What the kinetic energy of the projectile when it is shot from the gun?
Q. A spring-powered dart guy is un-stretched and has a spring constant 26.0 N/m. The spring is compressed by 4.0 cm and a 25.0 gram projectile is placed in the gyn. The kinetic energy of the projectile when it is shot from the gun is what?
Asked by Ari - Wed Oct 14 21:13:09 2009 - - 1 Answers - 0 Comments
A. Potential energy of the spring = Kinetic energy right after it is shot Ke=.5kx^2 Plug it in and get .0208 J
Answered by brucelee41562 - Wed Oct 14 21:32:08 2009
Q. A spring-powered dart guy is un-stretched and has a spring constant 26.0 N/m. The spring is compressed by 4.0 cm and a 25.0 gram projectile is placed in the gyn. The kinetic energy of the projectile when it is shot from the gun is what?
Asked by Ari - Wed Oct 14 21:13:09 2009 - - 1 Answers - 0 Comments
A. Potential energy of the spring = Kinetic energy right after it is shot Ke=.5kx^2 Plug it in and get .0208 J
Answered by brucelee41562 - Wed Oct 14 21:32:08 2009
What is kinetic linking and how is it achieved?
Q. I always hear about boxers using 'kinetic linking' to make their punches more powerful. What is it and how can I do it?
Asked by skaterboy854856 - Mon Jan 11 19:23:32 2010 - - 3 Answers - 0 Comments
A. look at this it explains it all.
Answered by blackman comin thru - Mon Jan 11 21:25:36 2010
Q. I always hear about boxers using 'kinetic linking' to make their punches more powerful. What is it and how can I do it?
Asked by skaterboy854856 - Mon Jan 11 19:23:32 2010 - - 3 Answers - 0 Comments
A. look at this it explains it all.
Answered by blackman comin thru - Mon Jan 11 21:25:36 2010
How do I get the kinetic energy from these?
Q. At 20 degrees Celcius (approximately room temperature) the average velocity of molecules in air is 469.4 meters/second. 1. What is the kinetic energy (in Joules) of an N2 molecule moving at this speed? 2. What is the total kinetic energy of 1 mol of N2 molecules moving at this speed? *I always give 10 points* Thanks.
Asked by `~-Cynic - Mon Oct 13 16:46:43 2008 - - 2 Answers - 0 Comments
A. Kinetic energy is the energy of motion. It can be calculated by KE = 0.5mv^2 In this problem, you are given the velocity, so you need the mass. Molecular nitrogen (N2) is 28.0 g/mol or 0.028 kg/mol. Using Avogadro's number, you can calculate the mass of a single molecule from this. m = (0.028 kg/mol)/(6.022e23/mol) = 4.65e-26 kg Now KE = (4.65e-26 kg)(469.4 m/s)^2 = 1.03e-20 kgm^2/s^2 A kg*m^2/s^2 is a J. So the average kinetic energy of each molecule is 1.03e-20 J. To get the total kinetic energy, you must multiply by Avogadro's number. (1.03e-20 J)(6.022e-23/mol) = 6180 J/mol or 6.18 kJ/mol
Answered by nddulac - Mon Oct 13 16:59:03 2008
Q. At 20 degrees Celcius (approximately room temperature) the average velocity of molecules in air is 469.4 meters/second. 1. What is the kinetic energy (in Joules) of an N2 molecule moving at this speed? 2. What is the total kinetic energy of 1 mol of N2 molecules moving at this speed? *I always give 10 points* Thanks.
Asked by `~-Cynic - Mon Oct 13 16:46:43 2008 - - 2 Answers - 0 Comments
A. Kinetic energy is the energy of motion. It can be calculated by KE = 0.5mv^2 In this problem, you are given the velocity, so you need the mass. Molecular nitrogen (N2) is 28.0 g/mol or 0.028 kg/mol. Using Avogadro's number, you can calculate the mass of a single molecule from this. m = (0.028 kg/mol)/(6.022e23/mol) = 4.65e-26 kg Now KE = (4.65e-26 kg)(469.4 m/s)^2 = 1.03e-20 kgm^2/s^2 A kg*m^2/s^2 is a J. So the average kinetic energy of each molecule is 1.03e-20 J. To get the total kinetic energy, you must multiply by Avogadro's number. (1.03e-20 J)(6.022e-23/mol) = 6180 J/mol or 6.18 kJ/mol
Answered by nddulac - Mon Oct 13 16:59:03 2008
What is the kinetic energy of the mass when mass reaches the bottom?
Q. A 2 kg mass is moving with a speed of 1 m/s at the top of a frictionless slide that is 3 m high. What is the kinetic energy of the mass when it reaches the bottom? I believe this is the equation I'm supposed to use: KE=1/2mv2. (The second 2 is a squared). 2 things that really confuse me in this problem: the frictionless part and I don't know where to put the height. Thanks for the help!
Asked by ChemHelp - Thu Jun 26 10:29:22 2008 - - 4 Answers - 0 Comments
A. Protencel energy Gain = mgh = 2*9.8*3 since frictionless slide this is Converted to KE KE = 2*9.8*3= 58.8J
Answered by Kaushalya Damitha - Thu Jun 26 10:38:12 2008
Q. A 2 kg mass is moving with a speed of 1 m/s at the top of a frictionless slide that is 3 m high. What is the kinetic energy of the mass when it reaches the bottom? I believe this is the equation I'm supposed to use: KE=1/2mv2. (The second 2 is a squared). 2 things that really confuse me in this problem: the frictionless part and I don't know where to put the height. Thanks for the help!
Asked by ChemHelp - Thu Jun 26 10:29:22 2008 - - 4 Answers - 0 Comments
A. Protencel energy Gain = mgh = 2*9.8*3 since frictionless slide this is Converted to KE KE = 2*9.8*3= 58.8J
Answered by Kaushalya Damitha - Thu Jun 26 10:38:12 2008
What is the kinetic energy of the moving piece?
Q. A space vehicle of mass m has a speed v. At some instant, it seperates into 2 pieces of mass/2. One of the pieces is at rest just after separation. What is the kinetic energy of the moving piece?
Asked by Aggie Love :) - Tue Jul 7 10:26:56 2009 - - 2 Answers - 0 Comments
A. We know that in an explosive separation the First Law of Thermodyanmics cannot be directly applied because of losses due to heat and acoustic energy therefore this problem must be solved by applying the conservation of momentum. Let's use the following Notation: M = mass of complete spacecraft before separation V = speed of the space craft prior to separation. m1 = mass of moving piece = M/2 v1 = speed of moving piece m2 = mass of stationary piece = M/2 v2 = speed of stationary piece = 0 So from conservation of momentum M * V = (M/2) * v1 + (M/2 * v2) M divides out and v2 = 0 so: V = v1/2 or v1 = 2 V Ek = 1/2 m v^2 so Ek = 1/2 (M/2) * (2V)^2 Ek= (M/4)*4*V^2 Ek = MV^2
Answered by Doug R - Tue Jul 7 11:06:46 2009
Q. A space vehicle of mass m has a speed v. At some instant, it seperates into 2 pieces of mass/2. One of the pieces is at rest just after separation. What is the kinetic energy of the moving piece?
Asked by Aggie Love :) - Tue Jul 7 10:26:56 2009 - - 2 Answers - 0 Comments
A. We know that in an explosive separation the First Law of Thermodyanmics cannot be directly applied because of losses due to heat and acoustic energy therefore this problem must be solved by applying the conservation of momentum. Let's use the following Notation: M = mass of complete spacecraft before separation V = speed of the space craft prior to separation. m1 = mass of moving piece = M/2 v1 = speed of moving piece m2 = mass of stationary piece = M/2 v2 = speed of stationary piece = 0 So from conservation of momentum M * V = (M/2) * v1 + (M/2 * v2) M divides out and v2 = 0 so: V = v1/2 or v1 = 2 V Ek = 1/2 m v^2 so Ek = 1/2 (M/2) * (2V)^2 Ek= (M/4)*4*V^2 Ek = MV^2
Answered by Doug R - Tue Jul 7 11:06:46 2009
What is the kinetic energy change when mass and speed are changed?
Q. 6) A truck has four times the mass of a car and is moving with twice the speed of the car. If Kt and Kc refer to the kinetic energies of truck and car respectively, it is correct to say that A) Kt = 16Kc B) Kt=4Kc. C) Kt = 2Kc. D) Kt=5Kc. E) Kt= Kc.
Asked by Science456 - Thu Oct 15 03:29:59 2009 - - 1 Answers - 0 Comments
A. Remember the equation that defines kinetic energy, K = 0.5 mv . For the truck: Kt = 0.5(mt)(vt) For the car: Kc = 0.5(mc)(vc) . "truck has 4 times the mass of a car" means mt = 4mc "moving with twice the speed of the car" means vt = 2vc. Substituting this information into the Kt equation gives Kt = 0.5(4mc)(2vc) Kt = 0.5*4(mc)2 (vc) Kt = 4*4*0.5(mc)(vc) Kt = 16*0.5(mc)(vc) Kt = 16Kc
Answered by KiwiJoey - Thu Oct 15 05:19:53 2009
Q. 6) A truck has four times the mass of a car and is moving with twice the speed of the car. If Kt and Kc refer to the kinetic energies of truck and car respectively, it is correct to say that A) Kt = 16Kc B) Kt=4Kc. C) Kt = 2Kc. D) Kt=5Kc. E) Kt= Kc.
Asked by Science456 - Thu Oct 15 03:29:59 2009 - - 1 Answers - 0 Comments
A. Remember the equation that defines kinetic energy, K = 0.5 mv . For the truck: Kt = 0.5(mt)(vt) For the car: Kc = 0.5(mc)(vc) . "truck has 4 times the mass of a car" means mt = 4mc "moving with twice the speed of the car" means vt = 2vc. Substituting this information into the Kt equation gives Kt = 0.5(4mc)(2vc) Kt = 0.5*4(mc)2 (vc) Kt = 4*4*0.5(mc)(vc) Kt = 16*0.5(mc)(vc) Kt = 16Kc
Answered by KiwiJoey - Thu Oct 15 05:19:53 2009
What is the kinetic every, potential energy and acceleration of gravity in this experiment?
Q. If you drop a ball from a certain height, were is the kinetic energy and potential energy? And how about the acceleration energy?
Asked by snowflake - Sat Oct 6 17:44:16 2007 - - 1 Answers - 0 Comments
A. Total energy of anything, any time, anywhere can be written down as TE = KE + PE + WE; where KE is kinetic energy, PE is potential energy, and WE is work energy. Thus TE = constant for a given system and what changes are KE, PE, and WE so that the sum of the three terms always remains TE = constant. All this comes from the conservation of energy law, which says energy is not created or destroyed, but can be converted from one type to another type of energy. Let's look at your ball in this light. TE(h) = total energy of the ball at height h before dropping it = KE + PE + WE. But KE = 0 because the ball is not moving and WE = 0 because it is neither doing work or being worked on. So that leaves us with TE(h) = PE(h) = mgh; where m is… [cont.]
Answered by oldprof - Sat Oct 6 18:28:28 2007
Q. If you drop a ball from a certain height, were is the kinetic energy and potential energy? And how about the acceleration energy?
Asked by snowflake - Sat Oct 6 17:44:16 2007 - - 1 Answers - 0 Comments
A. Total energy of anything, any time, anywhere can be written down as TE = KE + PE + WE; where KE is kinetic energy, PE is potential energy, and WE is work energy. Thus TE = constant for a given system and what changes are KE, PE, and WE so that the sum of the three terms always remains TE = constant. All this comes from the conservation of energy law, which says energy is not created or destroyed, but can be converted from one type to another type of energy. Let's look at your ball in this light. TE(h) = total energy of the ball at height h before dropping it = KE + PE + WE. But KE = 0 because the ball is not moving and WE = 0 because it is neither doing work or being worked on. So that leaves us with TE(h) = PE(h) = mgh; where m is… [cont.]
Answered by oldprof - Sat Oct 6 18:28:28 2007
What is the maximum kinetic energy of the ball?
Q. A 1-lb rubber ball bounces down a flight of stairs, striking each step in turn and rebounding to the height of the step above. If the height of each step is 1 ft, find the coefficient of restitution. What is the maximum kinetic energy of the ball? Show the computations please... Thanks!
Asked by macedon - Mon Feb 11 04:15:55 2008 - - 2 Answers - 0 Comments
A. Fuuck off do your own homework.
Answered by AdrianaP - Mon Feb 11 04:25:43 2008
Q. A 1-lb rubber ball bounces down a flight of stairs, striking each step in turn and rebounding to the height of the step above. If the height of each step is 1 ft, find the coefficient of restitution. What is the maximum kinetic energy of the ball? Show the computations please... Thanks!
Asked by macedon - Mon Feb 11 04:15:55 2008 - - 2 Answers - 0 Comments
A. Fuuck off do your own homework.
Answered by AdrianaP - Mon Feb 11 04:25:43 2008
What is the book's total kinetic energy at the instant just before it strikes the floor?
Q. Imagine a book that is falling from a shelf. At a particular moment during its fall, it has a kinetic energy of 13 J and a potential energy with respect to the floor of 75 J. I know that E_k = 1/2mv^2 and E_p = mgh but I have no idea how to even approach the solution to this question. Q: What is the book's total kinetic energy at the instant just before it strikes the floor?
Asked by Manjui - Mon Mar 9 10:27:30 2009 - - 2 Answers - 0 Comments
A. The book's potential energy will change into kinetic energy, until its kinetic energy is a maximum which occurs when its potential energy is a minimum. So the final kinetic energy will be 75+13 = 88J
Answered by Garrett O - Mon Mar 9 10:33:19 2009
Q. Imagine a book that is falling from a shelf. At a particular moment during its fall, it has a kinetic energy of 13 J and a potential energy with respect to the floor of 75 J. I know that E_k = 1/2mv^2 and E_p = mgh but I have no idea how to even approach the solution to this question. Q: What is the book's total kinetic energy at the instant just before it strikes the floor?
Asked by Manjui - Mon Mar 9 10:27:30 2009 - - 2 Answers - 0 Comments
A. The book's potential energy will change into kinetic energy, until its kinetic energy is a maximum which occurs when its potential energy is a minimum. So the final kinetic energy will be 75+13 = 88J
Answered by Garrett O - Mon Mar 9 10:33:19 2009
What is the final kinetic energy of the cart?
Q. A horizontal force of 200 N is applied to move a 55-kg cart (initially at rest) across a 10 m level surface. What is the final kinetic energy of the cart?
Asked by lovety89 - Wed Sep 3 09:40:54 2008 - - 1 Answers - 0 Comments
A. f = ma 200=55xa a=3.64m/s^2 v^2 = u^2 + 2as v^2 = 0^2 + 2 x 3.64 x 10 v^2 = 72.73 v = 8.53ms KE=1/2mv^2 KE=1/2 x 55 x 8.53^2 KE=2000J=2kJ
Answered by beattme - Wed Sep 3 09:48:14 2008
Q. A horizontal force of 200 N is applied to move a 55-kg cart (initially at rest) across a 10 m level surface. What is the final kinetic energy of the cart?
Asked by lovety89 - Wed Sep 3 09:40:54 2008 - - 1 Answers - 0 Comments
A. f = ma 200=55xa a=3.64m/s^2 v^2 = u^2 + 2as v^2 = 0^2 + 2 x 3.64 x 10 v^2 = 72.73 v = 8.53ms KE=1/2mv^2 KE=1/2 x 55 x 8.53^2 KE=2000J=2kJ
Answered by beattme - Wed Sep 3 09:48:14 2008
What is the maximum kinetic energy of the photoelectrons?
Q. Light of 1.5 1015 Hz illuminates a piece of tin, which has a work function of 4.97 eV. (a) What is the maximum kinetic energy of the photoelectrons? (b) What is the threshold frequency of tin? Light of 1.5 10^15 Hz illuminates a piece of tin, which has a work function of 4.97 eV. (a) What is the maximum kinetic energy of the photoelectrons? (b) What is the threshold frequency of tin?
Asked by Joe - Fri May 1 20:40:29 2009 - - 1 Answers - 0 Comments
A. W = 4.97 eV =4.97*1.6e-19 J hf = W + Ekmax, where h=6.626e-34 Js is Planck's constant a) Ekmax = hf - W = 6.626e-34*1.5e15 - 4.97*1.6e-19 = 1.987e-19 J = 1.24 eV b) hf = W ==> f = W/h = 4.97*1.6e-19/6.626e-34 = 1.2e15 Hz
Answered by rozeta53 - Sat May 2 09:40:19 2009
Q. Light of 1.5 1015 Hz illuminates a piece of tin, which has a work function of 4.97 eV. (a) What is the maximum kinetic energy of the photoelectrons? (b) What is the threshold frequency of tin? Light of 1.5 10^15 Hz illuminates a piece of tin, which has a work function of 4.97 eV. (a) What is the maximum kinetic energy of the photoelectrons? (b) What is the threshold frequency of tin?
Asked by Joe - Fri May 1 20:40:29 2009 - - 1 Answers - 0 Comments
A. W = 4.97 eV =4.97*1.6e-19 J hf = W + Ekmax, where h=6.626e-34 Js is Planck's constant a) Ekmax = hf - W = 6.626e-34*1.5e15 - 4.97*1.6e-19 = 1.987e-19 J = 1.24 eV b) hf = W ==> f = W/h = 4.97*1.6e-19/6.626e-34 = 1.2e15 Hz
Answered by rozeta53 - Sat May 2 09:40:19 2009
What is the average kinetic energy of a gas at zero degrees kelvin?
Q. What is the average kinetic energy of a gas at zero degrees kelvin?
Asked by divaness_06 - Tue Jul 22 20:00:31 2008 - - 3 Answers - 0 Comments
A. at 0 degrees Kelvin or absolute zero, all motion stops so the average kinetic energy is 0.
Answered by ninjamayn - Tue Jul 22 21:14:02 2008
Q. What is the average kinetic energy of a gas at zero degrees kelvin?
Asked by divaness_06 - Tue Jul 22 20:00:31 2008 - - 3 Answers - 0 Comments
A. at 0 degrees Kelvin or absolute zero, all motion stops so the average kinetic energy is 0.
Answered by ninjamayn - Tue Jul 22 21:14:02 2008
Suppose an automobile has 3300 J of kinetic energy. When it moves at twice the speed, what will be its kinetic?
Q. Suppose an automobile has 3300 J of kinetic energy. When it moves at twice the speed, what will be its kinetic energy? What's its kinetic energy at three times the speed?
Asked by Hghsd - Sat Dec 19 01:45:24 2009 - - 5 Answers - 0 Comments
A. Four times for twice the speed. Nine times for three times the speed. Since, kinetic energy =1/2*mv ,Then, velocity is increased "m" times kinetic energy becomes "m " times
Answered by kasab - Sat Dec 19 01:51:54 2009
Q. Suppose an automobile has 3300 J of kinetic energy. When it moves at twice the speed, what will be its kinetic energy? What's its kinetic energy at three times the speed?
Asked by Hghsd - Sat Dec 19 01:45:24 2009 - - 5 Answers - 0 Comments
A. Four times for twice the speed. Nine times for three times the speed. Since, kinetic energy =1/2*mv ,Then, velocity is increased "m" times kinetic energy becomes "m " times
Answered by kasab - Sat Dec 19 01:51:54 2009
How do you get a kinetic energy in joules and calories?
Q. Like if your given a molar mass and temperature, what formula can you use so you can get values of the kinetic energy into joules and calories. The only formula I can think of is root((3RT)/M) but that gives meters per second.
Asked by Peter P - Mon Feb 5 17:19:13 2007 - - 1 Answers - 0 Comments
A. Not sure what you're asking...but I know that 1 calorie= 4.18 Joules.
Answered by unknown - Mon Feb 5 18:49:52 2007
Q. Like if your given a molar mass and temperature, what formula can you use so you can get values of the kinetic energy into joules and calories. The only formula I can think of is root((3RT)/M) but that gives meters per second.
Asked by Peter P - Mon Feb 5 17:19:13 2007 - - 1 Answers - 0 Comments
A. Not sure what you're asking...but I know that 1 calorie= 4.18 Joules.
Answered by unknown - Mon Feb 5 18:49:52 2007
How do I find the percentage of the kinetic energy of an object?
Q. A sled is being pulled across a horizontal patch of snow. Friction is negligible. The pulling force points in the same direction as the sled's displacement, which is along the +x axis. As a result, the kinetic energy of the sled increases by 31.0 percent. By what percentage would the sled's kinetic energy have increased if this force had pointed 68.0 above the +x axis?
Asked by ~Miizz RiVeRa!~ - Wed Feb 17 21:54:29 2010 - - 1 Answers - 0 Comments
A. Let the force is F and the initial energy of sled is 100 J if it move s meter in the direction of force =>F x s = 31 ---(i) In the second case The force component in the direction of displacement will be Fcos68 =>Delta E = W = Fcos68 x s = F x s x cos68 = 31 x cos68 = 31 x 0.38 = 11.78 J =>Delta E = 11.78%
Answered by akm69 - Wed Feb 17 22:14:27 2010
Q. A sled is being pulled across a horizontal patch of snow. Friction is negligible. The pulling force points in the same direction as the sled's displacement, which is along the +x axis. As a result, the kinetic energy of the sled increases by 31.0 percent. By what percentage would the sled's kinetic energy have increased if this force had pointed 68.0 above the +x axis?
Asked by ~Miizz RiVeRa!~ - Wed Feb 17 21:54:29 2010 - - 1 Answers - 0 Comments
A. Let the force is F and the initial energy of sled is 100 J if it move s meter in the direction of force =>F x s = 31 ---(i) In the second case The force component in the direction of displacement will be Fcos68 =>Delta E = W = Fcos68 x s = F x s x cos68 = 31 x cos68 = 31 x 0.38 = 11.78 J =>Delta E = 11.78%
Answered by akm69 - Wed Feb 17 22:14:27 2010
How do you find the kinetic coefficient of friction for this problem?
Q. A book is kicked on a level floor. It has an initial velocity of 2 m/s. The book slides 1.4 m before stopping. Determine the kinetic coefficient of friction between the book and the floor. Can someone please help me with this process? Thanks so much!
Asked by Kay - Thu Dec 10 20:24:41 2009 - - 2 Answers - 0 Comments
A. The force due to friction is F = *m*g. The amount of energy the book initially had is its kinetic energy = 1/2*m*v^2. The amount of energy the book has at the end is zero. So we know the book lost 1/2*m*v^2 of energy. Work done on an object is F*d (force*distance). The amount of work the friction did on the book (in the opposite direction of its sliding) is how the book lost its energy. So, *m*g*d = 1/2*m*v^2. Divide both sides by m, then solve the equation for . = v^2/(2*g*d) = ((2 m/s)^2) / (2 * (9.8 m/s^2) * (1.4 m)) = .15 (no units).
Answered by Joelogs - Thu Dec 10 20:58:08 2009
Q. A book is kicked on a level floor. It has an initial velocity of 2 m/s. The book slides 1.4 m before stopping. Determine the kinetic coefficient of friction between the book and the floor. Can someone please help me with this process? Thanks so much!
Asked by Kay - Thu Dec 10 20:24:41 2009 - - 2 Answers - 0 Comments
A. The force due to friction is F = *m*g. The amount of energy the book initially had is its kinetic energy = 1/2*m*v^2. The amount of energy the book has at the end is zero. So we know the book lost 1/2*m*v^2 of energy. Work done on an object is F*d (force*distance). The amount of work the friction did on the book (in the opposite direction of its sliding) is how the book lost its energy. So, *m*g*d = 1/2*m*v^2. Divide both sides by m, then solve the equation for . = v^2/(2*g*d) = ((2 m/s)^2) / (2 * (9.8 m/s^2) * (1.4 m)) = .15 (no units).
Answered by Joelogs - Thu Dec 10 20:58:08 2009
What is the coefficient of kinetic friction between the ice and the puck?
Q. A hockey player hits a puck with her stick, giving the puck an initial speed of 5m/s. If the puck slows uniformly and comes to rest in a distance of 20m, what is the coefficient of kinetic friction between the ice and the puck?
Asked by ME - Sun Oct 4 07:04:17 2009 - - 1 Answers - 0 Comments
A. U = 5m/s s = 20m V= 0 V^2 = U^2 - 2as a = U^2/2s = 25/2x20 = - 0.625 m/s^2 coefficient of friction = (mu) = f/N = f/W The frictional force = - f = ma = (W/g)a a = - gf/W = - g (mu) (mu) = a/g= .625/9.8 = 0.0638
Answered by Bandagadde S - Sun Oct 4 07:34:42 2009
Q. A hockey player hits a puck with her stick, giving the puck an initial speed of 5m/s. If the puck slows uniformly and comes to rest in a distance of 20m, what is the coefficient of kinetic friction between the ice and the puck?
Asked by ME - Sun Oct 4 07:04:17 2009 - - 1 Answers - 0 Comments
A. U = 5m/s s = 20m V= 0 V^2 = U^2 - 2as a = U^2/2s = 25/2x20 = - 0.625 m/s^2 coefficient of friction = (mu) = f/N = f/W The frictional force = - f = ma = (W/g)a a = - gf/W = - g (mu) (mu) = a/g= .625/9.8 = 0.0638
Answered by Bandagadde S - Sun Oct 4 07:34:42 2009
What is the coefficient of kinetic friction between the puck and the surface?
Q. In a game of shuffleboard (played on a horizontal surface), a puck is given an initial speed of 6.0 m/s. It slides a distance of 9.0 m before coming to rest. What is the coefficient of kinetic friction between the puck and the surface?
Asked by webs2slow4me - Wed Feb 4 09:54:17 2009 - - 1 Answers - 0 Comments
A. determine acceleration F = ma - uN = ma - umg = ma a = - ug apply kinematics equation a = (vf - vi )/(2d) - ug = (0 - vi )/(2d) u = vi /(2gd) u = 0.20
Answered by unknown - Wed Feb 4 10:32:56 2009
Q. In a game of shuffleboard (played on a horizontal surface), a puck is given an initial speed of 6.0 m/s. It slides a distance of 9.0 m before coming to rest. What is the coefficient of kinetic friction between the puck and the surface?
Asked by webs2slow4me - Wed Feb 4 09:54:17 2009 - - 1 Answers - 0 Comments
A. determine acceleration F = ma - uN = ma - umg = ma a = - ug apply kinematics equation a = (vf - vi )/(2d) - ug = (0 - vi )/(2d) u = vi /(2gd) u = 0.20
Answered by unknown - Wed Feb 4 10:32:56 2009
What is the Difference between Kinetic Energy and Gravitational Potential Energy?
Q. And how do you know when an object has kinetic energy, gravitational potential energy, or both?
Asked by Gilda - Sun Dec 13 18:47:26 2009 - - 1 Answers - 0 Comments
A. Kinetic energy is energy of motion. If an object is moving, it has kinetic energy. Gravitational potential energy is energy due to the gravitational interaction of two or more objects (on earth, it's equal roughly to mgh). If there is a change in height (or the radius) between the objects, there's a change in potential energy.
Answered by Kath D - Sun Dec 13 19:01:09 2009
Q. And how do you know when an object has kinetic energy, gravitational potential energy, or both?
Asked by Gilda - Sun Dec 13 18:47:26 2009 - - 1 Answers - 0 Comments
A. Kinetic energy is energy of motion. If an object is moving, it has kinetic energy. Gravitational potential energy is energy due to the gravitational interaction of two or more objects (on earth, it's equal roughly to mgh). If there is a change in height (or the radius) between the objects, there's a change in potential energy.
Answered by Kath D - Sun Dec 13 19:01:09 2009
What is the Kinetic energy and speed of an electron?
Q. What is the kinetic energy and speed of an electron ejected from sodium with work function 2.28eV, when illuminated with a light of wavelength 550nm? I got the Kinetic energy as being 3.6 * 10^-28 J but do not know if this is right! Also have no idea how to find the speed...!!! Please help...:) Thanks!
Asked by Sorch - Wed Feb 3 11:22:30 2010 - - 1 Answers - 0 Comments
A. I don't have time right now to do the math, but the order of magnitude sounds correct (maybe a little small?), so it's more than likely correct. You find velocity from KE=.5mv^2, where KE is Kinetic Energy, m is the mass of an electron, and v is the velocity you're solving for. It can be rearranged to sqrt(2KE/M), plug'n chug
Answered by Michael - Wed Feb 3 11:33:01 2010
Q. What is the kinetic energy and speed of an electron ejected from sodium with work function 2.28eV, when illuminated with a light of wavelength 550nm? I got the Kinetic energy as being 3.6 * 10^-28 J but do not know if this is right! Also have no idea how to find the speed...!!! Please help...:) Thanks!
Asked by Sorch - Wed Feb 3 11:22:30 2010 - - 1 Answers - 0 Comments
A. I don't have time right now to do the math, but the order of magnitude sounds correct (maybe a little small?), so it's more than likely correct. You find velocity from KE=.5mv^2, where KE is Kinetic Energy, m is the mass of an electron, and v is the velocity you're solving for. It can be rearranged to sqrt(2KE/M), plug'n chug
Answered by Michael - Wed Feb 3 11:33:01 2010
From Yahoo Answer Search: 'kinetic'
Wed Mar 10 06:10:23 2010 [ refresh local cache ]
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The negative space between the letters also makes Bandoleer seem kinetic , which increases its intrinsic eye-catchiness. Eben Sorkin's No Step, ...
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April 2008 Kinetic review on BikeRadar See the full review of the new 2008 singletrack monster the Brodie Kinetic over on BikeRadar Click here to see the review
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Wed, 10 Feb 2010 07:00:26 GM
Adele Peters might just be the winner at the Greener Gadgets Design Competition this year. One of 18 entries that made it into the competition...
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