Calculate the acceleration of a rocket of mass 1.2x 10^6 kg if?
83305
I suppose the problem was already due so I'll finish it for you now.\n\nF=m*a\na=F/m= 2*10^6N/1.2*10^6kg = 1.67m/s^2 * g/9.81m/s^2 = 0.17g
989843
net force IN THE DIRECTION OF DISPLACEMENTis (1.7*10^6 * cos 14)*2 \n= 3.3* 10 ^ 6 N\nwork done = 3.3*10^6*860=2.838 * 10^9 Joules
289324
simple F=m*a\nm=0.8\na= the acceleration due to gravity (9.8 ms^2)\nWeight is 7.84 N\nthe conversion should be in your textbook from newtons to lbs and from lbs to oz
188896
K = mv^2 = (10kg)(4.5 m/s)^2 = 202.5 J\nI would suggest you try doing your homework, asking for all the answers will only hurt you in the future.
841719
sorry for laughing but I know everything can be used with math. Those rockets, how long is goes up, where it lands the wind , angle bla/bla
824196
conservation of linear moment:\nm1*v1 = m2*v2\nm1*v1/m2 = v2\n(100.0 kg)(3.0 m/s)/(175.0 kg) = v2 = 1.71 m/s
884244
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.
59053
actually... although the first answerer did the correct calculation, he left out a step. that is the acceleration produced by the rocket, however you must take gravity into account. \nF=ma, a=F/m= 23.3 is the acceleration produced by the thrust of the rocket, but the rocket is not actually going to accelerate that much because an acceleration of 9.8(due to gravity) is pulling it down in the opposite direction. so 23.3-9.8= 13.53\nso the initial acceleration of the rocket is:\n13.53m/s^2
307097
Ksp = [Ag+]^2[CrO4 2-]\nsince [Ag+] = 2*[CrO4 2-]\nKsp = 2*[Ag+]^3\nsolve for [Ag+] then plug it in this equation\n[Ag+] = mass /0.010L H2O
618594
The thrust of the burning gasses out of the rear of the rock causes a force against the rocket. If the resulting force is more than the force of gravity, the rocket accelerates according to the formula A= F/M. Where M= the mass of the rocket and F=the net force of the burning gasses minus the force of gravity. As the rock rises, it looses mass due expending its burnt fuel. That causes it to accelerate faster. Also as it rises, the force of gravity becomes less increasing the net force impelling the rocket, thus causing it to accelerate faster.
565790
g=m*a\nGravitational Force=Mass * Acceleration
767928
1. its final velocity depends on the direction it was heading to start off with. If it was flying in the x direction (no components in y or z) then after 4.3 seconds the x component has increased by 8.92*4.3  and the y component has increased (from 0 this time) by 1.21*4.3.\nso now you can work out the new direction (in the x direction but slightly downwards by whatever angle has a tangent of 1.21*4.3/(152+8.922*4.3) \nand using pythagoras to determine the speed.\n\n2. You can ignore the gravity as it is << the precision with which the acceleration due to the electric field is stated (I'm assuming the 106 is 10^6 = 1000000) \nso the final velocity is 3.80*2.45*1000000 m/s upwards.
879746
Ek = 1/2.m.v2   (1/2 mass times velocity squared) - \n\nsorry can't do superscript on this site
731455
Something that may be fun is a project with model rockets. Give them the necessary information to calculate the height of the rocket, based on the rocket engine and mass of the rocket as homework.\n\nThen at school, fire a few of them and using basic trig calculate the height and see how close they are.\n\nRocket engines are typically rate something like: A8-3 which breaks down as follows:\nA --> 2.50 Newton-seconds of power\n8 --> average thrust (in newtons)\n3 --> seconds from end of thrust to ejection charge (parachute)\n\nThe engines also have a time-thrust curve that shows how the power is delivered over time. Check the following link below for some more info:\n\n------------------------------------------------------------------------\n\nHere is another one:\nNot sure how readily available props are for you but if you can hook up a camera in a room with a strobe light/flash and then set up a collision (air hockey table or pool table come to mind) then you can calculate the velocity of each object, measure its mass and then demonstrate the conservation of momentum using vectors. Because you know the frequency of the strobe, and the distance traveled, you can compute the vector velocity.\n\nOr how about:\nFirst, fire an arrow down a football field. Given the amount of time aloft calculate the initial velocity (ignore air resistance/friction) Then fire an arrow at a bale of hay (extend the bow by the same amount) and measure how far into the bale the arrow went. Using the same initial velocity and distance into bale, calculate the rate of deceleration and coefficient of friction of the hay bale.
283082
An example is the rocket equation:\nV = Vex * log(m0/m1)\nwhere \nV is final velocity\nVex is the rocket exhaust velocity\nm0 is liftoff mass\nm1 is burnout mass\nThis equation says (e.g.) that you need a ratio of liftoff to burnout weights = e (2.718, base of natural logarithms) to achieve a V = Vex.\nNow suppose you know the final velocity you want and you know the mass of the object you want to impart the velocity to, which is a typical design situation.  Then you turn the equation around:\nM0=M1*e^(V/Vex)\n(The results can be surprising if you're unfamiliar with rocket propulsion.  With a state-of-the-art engine like the Shuttle main engines, Vex is about 15000 ft/sec. If you want 1000 pounds (empty rocket plus any payload) to reach 35000 ft/sec (~earth escape velocity) with an engine of equal efficiency, you'd need a liftoff weight of about 10000 lb.)
738706
Both questions use the formula F=ma (Force = Mass * Acceleration):\n\n1) F = ma\n1310 = 214*a\na = 1310/214 = 6.1215 m/s^2\n\n2) F = ma\n3.6 = m*9.8\nm = 3.6/9.8 = 0.367 kg
1163903
Set m1 = 0.145 kg, v1i = 0.720 m/s\nand m2 = 0.295 kg, v2i = 2.22 m/s\n\nConservation of linear momentum:\nm1v1i + m2v2i = m1v1f  + m2v2f\nSimplified\nm1(v1i-v1f) = -m2(v2i-v2f)\n\nConservation of energy (kinetic only):\n(1/2)m1v1i^2 + (1/2)m2v2i^2 = (1/2)m1v1f^2 + (1/2)m2v2f^2\nSimplified:\nm1(v1i - v1f)(v1i + v1f) = -m2(v2i - v2f)(v2i + v2f) \n\nSolving the two simplified equations:\nv1f =((m1-m2)/(m1+m2))v1i + ((2m2)/(m1+m2))v2i\nand\nv2f =((m2-m1)/(m1+m2))v2i + ((2m1)/(m1+m2))v1i\n\nTherefore,\nv1f = 3.315 m/s\nv2f = 1.23 m/s