Monday
April 21, 2014

# Search: M1, Mechanics

Number of results: 3,198

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

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

Physics Classical Mechanics Help ASAP
now use conservation of energy "assuming no energy is lost due to this process" hence you should have Potential of m1 = Kinetic m1 + Kinetic m2 or to elaborate m1gR = 1/2(m1)*(v1)^2 + 1/2(m2)*(v2)^2 given all those formula, you should be able to find an expression for v1 :) ...
Friday, January 10, 2014 at 8:03am by C++MeetsJava

Mechanics M1
Vb=a*t=.8*3 m/s
Wednesday, March 7, 2012 at 7:42am by bobpursley

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 m1 ...
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 m1 ...
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

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
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-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
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
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

Classical Mechanics - Need Help Please
merry is just a little conservation of momentum.. m1: mass merry-go-round m2: mass sledgehammer v1: velocity merry-go-round before collision (=0) v2: velocity sledgehammer before collision v1': velocity merry-go-round after collision v2': velocity sledgehammer after collision...
Sunday, December 8, 2013 at 3:38am by bw

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
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

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 (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
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
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 that to ...
Monday, October 31, 2011 at 9:36pm by drwls

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
(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
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

Physics
If it is elastic, not only momentum is the same before and after but also kinetic energy If they stick together, then only momentum is conserved but the single mass after collision is m1 + m2 part a m1 Vi = m1 u + m2 v (1/2)m1 Vi^2 = (1/2)m1 u^2 +(1/2)m2 v^2 part b m1 Vi = (m1...
Sunday, January 12, 2014 at 1:14pm by Damon

Physics Classical Mechanics
equation is fine..check your parenthesis write ti down this way Sqrt(5*G*m1/4*r1)
Sunday, December 8, 2013 at 4:48am by Greco

Physics Classical Mechanics
A pulley of mass mp , radius R , and moment of inertia about the center of mass Ic=12mpR2 , is suspended from a ceiling. The pulley rotates about a frictionless axle. An inextensible string of negligible mass is wrapped around the pulley and it does not slip on the pulley. The...
Thursday, November 7, 2013 at 1:53pm by A

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•v1x+m2•v2x)/(m1+m2) =(2•2.1+4.5•4.6)/6.5 = 3....
Tuesday, June 12, 2012 at 12:14pm by Elena

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
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 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= (m1+m2+m) • v3 v3 = (m1+m)•v2/(m1+m2+m) (d) (m1+m2+m)•v3²...
Monday, October 15, 2012 at 5:37pm by Elena

Math
m1 = 3h1 m2 = m1-7 h2 = h1+9 m2 = h2 now make some substitutions looking for m1 and h1 m1 = 3(h2-9) = 3(m2-9) = 3(m1-7-9) m1 = 3m1 - 48 2m1 = 48 m1 = 24 so, h1 = 8 check: m2 = 24-7 = 17 = 8+9 = h2
Tuesday, September 24, 2013 at 9:53pm by Steve

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 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 to do a...
Thursday, July 30, 2009 at 4:21pm by physics (the algebra bit)

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 system for acceleration a, we obtain a= (m1-m2)•g/(m1 +m2+m/2) =(3-2)/(3+2+10/2) = 0.1 m/s^2. ...
Wednesday, April 18, 2012 at 10:15am 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

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
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•m1•t Q3 = c•m2 •(t2-t) . Q1+Q2 = ...
Monday, April 9, 2012 at 12:25am by Elena

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. Double ...
Monday, January 18, 2010 at 11:02am by drwls

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

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
F=G (M1)(4-M1)/.21^2 solve for M1, then 4-m1
Tuesday, April 6, 2010 at 12:09am by bobpursley

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
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

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
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

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 previous direction (away from the station) but at ...
Wednesday, April 25, 2012 at 11:34pm 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 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 ∆K = Kf −Ki =(m1•v²/2 +m2•v²/2) - (m1 +m2...
Wednesday, October 17, 2012 at 11:03am 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 +m2) •v²/2. mechanical energy -> thermal energy, sound, etc
Wednesday, October 17, 2012 at 11:03am by Elena

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-T- μ•N2= F-T- μ•m2•g a•(m1+m2) = T- μ•m1•g+ F...
Wednesday, October 3, 2012 at 8:07pm by Elena

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-T- μ•N2= F-T- μ•m2•g a•(m1+m2) = T- μ•m1•g+ 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. Law of conservation of energy KE2=PE2 (m1+m2)•u²/2=(...
Thursday, July 12, 2012 at 1:33pm by Elena

physics
m1•v1+m1•v2=(m1+m2)•v
Monday, September 3, 2012 at 8:46pm 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
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

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 friction is necessary for the second part of this ...
Saturday, April 7, 2012 at 8:40pm 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
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
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 - 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

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 Classical Mechanics
An instrument-carrying projectile of mass m1 accidentally explodes at the top of its trajectory. The horizontal distance between launch point and the explosion is xm. The projectile breaks into two pieces which fly apart horizontally. The larger piece, m3, has three times the ...
Monday, October 28, 2013 at 6:10am by Anonymous

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

Physics Classical Mechanics Help ASAP
A small cube of mass m1= 1.0 kg slides down a circular and frictionless track of radius R= 0.6 m cut into a large block of mass m2= 4.0 kg as shown in the figure below. The large block rests on a horizontal and frictionless table. The cube and the block are initially at rest, ...
Friday, January 10, 2014 at 8:03am by Anonymous

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
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
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 +4m) = 32.8 m/s
Wednesday, March 21, 2012 at 3:53pm by Elena

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 the...
Sunday, July 12, 2009 at 12:06pm by Physics

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

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) = L/sqrt(2•h/g). v1= L•(m1 + m2)/m1• sqrt(2•h/g) = = 2.4•(4.4•10^-3+21...
Wednesday, May 16, 2012 at 6:43pm by Elena

physics
Momentum = m1*V1 - m2*V2= 1*10-3*2=4 in the direction of m1. m*V = 4 V = 4/m = 4/(1+3) = 1 m/s. in the direction of m1.
Tuesday, January 7, 2014 at 7:26pm by Henry

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
Two blocks with masses M1=8.50kg and M2=1.40kg are attached by a thin string which goes over a frictionless, massless pulley. M1 slides on an incline and there is friction between M1 and the incline. The incline is at an angle of 23.0 degrees from horizontal. M2 travels down ...
Tuesday, June 18, 2013 at 8:40pm by Sam Cole

Physics Classical Mechanics
A pulley of mass mp , radius R , and moment of inertia about its center of mass Ic , is attached to the edge of a table. An inextensible string of negligible mass is wrapped around the pulley and attached on one end to block 1 that hangs over the edge of the table. The other ...
Thursday, November 14, 2013 at 7:11am by Anonymous

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

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•sinα) +m1•v²/2 {m2•g(h+s) +m2•v²/2 +ks²/2. From ...
Thursday, October 25, 2012 at 1:31pm by Elena

physics
m2a=m2g-m1a m2a+m1a=m2g a= m2g/(m1+m2) I got this m2a=m2g-m1a m2a + m1 a = m2g-m1a + m1 a m2 a + m1 a = m2g (m2 a + m1 a = m2g(m2 + m1)^-1 = a + a = (m2 + m1)^-1 m2g a + a = (m2 + m1)^-1 m2g 2a = (m2 + m1)^-1 m2g (2a = (m2 + m1)^-1 m2g)2^-1 = a = (2(m2 + m1))^-1 m2g a = (2(m2...
Tuesday, July 14, 2009 at 4:26pm by physics

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

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

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) = L/sqrt(2•h/g). v1= L•(m1 + m2)/m1• sqrt(2•h/g) = …
Wednesday, October 24, 2012 at 8:48pm by Elena

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 frequency...
Friday, April 6, 2007 at 2:56pm by Mary

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
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

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

STATISTICS
The probabilities of colors of each pill are independent. Let m[n] = C, be the event that the [n]th pill is a color (C). eg: P(m1=Blue) = 0.12 (1) P(m1=Blue and m2=Blue) = P(m1=Blue) P(m2=Blue) (2) P(m1<>Red and m2<>Red) = (1-P(m1=Red))(1-P(m2=Red)) (3) P(m1=Red or...
Wednesday, July 24, 2013 at 12:34am by Graham

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

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 Classical Mechanics
For question 5 ballistic missile m1= 5*10^24kg; r1=6000km; m2<<m1; alpha=30 degrees; r2=(5/2)r1=15000; G=6.674*10^-11. What is the initial speed of the projectile? My solution: Vo^2=g*range/sin(theta) Vo^2=6.674*10^-11*(r1+r2/sin30degrees) Vo^2=(6.674*10^-11)*(21000/0.5...
Saturday, December 7, 2013 at 6:51am by Mets

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
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
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•cosα, .....(1) m1•u1•sin20⁰ = m2•u2•sinα ...............(2) Divide (2) by (1) tan α = ...
Thursday, December 20, 2012 at 10:24pm by Elena

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
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•sinα)/(m1 +m2), T = m1•(g -a).
Tuesday, May 15, 2012 at 2:28pm by Elena

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
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•g – a(m1+m2)]/m1•g = [5•9.8 – 1.4•15]/10•9.8 = 0.27. 2. 0 = m2•g –T 0=T-F1(fr) = T- μ(s) •m1•g m2•g...
Monday, October 8, 2012 at 1:07pm by Elena

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
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
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••a = m2g – T. F(fr) = k•N. Solving for acceleration a, we obtain a = g• (...
Thursday, April 19, 2012 at 6:46am by Elena

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