a car devoloping 4kw of power is travelling with aconstant velocity of 50kmph on alevel road.what is the force
494540
Haha whats with all these answers, if you know basic physics, One of Newtons laws of motion is that:\n\n"for an object that is stationary or travelling at constant velocity, no force acts."\n\nso you ask what is the force?\nthere is no resultant force, the driving force and resistance are equal and cancel each other out.\n\n\nAnd additionally the basic equation to find force is:\nF=ma (force = mass x accelleration), this is evidence that an accelleration needs to be taking place for a force to act.\n\nLets say you want the force for a 50kg object accellerating at 3ms-2. F=ma  ->  F=50x3 ->  F=150N.\n\nNow lets say you want the force for a 50kg object travelling at constant velocity, no accelleration.\nF=ma  ->  F=50x0  ->  F=0N\n\n\nThere you go, wasnt that easy
904741
The change in momentum of the automobile is equal to the impulse of the applied force.\n\nInitial momentum = mv = 1200*50 = 60000 km-kg/h\nFinal momentum = 0\n\nWe need to convert the 60000 km-kg/h to useful units\n\n60000 km-kg/h * 1000 m/km / 3600 sec/h = 16667 m-kg/sec\n\nImpulse = force * time = F*0.14 sec\n\nSo F*0.14 = 16667\n\nForce F = 16667/0.14 = -120 KNewtons\n\nNote that the applied force is against the original direction of motion, so the force is negative...
336419
s = ut +1/2 at^2\n50 = 1/2a(5)^2\n50 = 1/2*a*25\nSo, a=4m/s^2  \n\nNow, F = ma\n     F = 60 * 4\n       = 240 N\n\nW = Fd\nW = 240 * 50\n  = 12000 J.\n\nP = W/t\n  = 12000/5\n  = 2400W = 2.4 KW.\n\nFinished! Hope u understan. I think u'll give 10 points. Isn't it?
311499
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
934211
Ok, so you have two simple equations to keep in mind:\n\n1)  KE = (1/2)*m*v^2\n(kinetic energy equals one half mass times velocity squared)\n\n2) Paverage = E/t\n(average power equals energy divided by time)\n\nFirst you find the kinetic energy (convert that km/hr speed into m/s).  Now you have the kinetic energy for the car after the acceleration period (in Joules) which can then be divided by the time to get the average power (in J/s or Watts) required to get that car up to the given speed.\n\nThen we sometimes like to convert mechanical power from Watts into horsepower.  The conversion happens to be 745.7 W/hp.\n\nI will leave the calculations up to you, but those are the equations/concepts you need to know and apply.  Also, take note that this is an idealized solution that does not take into account wind resistance, rolling friction and various other factors that would require more power output from the engine of the vehicle.
900616
If the rock is not being slowly pushed into the ground , then the net force is 0 because the force down is equal to the normal force upwards exerted by the ground.\n\nif the car is really heavy and pushes the rock into the ground, then the net force is greater than 0, in a downwards direction.
404136
Calculate the total energy of the system before and after and make an energy balance.\n\nFor example, lets say you have an inelastic collision between two cars of equal mass m:\n\nInitial State:\nCar 1 moves at v1 to the right\nCar 2 moves at v2 to the left\n\nAssume v1 > v2\nThey hit and stick together completely.  Use conservation of momentum to find the total resultant velocity:\n\np1 + p2 = ptotal (p is the momentum of car 1 and car 2 respectively)\n\nm*v1 - m*v2 = (m+m)*vf\n\nvf is the final velocity of the clump\nthe minus sign is because I am choosing the direction "right" as positive (momentum is a vector)\n\nvf = (v1-v2)/2\n\nNow do an energy balance:\n\nEbeforecollision = 1/2*m*(v1^2) + 1/2*m*(v2)^2\nEaftercollision = 1/2*(m+m)*(vf)^2 + Efrictionlosses\n\nEbefore = Eafter (this is conservation of energy, and it is true as long as you account for all forms of energy...it wont be true if you forget about the frictional losses in Eafter)\n\nNow solve for Efrictionlosses\n\nTo get the total fractional energy loss, take the ratio Efrictionlosses/Ebefore
860183
This is a momentum question.\n\nMomentum = mass x speed\n\nSo, the original momentum is contained solely withing the moving engine and equals 4 units of mass times 5 km/h.  This equals a 'momentum' of 20.\n\nThe final momentum equals the combined mass of the engine and freight car, which is 4 units plus 1 unit for a total of 5 units of mass.\n\nSo, if Momentum = mass x speed, then since our final momentum is 20 (the same as the original momentum) and our final mass is 5, then 20 = 5 x Speed\n\nDividing both sides by 5 yields a final speed of 4 km/h.\n\nBest wishes and good luck.
165215
High speed Electrical Car, which can reach the speed of 370 KM/H
762388
take  g = 9.8 m/s^2  and let us assume the load is being lifted vertically upwards. This should have been mentioned in the question or the force against which it is being moved should have been mentioned. In the absence of that we make these simplifying assumptions.\n\nThe motor does 45x9.8x200 joules of work in 90 seconds so the power with which the machine works is \n\n(45x9.8x200)/90 = 980 J/s or 980 watt
841492
a)The acceleration a = Force / mass = -6070 / 2600 = -2.33 m/s^2\nInitial velocity u = 34.9 m/s\nThe velocity at any time is given by v = u+a*t = 34.9 - (2.33*1.76) = 30.8 m/s\nb) v^2 - u^2 = 2*a*s, where s is the distance travelled in that time. \ns = ( v^2 -u^2)/2a = 57.8 m\nc) The car stops when u + a*t = 0.\nt = 34.9/2.33 = 15 s
14955
(a) it's an inelastic collision.  An inelastic collision is when they stick together, and an elastic collision is when they bounce off of each other\n(b) p=mv  momentum= mass x velocity.  p= 30 m/s x 500kg = 15,000 kg*m/s\n(c) well, momentum is conserved, so the we would use the formula from before and the momentum(p) from before, except that now the mass is different because it is the combined masses of the two vehicles (300+500=800) , so you can solve for the new velocity.\np=mv  15000 = 800 x v\n\ndivide both sides by 800 to get  v=15000/800 = 18.57 m/s
262728
Simple.\nThe car tyres rotate thus the force is called centripetel force(f).It is equal to v^2/r where r is the radius of tyre.\n\nforce at 15m/s = 15*15/r\nforce at 30m/s = 30*30/r\n\nWhen we divide them we get 1:4 the answer\nit is assumed that car tyre is of same radius in both cases.
1140843
Force with bike=100*sin30*10=500N \nForce without bike= 0.9*500=450N\nThus, walking will save him 50*300=15000 joules.\nBut, don't forget that the bike has a bigger mechanical advantage or force multiplier that reduces the work. Suppose the MA is 4,then the force with bike will be 500/4=125N and hence work will be 15000/4=3750 J only. Therefore, riding is preferable.
677621
It's 220mph, near enough, so you're looking at perhaps a bullet, or a jet. It might be worth dropping it to 49 m/s (110mph) or even 24.5 m/s (55mph) to bring it within the range of a sports car or ordinary motor.
537140
Think of it this way.  Your car will slow down if I drop an elephant in your trunk while your driving.  \nA high load drags the voltage down.  \nThe power supply can't keep up with the amount of energy the load is wanting.
12602
If I understand the question correctly, I think the force you're looking for is inertia. Your body is moving at the speed of the car - then the car stops. Now the car has no speed, but you're still moving forward at that speed until something (hopefully the seat belt) acts to stop you.\n\nIf you're referring to the force provided to stop you - that's the Newtonian counter-force to your inertia moving forward.