Thursday

April 17, 2014

April 17, 2014

Number of results: 105,679

**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**

Two particles of mass m1 = 1.0 kg and m2 = 2.6 kg are traveling with velocities v1 = 4.9, and v2 = 2.5 as shown in the figure below. What is the total momentum of this system? Be sure to give the magnitude and direction of the momentum. magnitude 1 kg · m/s direction 2° above ...
*Tuesday, November 9, 2010 at 10:34pm by cody*

**physics**

Two particles of mass m1 = 1.0 kg and m2 = 2.6 kg are traveling with velocities v1 = 4.9, and v2 = 2.5 as shown in the figure below. What is the total momentum of this system? Be sure to give the magnitude and direction of the momentum. magnitude 1 kg · m/s direction 2° above...
*Wednesday, November 10, 2010 at 9:35am by cody*

**physics**

A tractor T is pulling two trailers, M1 and M2, with a constant acceleration. T has a mass of 200 kg, M1 has a mass of 100 kg, and M2 has a mass of 150 kg. If the forward acceleration is 0.60 m/s2, then the horizontal force on the tractor due to the attachment to M1 is ?
*Wednesday, October 5, 2011 at 4:57pm by wahida*

**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**

m1=1.7 kg, m2 =2.4 kg, L=1.29 m. PE=KE m1gL=m1v²/2 v=sqrt(2gL) 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 **

In figure, the mass m2 = 10 kg slides on a frictionless shelf. The coefficients of static and kinetic friction between m2 and m1 = 5 kg are ΅s = 0.6 and ΅k = 0.4. (a) What is the maximum acceleration of m1? (b) What is the maximum value of m3 if m1 moves with m2 without ...
*Monday, October 17, 2011 at 10:32pm by Eric*

**physics**

M1 is a spherical mass (38.6 kg) at the origin. M2 is also a spherical mass (10.3 kg) and is located on the x-axis at x = 59.8 m. At what value of x would a third mass with a 20.0 kg mass experience no net gravitational force due to M1 and M2?
*Tuesday, March 11, 2014 at 7:22pm by Marie*

**physics**

In the Atwood's machine of Figure 8-23, the two masses shown are initially at rest at the same height. After they are released, the large mass, m2, falls through a height h and hits the floor, and the small mass, m1, rises through a height h. The mass m2 remains at rest once ...
*Thursday, March 31, 2011 at 7:33pm by jennifer*

**URGENT physics**

M1 is a spherical mass (38.6 kg) at the origin. M2 is also a spherical mass (10.3 kg) and is located on the x-axis at x = 59.8 m. At what value of x would a third mass with a 20.0 kg mass experience no net gravitational force due to M1 and M2?
*Tuesday, March 11, 2014 at 11:36pm by Marie*

**Physics help**

Two masses m1 = 2 kg and m2 = 4 kg are connected by a light string. Mass m2 is connected to mass M = 14 kg by a string that passes over an ideal pulley. Determine the following quantities when the system is released from rest. (a) The acceleration of the masses m/s2 (b) The ...
*Friday, October 26, 2012 at 3:35am by arcos15*

**Physics**

Three objects with masses m1= 36.5 kg m2= 19.2 kg m3= 12.5 kg Are hanging from massless ropes that run over a frictionless pulleys. If the system is released from rest, what is the magnitude of the acceleration in m/s^2 of mass m1? ___________________ O O O I I I I I I I I I I...
*Thursday, January 16, 2014 at 7:20pm by Holly*

**Physics**

Three objects with masses m1= 36.5 kg m2= 19.2 kg m3= 12.5 kg Are hanging from massless ropes that run over a frictionless pulleys. If the system is released from rest, what is the magnitude of the acceleration in m/s^2 of mass m1? ___________________ O O O I I I I I I I I I I...
*Thursday, January 16, 2014 at 7:20pm by Holly*

**Physics**

m1=2250 kg, m2 =330 kg, R=550 N, a= 2 m/s² m1a=F-R-T m2a =T- m2g a(m1+m2)=F-R-T+T-m2g F= a(m1+m2) +m2g+R=... T=m2(a+g)=
*Friday, October 19, 2012 at 1:20pm by Elena*

**Physics**

Two masses are held while suspended on a frictionless pulley.Their masses are 0.250 kg and 0.200 kg. a. Compute for the acceleration of the masses once they are released? b.Find the tension on the string m1 accelerates upward and m2 downward Let a be the acceleration of m1 ...
*Sunday, September 10, 2006 at 8:43am by fluffy*

**Physics**

Consider two masses m1 and m2 connected by a thin string. Assume the following values: m1 = 4.51 kg and m2 = 1.00 kg. Ignore friction and mass of the string.
*Tuesday, January 25, 2011 at 1:26pm by Lauren*

**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) =(m1x1+m2x2)/(m1+m2) =(22+4.54)/6.5 = 3.38 m, y(c) =(m1y1+m2y2)/(m1+m2) =(26+4.51)/6.5 = 2.54 m. v(cx) =(m1v1x+m2v2x)/(m1+m2) =(22.1+4.54.6)/6.5 = 3....
*Tuesday, June 12, 2012 at 12:14pm by Elena*

**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*

**Physics**

A cart of mass M1 = 6 kg is attached to a block of mass M2 = 3 kg by a string that passes over a frictionless pulley. The system is initially at rest and the table is frictionless. After the block has fallen a distance h = 1 m: What is the work Ws done on the cart by the ...
*Tuesday, June 13, 2006 at 12:30pm by Jamila*

**PHYSICS**

use the formula F=(G(m1)(m2))/r^2 G=universal gravitational constant= 6.67x10^-11 N m^2/kg^2 F=9.8 N (given) m1=1.0 kg (given) m2=mass of earth (variable) r=6.4x10^6 m (given) 9.8 N=((6.67x10^-11 N m^2/kg^2)(1.0 kg)(m2))/(6.4x10^6 m) x=6.02x10^24 kg
*Wednesday, December 8, 2010 at 9:58pm by Kaitlin*

**Please help!!!!!Physics**

Find the resultant force on (a) the mass m1 = 0.183 kg and (b) the mass m2 = 0.106 kg (the masses are isolated from the earth).. The first mass is 400kg and its 10 cm from the second mass which is 0.183 kg (m1). the second mass is 10 cm away from the third mass which is 0.106 ...
*Thursday, September 29, 2011 at 10:45pm by kellie*

**Physics 203 ( College)**

Four particles, one at each of the four corners of a square with 2.5-m-long edges, are connected by mass less rods. The masses of the particles are m1 = m3 = 2.0 kg and m2 = m4 = 7.0 kg. Find the moment of inertia of the system about the z axis.
*Monday, April 4, 2011 at 11:58am by Kayla*

**physics**

m1 =0.0112 kg, v =? m2 = 1.01kg, u = 4.4 m/s The law of conservation of linear momentum m1v + 0 = (m1+m2) u, (a) v =(m1+m2) u/m1. (b) ΔKE = m1v^2/2 (m1+m2) ^2/2.
*Monday, April 30, 2012 at 9:12pm by Elena*

**Physics**

For the system of two crates and a pulley in the figure below, what fraction of the total kinetic energy resides in the pulley? (Assume m1 = 14 kg, m2 = 10 kg and mpulley = 5 kg.) (( o )) | | | | [m1] | [m2]
*Tuesday, April 1, 2014 at 9:58pm by Sarah*

**physics**

In a Physics laboratory class, an object of mass 2.1 kg, attached by massless strings to two hanging masses, m1= 1.0 kg and m2= 4.0 kg, is free to slide on the surface of the table. the coefficient of kinetic friction between m2 and the table is 0.30. calculate the ...
*Sunday, December 13, 2009 at 12:16am by Ashley*

**Physics**

In a Physics laboratory class, an object of mass 2.1 kg, attached by massless strings to two hanging masses, m1= 1.0 kg and m2= 4.0 kg, is free to slide on the surface of the table. the coefficient of kinetic friction between m2 and the table is 0.30. calculate the ...
*Monday, October 15, 2012 at 7:30pm by Ashley*

**Physics**

A dynamics cart m1 attached to has a hanging mass m2 m1 = 0.20 kg, and m2 = 1.3 kg. The magnitude of the acceleration of this system will be what approximately?
*Thursday, December 8, 2011 at 7:57pm by Anonymous*

**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 = m1v1 (m+m2)u1 u1 = {m1v1 - (m+m1+m2) v}/{m+m2} = ={110 (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**

Consider two masses m1 and m2 connected by a thin string. Assume the following values: m1 = 4.18 kg and m2 = 1.00 kg. Ignore friction and mass of the string. 1) what is the acceleration of the 2 masses? 2)What should be the value of mass m1 to get the largest possible value of...
*Thursday, September 1, 2011 at 10:50pm by Logan*

**physics**

Two balls have their centers 2.0 m apart. One ball has a mass of m1 = 8.3 kg. The other has a mass of m2 = 5.9 kg. What is the gravitational force between them? (Use G = 6.67 10-11 N · m2/kg2.)
*Monday, December 3, 2012 at 3:20pm by Danny*

**Physics**

Two masses, m1 = 2.0 kg and m2 = 3.3 kg, are moving in the xy-plane. The velocity of their center of mass and the velocity of mass 1 relative to mass 2 are given by the vectors vcm = (−1,+2.2) m/s and vrel = (+4.5, +1.0) m/s. (a) Determine the total momentum of the ...
*Tuesday, October 11, 2011 at 9:52am by mymi*

**physics**

Use the Newton Universal Law of gravity. F = G*M1*M2/R^2 Look up the value of G if you don't know it. Make sure M1 and M2 are both in kg. The lead ball mass is 0.100 kg
*Wednesday, February 3, 2010 at 7:04pm by drwls*

**Physics**

A mass m1 = 4.1 kg rests on a frictionless table and connected by a massless string to another mass m2 = 5.9 kg. A force of magnitude F = 32.0 N pulls m1 to the left a distance d = 0.83 m. 1) How much work is done by the force F on the two block system? 2) How much work is ...
*Wednesday, January 2, 2013 at 1:27pm by Drew*

**physics**

m1 =1.5 kg, m2 = 7.5 kg, a=2.9 m/s² (a) The horizontal projections of the equations of motion for each block are m1a = T, m2a = T-F, F = (m1+m2) a = (1.5+7.5) 2.9 = 26.1 N, (b) T= m2a - F= 7.52.9 26.1 = 4.35 N.
*Monday, May 7, 2012 at 10:42pm by Elena*

**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 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*

**Physics**

m1 = 152 kg, m2 = 269 kg, m3 = 53.5 kg. d =0.45 m, r =d/2 =0.225 m, G = 6.67210^−11 N m2/kg2. m1
.
.<-------m3-------------->
..m2 F13 F23 F13 = Gm1m3/r^2, F23 = Gm2m3/r^2, F = F23 F13, m1
.
.<-------m3------->
..m2 F13 F23 F13 = F23, Gm1m3/(d-...
*Thursday, May 3, 2012 at 5:30pm by Elena*

**Physics!!**

m1 = 152 kg, m2 = 269 kg, m3 = 53.5 kg. d =0.45 m, r =d/2 =0.225 m, G = 6.67210^−11 N m2/kg2. m1
.
.<-------m3-------------->
..m2 F13 F23 F13 = Gm1m3/r^2, F23 = Gm2m3/r^2, F = F23 F13, m1
.
.<-------m3------->
..m2 F13 F23 F13 = F23, Gm1m3/(d-...
*Tuesday, May 8, 2012 at 3:42pm by Elena*

**physics**

Two masses m1 and m2 are suspended from the ceiling by two string segments. What is the tension in each string segment? Let m1 = 10 kg and m2 = 20 kg.
*Tuesday, December 7, 2010 at 5:45pm by Kay*

**physics**

two particles of mass m1 and m2 are joined by a massless spring of spring constant k and natural length L.Initially m2 is on table and m1 is above m2 at a vertical height h.At t=0 m1 moves with velocity v0.find the position of the masses
*Saturday, December 4, 2010 at 8:55pm by Anonymous*

**physics**

Two blocks with masses M1 = 3.9 kg and M2 = 5.0 kg are connected with a massless string over two massless and frictionless pulleys, as shown in the Figure. One end of the string is connected to M1 while the other end is fixed. What is the acceleration of mass M2
*Sunday, September 23, 2012 at 7:21pm by jameson*

**Physics**

M1 has a mass of 5.01 kg. It is on a horizontal surface, connected by a light string to a hook. Mass M2 can be increased smoothly by adding masses little at a time. The pulley has a negligible mass and no friction. When M2 is 2.40 kg it begins to accelerate downwards at a rate...
*Thursday, October 6, 2011 at 7:43pm by Bill Burger*

**physics**

Two blocks with masses m1 = 10.9 kg and m2 = 79.3 kg, shown in the figure, are free to move. The coefficent of static friction between the blocks is 0.66 but the surface beneath m2 is frictionless. What is the minimum force F required to hold m1 against m2?
*Monday, November 7, 2011 at 9:06pm by lisa*

**physics**

Two blocks with masses m1 = 20.7 kg and m2 = 55.5 kg, shown in the figure, are free to move. The coefficent of static friction between the blocks is 0.74 but the surface beneath m2 is frictionless. What is the minimum force F required to hold m1 against m2?
*Wednesday, February 8, 2012 at 10:58am by Katie*

**phsyics**

A mass, m1 = 9.36 kg, is in equilibrium while connected to a light spring of constant k = 112 N/m that is fastened to a wall. A second mass, m2 = 6.71 kg, is slowly pushed up against mass m1, compressing the spring by the amount A = 0.201 m The system is then released, and ...
*Thursday, November 5, 2009 at 11:43pm by Anonymous*

**physics**

m1=0.005 kg, m2= 1 kg, R=0.2 m. v=? Law of conservation of momentum m1v = m1(v/2) +m2u m1v/2= m2u, u= m1v/2m2. (1) Law of conservation of energy KE =PE +KE1 m2u²/2=m2g2R+m2u1²/2 u²/2= g2R+ u1²/2 . (2) At the highest point m2a=m2g m2u1²/R...
*Thursday, July 12, 2012 at 1:34pm by Elena*

**Physics URGENT!!!**

m=2 kg , m1=70+2=72 kg, m2= 50 kg v =9 m/s, Δh=4 m, h=20 m (a) m1v =(m1+m)v1 v1= m1/(m1+m) (b) KE1+ΔPE=KE2 (m1+m)v1²/2+(m1+m)gΔh=(m1+m)v2²/2 v2=sqrt(v1²+2gΔh) (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*

**Physics**

Four particles in the xy plane have the following masses and coordinates: 1 kg at (4 m, 4 m), 2 kg at (−3 m, 5 m), 3 kg at (5 m, −2 m) and 4 kg at (−2 m, −4 m). Find the total moment of inertia about: a) the x-axis: kg*m2 b) the y-axis: kg*m2 b) the z-...
*Tuesday, December 14, 2010 at 7:17pm by applebottom*

**Physics**

For m1: m1a=T1 (block m1=1 kg is on the surface ) For m2: m2a=m2g T2 (m2=9 kg) For pulley: Iε = M => Ia/R= (T2-T1)R => (T2-T1) = Ia/R² a(m1+m2+ I/R²)=m2g a= m2g/(m1+m2+ I/R²) =99.8/{1+9+(0.3/0.25)}=7.885 m/s² T1= m1m2g/(m1+m2+ I/R&#...
*Tuesday, November 6, 2012 at 4:02pm by Elena*

**physics**

Two masses M1 = 3.00 kg and M2 = 7.50 kg are stacked on top of each other as shown in the figure. The static coefficient of friction between M1 and M2 is μs = 0.520. There is no friction between M2 and the surface below it
*Sunday, September 16, 2012 at 11:09pm by Kylie*

**physics**

Two masses m1= 35.0 Kg and m2= 15.0Kg are connected by a mass less string over a smooth pulley (no friction), the mass m1 being on a horizontal frictionless table. The motion of the system involves m1 moving horizontally to the right and m2 vertically downward, both with ...
*Saturday, April 23, 2011 at 11:41am by Kayla*

**Physics **

The sum of the two cable tensions is T1 + T2 = (M1 + M2)g = 1176 Newtons M1 = board mass (50 kg) M2 = cleaner's mass (70 kg) For the tension T1,set the moment about the other end equal to zero. Let the cleaner be 2 m from cable 1. Take moments about cable 2, 3m from the man T1...
*Wednesday, May 18, 2011 at 7:03am 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/(kgK) is heat capacity of water Q1 = λm1 Q2 =cm1 (t-t1) =cm1t Q3 = cm2 (t2-t) . Q1+Q2 = ...
*Monday, April 9, 2012 at 12:25am by Elena*

**Physics**

all correct. One way to think on four is to stop considering "g" as acceleration as in three. In the equation gravitational force= mass*g consider g to be gravitational field strength, in newtons/kg. In Earths case, g at the surface is 9.8N/kg Using this thinking, then if the ...
*Sunday, July 18, 2010 at 3:36am by bobpursley*

**Physics**

The two masses (m1 = 5.0 kg and m2 = 3.0 kg) in the Atwood's machine shown in Figure 10-23 are released from rest, with m1 at a height of 0.87 m above the floor. When m1 hits the ground its speed is 1.4 m/s. Assume that the pulley is a uniform disk with a radius of 12 cm. ...
*Monday, March 8, 2010 at 8:10pm by Mary*

**Physics**

The two masses (m1 = 5.0 kg and m2 = 3.0 kg) in the Atwood's machine shown in Figure 10-23 are released from rest, with m1 at a height of 0.91 m above the floor. When m1 hits the ground its speed is 1.5 m/s. Assume that the pulley is a uniform disk with a radius of 12 cm. ...
*Monday, March 8, 2010 at 10:28pm by Jon*

**College Physics**

13. What is the magnitude of the gravitational force that acts on two particles? Assume that particle 1 (m1) is 12 kg and particle 2 (m2) is 25 kg, they are both separated by a distance of 1.2m
*Wednesday, October 17, 2012 at 5:50pm by Suzie*

** college physics**

a) The applied force of F = 46N acts on the system of all three blocks. You can ignore the internal foces (cord and contact) between blocks and apply F = (M1 + M2 + M3) a, where M1 = 1 kg, M2 = 2 kg and M3 = 3 kg. a = 46 N/6 kg = 23/3 m/s^2 b) Let f1 be the cord tension. Only ...
*Sunday, September 21, 2008 at 5:31am by drwls*

**physics**

An object with mass m1 = 5.00 kg, rests on a frictionless horizontal table and is connected to a cable that passes over a pulley and is then fastened to a hanging object with mass m2 = 9.0 kg, as shown in Figure P4.30. Find the acceleration of each object and the tension in ...
*Monday, June 9, 2008 at 2:11pm by Elisa*

**physics**

Two particles are in a uniform electric field whose value is +4000 N/C. The mass and charge of particle 1 are m1 = 1.0x10-5 kg and q1 = -15 ΅C, while the corresponding values for particle 2 are m2 = 3.0 x10-5 kg and q2 = +35.0 ΅C. Initially the particles are at rest. The ...
*Monday, October 4, 2010 at 1:25pm by Nate*

**Physics.**

A mass of m1 = 12 kg is placed on an incline with an inclination angle of 50 degrees, a coefficient of kinetic friction of 0.16, and a coefficient of static friction of 0.89. m1 is tied to a rope which runs parallel to the incline and runs over a pulley to a hanging mass of m2...
*Friday, October 21, 2011 at 2:20pm by Amanda *

**Algebra**

Things would be less messy if we do the algebraic manipulations before putting in numbers. Let: Mass Moon, M2 = 7.35 E 22 kg Mass Earth, M1 = 5.98 E 24 Kg Distance Earth to Moon, D = 384,403,000 and r the distance of space craft from the centre of the earth/moon. GM1m/r²...
*Wednesday, December 23, 2009 at 9:47pm by MathMate*

**Mechanics**

Two blocks with masses m1 = kg and m2 = kg are connected by a string that hangs over a pulley of mass M = kg and radius R = m as shown above. The string does not slip. Assuming the system starts from rest, use energy principle find the speed of m2 after it has fallen by m. ...
*Saturday, December 3, 2011 at 5:05pm by Costa*

**physics - HELP**

m1=62.9 kg m2=697 kg F=117 N t=0.977 s. v1 = velocity of astronaut after contact v2 = velocity of satellite after contact By the conservation of momentum, m1 v1 + m2 v2 =0 v2=-(m1 v1)/m2 ....(1) By the conservation of energies, F*t = (m1/2)v1² + (m2/2)v2² ....(2) ...
*Sunday, September 27, 2009 at 3:28pm by MathMate*

**Physics Class**

I don't understand this question. Three weights are attached to the corners of an equilateral triangle with a side length 10 cm. Measured from the bottom left corner, what is the center of mass of this object if M1 = 0.1 kg, M2 = 0.2 kg, and M3 = 0.3 kg? and M2 is on the left...
*Tuesday, February 22, 2011 at 9:37pm by Ann*

**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 *

**Physic**

Ok so this question was posted before but I'm confused by the answer. The problem is: The two masses (m1 = 5.0 kg and m2 = 3.0 kg) in the Atwood's machine shown in Figure 10-23 are released from rest, with m1 at a height of 0.87 m above the floor. When m1 hits the ground its ...
*Monday, March 8, 2010 at 8:47pm by Kelsey*

**physics**

The location of two particles at t = 0 is shown in the figure below. The particles are initially at rest. Each particle is subject to a constant force, as indicated in the figure. (Take m1 = 2.06 kg and m2 = 0.95 kg.) (a) Find the position of the center of mass at t = 0. xcm...
*Thursday, February 10, 2011 at 4:25pm by Anonymous*

**Physics Help**

Use conservation of linear momentum. M1*V1 + M2*V2 = (M1+M2)*V3 The only unknown is M2 in this case, so you can solve for it. 1.9*1.8 + M2*4.5 = (1.9 + M2)*3.5 3.42 + M2 = 6.65 M2 = ____ kg
*Wednesday, August 10, 2011 at 10:46pm by drwls*

**Physics**

A mass m1 on a horizontal shelf is attached by a thin string that passes over a frictionless pulley to a 2.5 kg mass (m2) that hangs over the side of the shelf 1.2 m above the ground. The system is released from rest at t = 0 and the 2.5 kg mass strikes the ground at t = 0.82 ...
*Sunday, February 26, 2012 at 8:30pm by J*

**Physics**

A mass m1 = 5.0 kg rests on a frictionless table and connected by a massless string to another mass m2 = 5.9 kg. A force of magnitude F = 40.0 N pulls m1 to the left a distance d = 0.77 m. 1)How much work is done by the force F on the two block system? 2)What is the final ...
*Monday, October 1, 2012 at 8:32pm by J.C.*

**Physics**

Two objects are connected by a light string that passes over a frictionless pulley as in the fi gure. One object lies on a frictionless, smooth incline. In the figure, m1 = 8.48 kg, m2 = 4.84 kg, and theta = 40.2 degrees . When we let go of the mass m1, it accelerates ...
*Sunday, January 27, 2013 at 11:46pm by Lindsey*

**Physics**

Two blocks with masses m1 = 3.9 kg and m2 = 6.7 kg are connected by a string that hangs over a pulley of mass M = 2.2 kg and radius R = 0.11 m as shown above. The string does not slip. Assuming the system starts from rest, use energy principle find the speed of m2 after it has...
*Monday, November 28, 2011 at 3:55pm by John*

**Science**

Two blocks with masses m1 = 3.9 kg and m2 = 6.7 kg are connected by a string that hangs over a pulley of mass M = 2.2 kg and radius R = 0.11 m as shown above. The string does not slip. Assuming the system starts from rest, use energy principle find the speed of m2 after it has...
*Wednesday, November 30, 2011 at 5:41pm by John*

**physics**

m1=10 kg, m2 = 4 kg, α = 40º. m1a= m1g -T, m2a = T m2gsinα, a(m1 +m2) = m1g T+T- m2gsinα = = g(m1 m2sinα), a= g(m1 m2sinα)/(m1 +m2), T = m1(g -a).
*Tuesday, May 15, 2012 at 2:28pm by Elena*

**Physics 121**

m1 =4 kg, m2 = 2 kg, m = 0.5 kg, R = 0.25 m, h= 0.75 m. Projections on the horizontal and vertical axis: m1a = T1 m2a =m2g-T2, Iε =M. If the pulley is the disk (cylinder) I =mR²/2 , M = torque = (T1-T2) R, ε = a/R, Iε =M => mR²a/2R =(T1-T2...
*Sunday, May 6, 2012 at 8:13pm by Elena*

**Physics**

Two blocks are connected by a light string that passes over a frictionless pulley as in the figure below. The system is released from rest while m2 is on the floor and m1 is a distance h above the floor. (a) Assuming m1 > m2, find an expression for the speed of m1 just as ...
*Monday, March 19, 2012 at 4:12am by Lilly*

**physics**

Two masses are hanging over a frictionless, massless pulley as shown blocks_pulley_aMass m1 is 5.0 kg and mass m2 is 7.0 kg
*Saturday, June 19, 2010 at 6:25pm by bill*

**Physics**

m1 =1,5 kg, m2 = 2 kg, m3 =0.0527 kg. v = 1.27 m/s u1 =75 m/s, u2 =? The law of conservation of linear momentum (m1+m2+m3) v = (m1+m2) u2 + m3u1, u2 = {(m1+m2+m3) v - m3u1}/(m1+m2) = ={(1.5+2+0.0527)1.27 0.052775}/(1.5+2)= =(4.51-3.95)/3.5 = 0.16 m/s (in the previous ...
*Monday, April 30, 2012 at 1:34pm by Elena*

**Physics help please**

the F of gravity is equal to (G*m1*m2)/(d^2) where g is the gravitational constant (6.673e-11 N*m^2/kg^2, notice the units are m not km), m1 is the mass of one object, m2 is the mass of the other object, and d= the distance between the centers of each object. in this situation...
*Friday, November 25, 2011 at 8:36pm by Cody*

**Physics**

m1=8.8 kg, v1=4.5 m/s m2=0.5 kg, v2=0 u1=? u1= (m1-m2)v1/)m1+m2), u2=2m1v1/(m1+m2).
*Monday, July 9, 2012 at 7:22pm by Elena*

**Physics 121**

m1 =4 kg, m2 = 2 kg, m = 0.5 kg, R = 0.25 m, h= 0.75 m. Projections of the equation according to the 2 Newton's law for two blocks on the horizontal (for the 1st block)and on the vertical (fot the 2nd block) axis: m1a = T1 m2a =m2g-T2, The equation of the pulley motion (2nd...
*Saturday, May 5, 2012 at 9:07pm by Elena*

**PHYSICS HELP PLEASE !!!!!**

A mass of m1 = 12 kg is placed on an incline with an inclination angle of 50 degrees, a coefficient of kinetic friction of 0.16, and a coefficient of static friction of 0.89. m1 is tied to a rope which runs parallel to the incline and runs over a pulley to a hanging mass of m2...
*Friday, October 21, 2011 at 5:40pm by Amanda *

**Physics**

m1+m2+m3 = 92.5 kg. m1= 0.31 kg m2+m3 =92.19 kg v=0.55 m/s v1= - 17 m/s v2=? (m1+m2+m3) v= - m1v1+(m2+m3) v2 v2= {(m1+m2+m3) v +m1v1}/(m2+m3)=
*Monday, October 22, 2012 at 8:40pm by Elena*

**physics**

1) Mass m1 = 0.7 kg moves with constant velocity v1i = 2.0 m/s along the x-axis and collides with mass m2 = 1.8 kg, which moves with velocity v2i = 3.4 m/s, as sketched below. After the collision, m1 and m2 stick together. Find the angle, θ, between the x-axis and the ...
*Saturday, October 13, 2012 at 5:54pm by Jim*

**physics**

1) Mass m1 = 0.7 kg moves with constant velocity v1i = 2.0 m/s along the x-axis and collides with mass m2 = 1.8 kg, which moves with velocity v2i = 3.4 m/s, as sketched below. After the collision, m1 and m2 stick together. Find the angle, θ, between the x-axis and the ...
*Saturday, October 13, 2012 at 5:55pm by Jim*

**physics**

What is the force of gravity between two spheres that have a mass of m1 =13 kg, m2 = 49 kg and are separated by 35 m?
*Friday, February 11, 2011 at 11:00am by Anonymous*

**physic**

Hi!An object with mass m1 = 5.00 kg, rests on a frictionless horizontal table and is connected to a cable that passes over a pulley and is then fastened to a hanging object with mass m2 = 10.0 kg, as shown in the figure below. Find the acceleration of each object and the ...
*Friday, October 9, 2009 at 3:19am by Chan*

**Physics**

m1 = 35 g = 0.035 kg, v1 =? m2 = 2.6 kg, v2 =0, u = 7.5 m/s m1v1 =(m1+m2) u, v1 =(m1+m2) u/m1 = ......
*Wednesday, May 9, 2012 at 9:09pm by Elena*

**physics**

Torque = cos 27.1*(M1*g*1.22m + M2*g*2.25m) M1 = 42.4 kg and M2 = 34.8 kg
*Tuesday, November 23, 2010 at 11:39am by drwls*

**physics**

Find the center of mass of the three mass system given that m1 1.90 kg and m2 1.30 kg Specify relative to the center of the left mass "1.00 kg "
*Wednesday, October 6, 2010 at 8:58pm by ami*

**phy**

Find the center of mass of the three mass system that m1 = 1.90 kg and m2 = 1.30 kg. Specify relative to the center of the left hand (1.00 kg) mass.
*Thursday, October 7, 2010 at 7:00pm by ami*

**physics**

Three blocks are connected on the table as shown below. The coefficient of kinetic friction between the block of mass m2 and the table is 0.400. The objects have masses of m1 = 3.00 kg, m2 = 1.45 kg, and m3 = 1.90 kg, and the pulleys are frictionless. Determine the ...
*Saturday, October 9, 2010 at 12:34am by Amber*

**physics**

Three blocks are connected on the table as shown below. The coefficient of kinetic friction between the block of mass m2 and the table is 0.370. The objects have masses of m1 = 5.00 kg, m2 = 1.40 kg, and m3 = 2.30 kg, and the pulleys are frictionless. Determine the ...
*Tuesday, March 6, 2012 at 12:11pm by Anonymous*

**Physics**

Two blocks with masses m1 = 1.40 kg and m2 = 3.00 kg are connected by a massless string, as shown in the Figure. They are released from rest. The coefficent of kinetic friction between the upper block and the surface is 0.390.In the figure m1 is on top of a horizontal table ...
*Wednesday, March 6, 2013 at 9:14pm by Heidi*

**physics**

?????Two books of mass m1 = 6.9 kg and m2 = 3.8 kg are stacked on a table as shown in the figure below. Find the normal force acting between the table and the book on the bottom. SO g*m1+g*m2=104.86 N = mgcos(θ) N= (104.86)(9.81)Cos? Where do i find cos? IS THIS RIGHT?
*Thursday, September 9, 2010 at 9:51pm by tommy *

**physics**

m1= 443 kg, m2=181 kg, F(fr) = F =Gm1m2/R², μm2g= Gm1m2/R², μg= Gm1/R², R =sqrt(Gm1/ μg).
*Thursday, December 13, 2012 at 9:51pm by Elena*

**physics**

m1= 443 kg, m2=181 kg F(fr) = F =Gm1m2/R² μm2g= Gm1m2/R² μg= Gm1/R² R =sqrt(Gm1/ μg)
*Thursday, December 13, 2012 at 9:29pm by Elena*

**Physics**

The particles in the figure below (m1=1.6 kg and m2=3.5 kg) undergo an elastic collision in one dimension. Their velocities before the collision are v1i=12 m/s and v2i=-7.5 m/s. Find the velocities of the two particles after the collision. (Indicate the direction with the sign...
*Saturday, October 27, 2012 at 10:18pm by David*

**phy**

Two crates, of mass m1 = 60 kg and m2 = 130 kg, are in contact and at rest on a horizontal surface (Fig. 4-54). A 620 N force is exerted on the 60 kg crate. The coefficient of kinetic friction is 0.15.
*Sunday, September 12, 2010 at 3:50pm by ami*

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