You and a friend together apply a force of 1,000 N to a 3,000 N automobile to make it roll 10 m in 1 minute and 40 seconds. \nhow much work did you and your friend do together?\nand what was your combined power.\n\n\nPELASE help!! thanks soo much!
130796
Work = force * distance\n= 1000 N * 10 m\n= 10000 Nm\n\nYou wouldn't take into consideration the mass of the car because the force you are applying is horizontal (assuming that you are pushing straight on the car and not down at an angle).  The only time you would use the 3000 N is if you took friction into consideration.  In that case, you would have a friction force opposing your 1000 N, thus reducing it.  (Force of friction = coefficient of friction * normal force)\n\nPower = work / time\n= 10000 Nm / 100 s\n= 100 Nm/s
215948
the answer to the first question is 5 m/s.\nF=ma; a=F/m=200N/40kg=5m/s\nthe second answer is 1m/s.\nF=ma; a=F/m=2N/2kg=1m/s\nthe third answer is 1:4.\nF=ma; m=F/a\n(1) M=f/a=F/a=m\n(2) M=f/a=4F/a=4m
403122
I solved the problem by stating that P would travel distance d in time t\nd=20*t+.5*.3*t*t\n\nand Q would travel d+20 in time t\nt is the time to collision\n\nd+20=30*t\nd=30*t-20\nSubtract the equations and solve for t:\n\n0=.15*t*t-10t+20\nt=2.064 seconds\n(the second root is superfluous)\nd=41.9m\n\nWhich proves collision\n\nj
151459
well ok. for part a, the acceleration is in the horizontal direction, because its moving horizontally. so the force is just mass x acceleration which is 70*6=420N. but thatz only in the horizontal direction.\n\nfor part b, you have to account for gravity. so gravity is pulling down at 9.45m/s^2, but youre accelerating upwards at 6m/s^2. so that means that youre overcoming the gravity and accelerating at 6m/s^2 MORE than the gravitational acceleration.\nso 9.45+6 = 15.45. this is your actual acceleration.\nso now you just do F=ma again and you get 70*15.45=1081.5N. \n\nhope this helps. let me know if you need more help. and this isnt thermodynamics.
266454
Newton's second law is \nforce = mass x acceleration\n\nHow about this? Two men were walking on a hill when they see a bar of chocolate at the bottom. Both are eager to get it. The first man, a skinny little guy of 120 lbs, trips and begins rolling down the hill. The second man, a fatty of 250 lbs, trips at the same time and begins rolling down the hill. Because the fat man has more mass, he is able to produce more force to accelerate himself down the hill faster than the skinny man. The happy fat man reaches the bottom of the hill first, and eats the chocolate.\n\nIllustrations shouldn't be too hard.
545339
Relative to a woman in the car:\n\n(a) the translational velocity is ZERO since the woman is moving at the same translational velocity as the car relative to the ground.\n\n(b) the magnitude of the translational acceleration of the center of the wheel is also ZERO because the center of the wheel is NOT moving and NOT rotating relative to the woman.\n\n(c) v, velocity of the point at the top of the tire is = 80 km/h because that is how fast the car is moving, thus the wheel must also be rotating at the same speed.\n\n(d) a, acceleration of the point at the top of the tire is given by:\n\na = v^2/r = (22.22)^2/(0.33) = 1,496.45 m/s^2\n\n(e) v, velocity of the point at the bottom of the tire is = -80 km/h because at the bottom of the tire, the tire is moving in the opposite direction\n\n(f) a, acceleration of the point at the bottom of the tire is the same as (d) a = v^2/r = 1,496.45 m/s^2\n\nNow relative to a hitch hiker on the ground:\n\n(g) v at the center of the wheel is = 80 km/h or 22.22 m/s, since the car and the center of the wheel (which is not moving relative to the car) are moving at the same speed relative to the ground.\n\n(h) a at the center of the wheel is = v^2/r = 1,496.45 m/s^2, since the center of the wheel is rotating about the bottom point of the tire as the tire rolls on the ground.\n\n(i) v at the top of the tire is v(top) = ωd, where ω is the angular velocity of the rotating tire and d is the diameter of the wheel. The reason we use diameter of the wheel is because the top of the tire is rotating abou the bottom of the tire, not the center.\n\nω = v/r = (22.22)/(0.33) = 67.33 rad/sec\nv(top) = ωd = (67.33)(0.66) = 44.44 m/s\n\n(j) a at the top of the tire is = v(top)^2/d\n    = (44.44)^2/(0.66) = 2,992.29 m/s^2  \n\n(k) v at the bottom of the tire is equal to ZERO because the bottom of the tire is not moving relativ to the hitch hiker or the ground.\n\n(l) a at the bottom of the tire is also equal to ZERO since the velocity is zero.
551349
310 times faster\n\nif A is going 10mph, and B is going 5mph, the 10/5=2 or twice as fast. so 9.3X10^10 / 3.0X10^8 = 310 times faster.\n\nbut nothing can't go faster than 3X10^8m/s!!!! The normal speed of light means it's going though a vacum (nothing), so if it's going though a gas, it will slow down!! Not speed up! Since the energy will hit atoms and slow down. \n\nIf an object is going 3E8 (short way of 3X10^8m/s) releative to the ground and an object passes it at 3E8, then the object passing the first objects speed was 3E8 releative to the ground... so both were going the same speed, even if some was twice as fast as it!
222354
None. It takes no power to maintain that speed (on level ground, no friction). The object has kinetic energy (0.5 * m * v^2). Power was required to put it in motion and power can be recovered by decelerating it. If you do the math, the energy is 3551.86 ft-lb, equivalent to 1 horsepower for 6.458 seconds. Answer 1 neglected the fact that weight must be converted to mass by dividing by g, and energy is proportional to v^2.
384970
Friction force:\nF = m g c\n\nm = mass of car\ng = gravitational accel\nc = coeff. of friction\n\nCentripetal force:\nF = m v^2 / r\n\nm = mass of car\nv = velocity\nr = radius of curve\n\nSet the two forces equal to each other:\nm g c = m v^2 / r\n\nCancel m, plug in given values:\n9.8 m/s^2 * c = (12 m/s)^2 / 52 m\nc = 0.28257\n\nSo, c has to be at least 0.28257 to keep the car from slipping.
406873
It's OK, you only forgot to put the angle, the other guy missed the 11 kg.\n\nfirst, let's find F_frict, \n\nF_normal=m*g*cos21=11*9.81*cos(21) =100.74 N\n( force of gravity normall to incline )\n\nand F_parallel=11*9.81*sin(21) = 38.67 \n( force of gravity parallel to incline )\n\nF_frict=0.347*F_normal = 34.96 N\n\nso F_incline= 150 - F_parallel - F_frict = 76.37 N \n\na)\n\nChange of KE = Work = F_incline*d = 76.37*7.46 = 569.7202 J\n\nb)  \n\n0.5*m*v^2 = 569.72, where V is change in speed\nv=10.17769397 m/s\n\nso the new speed is= 10.17769397 + 1.53 = 11.7 m/s
459303
1) I think your 2nd c.f. should say c. acceleration; that's the units you're using------ the c force of a mass m with acceleration a is f = ma, so youhave 1.5 kg*18 m/s² = 27 N\n\n2) First find the velocity: v = d/t = 2πr/t = 2π*.75m/.8sec = 5.89 m/s\nThen, acceleration: a = v²/r = 5.89²/.75 = 46.256 m/s²\nThen, the tension: T = ma = .1kg*26.256m/s² = 2.626 N\n\n3)  v = 2πr/t = 2π*90/12.3sec = 45.974m/s\na = v²/r = 45.97²/90 = 23.485 m/s²\nF = ma = 900*23.485 = 21,136.5 N\nThe lateral friction of the tires on the track
460976
a) The disk increased from 10 rev/s to 15 rev/s. Acceleration was constant so the average speed was 12.5 rev/s. At that average speed, how many seconds did it take to make 60 revs?\n60 revs * 1 s/12.5 revs = 4.8 seconds. Round to 5 seconds.\n\nTherefore the acceleration was \n(delta rev/s) / (delta time) = (5 rev/s) / 5 s = 1.0 rev/s^2\n\nb) See above, that was 5 seconds\n\nc) An acceleration of 1 rev/s^2 means each second the speed increases 1 rev/s. So it took 10 seconds to reach 10 rev/s.\n\nd) The average speed from start to the time speed reached 10 rev/s was 5 rev/s. In 10 seconds at that speed it would make 50 revolutions.
459691
The braking force was enough to dissipate the kinetic energy.\n\nForce x distance = work = kinetic energy\n\nma x 160 = mv^2/2 , notice that m cancels out\n\n160a = v^2/2, solve for v...
507822
If everything were lined up exactly inline and marble 1 hits 2 which hits 3, then both marbles 2 and 3 would move together at the same rate as marble 1.    (1~~>2~~>3)   You have probably seen that demonstrated with the newton's pendulum toy that has 5 chrome balls hanging by strings.   When you pull 1 of the balls back and let it swing into the other 4, 1 ball swings off the other end.  2 balls cause 2 to swing off the other end.\n\nIn your experiment, the 1st marble will hit the other 2 and release all of it's energy.  They will then go at the same rate as the first marble.\n\nV1 = 0\nV2 = 15.5 in a direction 60 degrees off to left.\nV3 = 15.5 in a direction 60 degrees off to the right.
175947
Part 1: Mike\nWork=Force x Distance(moved) x cos (theta) (angle between work and distance)\nF=215\nD=64.7\nTheta=35 degrees\n\nW=215*64.7*cos35\nW=11,394.815 Joules\n\nPart 2: Chris\nF=8N\nD=3.9\nTheta=90 degrees\nW=8*3.9*cos90\nW=0\n\nThats all you get...good luck
478930
a. 1.10 MJ\n\nEnergy = mass * acceleration * distance = m * a * s\n\nm = 1600 kg\nAssume that the car is accelerating at 1 m/s^2 =>  a = 1\n\nIt would take 25 seconds for a car accelerating at 1 m/s^2 to go from 15 m/s to 40 m/s.\n\nThe average velocity during the acceleration is (15 + 40)/2 = 27.5 m/s.\n\nThe distance, s, is 27.5 m/s * 25 seconds = 687.5 m\n\nm*s*a = 1600 kg * 1 m/s^2 * 687.5 m = 1.1 e6 J = 1.1 MJ\n\nIf you assume another acceleration (let's say 5 m/s^2) you will get the same because the same amount of work is being done.
410420
Ok.  From the hint start by figuring out the ratio of wheel speeds\nCircumference of A =2*pi*10\nCircumference of C = 2*pi*25\nso C/A=2.5  ie A must go around 2.5 times to make C go around 1 time.\nThere are 2*pi radians to 1 revolution so 100 rev/min =200*pi rad/min.  A must go 2.5 times as fast as this or 2.5*200*pi=500 rad/min\nAccelleration is the derivative of velocity which means that we must integrate the angular acceleration to obtain the speed at any given point of time.\nIntegrating 1.6rad/s^2 with respect to time and setting velocity equal to zero at time equals zero we obtain angular velocity= 1.6*time elapsed (rad/sec).  Note that this speed is in "per seconds" and we need it to be in "per minutes" so we must multiply by 60 for the units conversion.\nThis gives us the angular speed in rad/min for any chosen passage of time.  We shall set this equal to the goal speed of 500*pi rad/min as mentioned above. So\n1.6*time elapsed*60 =500*pi\nSolving for time elapsed we obtain a time of 1.63 minutes.