coefficient g = 9.8 m/s^2 Solve for U
Tuesday, February 21, 2012 at 5:16pm by drwls
AQA Mechanics (M1), Statics and forces,please help
Perform vertical and horizontal force balances (static equilibrium) and solve for the two unknown forces. Assume that both forces are normal to the smooth plane surfaces, and pass through the cylinder radii.
Monday, March 19, 2012 at 5:44am by drwls
Physics
m1•v1+0 = m1•u(1x) +m2•u(2x) 0=-m1•u(1y) + m2•u(2y) m1•v1 = m1•u1•cos20⁰+m1•u2•cosα 0= - m1•u1•sin20⁰ +m2•u2•sinα . m1•(v1 -u1•cos20⁰)=m1•u2&#...
Thursday, December 20, 2012 at 10:24pm by Elena
Physics- Help! Due soon!
m1•a=m1•g –T m2•a=T-m2•g•sinα a(m1+m2)= m1•g –T+ T-m2•g•sinα, a=g•(m1+m2•sin α)/(m1+m2) = ... T= m1•(g-a)=...
Wednesday, October 10, 2012 at 12:08am by Elena
Science
m2 = (5/2) * m1 = 2.5 m1 m = m1 + m2. Replace m2 with 2.5mi: m = m1 + 2.5m1 = 3.5m1. a = (1mi/3.5m1) * 5 = 1.43 m/s^2.
Saturday, December 1, 2012 at 5:24am by Henry
maths geometry
Find and write down a proof that the product of the gradients of two perpendicular lines is -1 Use trigonometry. Let the slope of a line from A to B be m1. Let the slope of a perpendicular line from A to C be m2. The tangent (or slope) of the line between these two lines is: ...
Sunday, February 25, 2007 at 6:45am by jigna
M1 mechanics
A particle of mass 5 kg is being towed at a constant speed of 6 m/s on a rough horizontal plane with coefficient of friction 0.2. At a certain point the towing force is reversed in direction. Find the distance that the particle will travel before coming to rest and explain ...
Monday, April 30, 2012 at 9:00am by Jat
physics
Since there is no motion of the vertical string up or down, the weight of m2 must balance the centripetal force of m1. m1*V^2/R = m2*g m2 = m1*V^2/(R*g) = 3.67 kg That rounds off to answer (e)
Thursday, April 12, 2012 at 5:01pm by drwls
physics
The acceleration of the masses in an Atwood's machine is a = g*(m1-m2)/(m1+ m2) As m1 -> m2' , a approaches zero. If m2 >> m1, a --> g
Tuesday, July 17, 2012 at 10:05pm by drwls
Physics
Do the calculations separately in the x and the y directions. Let each of the 15cm segments lie along the x, and y-axes, with the bend at the origin. Apply ([m1]x+[m2]x)/([m1]+m2]) to get, along the x-axis: m1=15, m2=15, x1=7.5, x2=0 x0=(m1x+m2x)/(m1+m2) =(7.5*15+0)/(15+15) =3...
Monday, November 22, 2010 at 9:12pm by MathMate
Physics
What is the net pulling force? Answer m1(g-a). What is the acceleration of the system? Answer: netforce/mass= (m1(g-a))/(m1+M2) Observation: the acceleration each block is the same
Tuesday, July 14, 2009 at 1:26pm by bobpursley
Physics
well, tension is 2.57g up the slide static friction is .605*M1*g*cosTheta, down the slide gravity down the slide=M1*g*sinTheta so, 2.57g=.605M1*g*cosTheta-M1*g*sinTheta solve for M1 check my thinking and typing.
Thursday, November 4, 2010 at 8:32pm by bobpursley
physics
The equations of the motion in vector form are m1•⌐a=m1•⌐g+⌐N +⌐F(fr) +⌐T, m2•⌐a = m2•⌐g + ⌐T, Projections on x- and y- axes: m1•a= T – F(fr), m1•g = N, m2•...
Thursday, April 19, 2012 at 6:46am by Elena
Physics
m1 =3 g, v1 =12 cm/s, m2=?, v2 = 0, u1 = -10cm/s, u2 =1.5 cm. u1 =(m1-m2) •v1/(m1+m2), u2 = 2•m1•v1/(m1+m2), u1/u2 = (m1-m2) •v1/2•m1•v1. (-10)/1.5 = (3 –m2) •12/2•3•12. Solving for m2 we obtain m2 =43 g.
Wednesday, May 9, 2012 at 9:08pm by Elena
Physics
We are presumably talking about gravity F = G M1*M2 / d^2 F' = G M1*M2/(2d)^2 = G M1*M2/4 d^2 =(1/4) F
Thursday, June 5, 2008 at 1:05pm by Damon
Mechanics M1
Q.1 Find the force needed to accelerate a 2 kg block at 3m/s^2 up a rough plane (coefficient of friction 0.2) inclined at 25 degree to the horizontal if the force is a) Parallel to the slope b) horizontal c) at 45 degree to the upward verticle
Sunday, April 29, 2012 at 5:58am by Jat
Physics
m1•v1=(m1+m2)v v1=(m1+m2)v/m1=(0.01+3.99)0.5/0.01=200 m/s
Wednesday, November 21, 2012 at 9:17am by Elena
Physics
For m1: m1•a=T1 (block m1=1 kg is on the surface ) For m2: m2•a=m2•g – T2 (m2=9 kg) For pulley: I•ε = M => I•a/R= (T2-T1)R => (T2-T1) = I•a/R² a(m1+m2+ I/R²)=m2•g a= m2•g/(m1+m2+ I/R²...
Tuesday, November 6, 2012 at 4:02pm by Elena
physics
m1=2 kg, v = 10 m/s, m2 = 3 kg, α =30º, u2 = 4 m/s. Find β and u1. Take the x axis to be the original direction of motion of 2kkg ball. The equations expressing the conservation of momentum for the components in the x and y directions separately are ...
Sunday, May 6, 2012 at 8:33pm by Elena
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