If a lady bird travels up a plant that is 30cm high at a rate of 3cm each day but then falls 2cm the next day and so on how long will it take the lady bird to reach the top?
132811
55 days.\n\n"3cm each day but then falls 2cm the next day"\n\nDay 1 = 3cm\nDay 2 = 1cm\nDay 3 = 4cm\nDay 4 = 2cm\nDay 5 = 5cm\nDay 6 = 3cm\netc\n\nSo it travels 1cm every two days. However, it only needs to get as high as 27cm, as the day after that it will reach 30cm before it falls back down.\n\n27x2 = 54\n54 + 1 = 55\n\n55 days.
142473
Let x = Eloise's time\nChristian's time = x-3\nsince Chistian's time is three hours less than Eloise's time\nNow distance equals rate times time or\nd = r*t\nFor Eloise r = 15, so d = 15x\nFor Christian, r = 30, so d = 30(x-3)\nThese distances are equal so\n15x = 30(x-3) \nDistribute 30\n15x = 30x - 90  \nSubtract 30x from each side\n-15x = -90 \nDivide both sides by -15\nx = 6\nThus, they traveled\nd = 15x\nd = 15(6)\n= 90 miles
300626
This is your garden variety related rates problem.\n\nAt any instant when the monkey is on the wall, you have yourself a right triangle, with the taut rope being the hypotenuse, the distance from fastening point to the wall being the horizontal leg, and the distance the monkey climbs vertically being the vertical leg. The one equation which relates all three of those quantities together is the Pythagorean Theorem: x^2 + y^2 = z^2 (x and y being the legs and z being the hypotenuse).\n\nSince this is a related rates problem, you need to differentiate this equation with respect to time (t). The result is 2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt). All of the d*/dt terms represent the rate * is changing over time.\n\nSo, now it's a matter of plugging in the appropriate values given in the word problem. According to the word problem, x=8 ft, dy/dt=4 ft/s, y=6 ft (at that instant). From this info, one can derive that z=10 ft (via Pythagorean Theorem) and dx/dt=0 ft/s (because the distance between the fasten point and the wall is fixed). The problem is asking for dz/dt. Now, plug in:\n2(8)(0) + 2(6)(4) = 2(10)(dz/dt)\nSolving for dz/dt, your answer should be 2.4 ft/s (if the math is correct).
514801
Something's wrong with the writeup for the first question.  I'm guessing it should have said THR = RHR + 0.65(MHR-RHR), because otherwise you're gonna get a THR of, like, 10000 beats per minute.\n\nProceeding from that assumption, the THR is RHR + 0.65(MHR-RHR).  You're told that the RHR is 54 bpm.  The MHR is 220 - the age(43), so it's 177.  Therefore the THR is 54 + 0.65(177-54) = 54 + 0.65(123) = 54 + 79.95 = 133.95 bpm.\n\nIn the second question, you didn't provide the "following equations," but think of it this way.  The airplane flew for a total of 15 hours at an average speed of 350 miles per hour.  That means it went 5250 miles.  In the final 7 hours it went 2275 miles.  So in the first 8 hours it had to go 2975 miles.  2975 miles/8 hours = 371.875 mph.  I don't exactly know how to write that in an equation, but now that you understand the reasoning, hopefully you'll be able to recognize the correct equation.  If all else fails, plug the data into each equation and see which one returns 371.875.
146272
You mean find "g" as if you didn't know the constant? Okay.\n\nPretty simply really. Have the height measured that you're going to drop it from to where it's going to land (this will be your "h"). Make it a high one (one meter is enough). Measure the mass of the object (that's the "w"). Then when you do the experiment, time it. Start it at the instant you let it drop and stop it once it touches where ever it lands. That's enough information to find "g." Just solve for "g" then that would be the answer.\n\nActually, for an experiment, you should solve for "g" as a formula before you start doing anything else. So the formula would be like this:\n\ng = (2h)/[(t^2)(w)]\n\n(t^2 is "t squared")\n\nSo just have that formula then do the experiment: measure height of the drop, the mass of the object, then time it when you drop it (you need good reflexes when you drop it, as it falls fast).\n\nPS Penguins!!!
137884
Let t be the time it takes for Denise to catch Gloria expressed in hours.\n\nGloria will have been walking for (t + 1/2) hours, at 3 mph\nso she will have gone (t+1/2) * 3 miles.\n\nDenise will have been jogging t hours at 5 mph, so she will have gone 5t milse.\n\nSince they are at the same place, the distances will have to be the same: 3(t+1/2) = 5t\n\nSolve for t:\n3t + 3/2 = 5t\n\n3/2 = 2t\n\n3/4t = t\n\nCheck Denise goes 3/4 * 5 = 15/4 miles\n\nGloria goes (3/4 + 1/2) * 3 = (5/4) * 3 = 15/4 miles\n\n\n\nIf you have a square, and you triple the length of the side,  the resulting square has an area that is 9 times that of the original square.  The same thing works for all figures in the plane. If you enlarge them by a factor of 3, the area grows by a factor of 9.   \n\nIf you were to break the star down into a collection of triangles, and analyzed the area of each one, you would see that you were multiplying the base and altitude of each triangle by a factor of three, so you are multiplying the area of every one of the pieces by 9. So 90 is your correct answer.
512423
I would say use conservation of momentum.\n\nmb*vb + mp*vp = 0\n\n(x*0.05)*(10) + 10000*(-125) = 0\n\nIt equals zero since the plane has a final velocity of zero (as would the birds, I'd imagine!) "x" is the number of birds.
539898
Apply v^2 = u^2 + 2as vertically;\nwhere v=end velocity, u=start velocity, a=acceleration, s=distance\nWhen the object reaches the height of its path, vertical velocity is zero.\nhence: 0=(30Sin30)^2 + 2(-10).s\ns = 225/20 = 11.25m. \nHence it reaches a total height of 11.25m+1.2(height from which it is thrown).\n.\n.\nTime of flight:\nApply: S=ut + 1/2 at^2 vertically, where S=distance, u = start velocity, a=acceleration, t=time\nSince the ball is caught at the same height at which it is thrown, the overall vertical distance travelled is zero.\n0 = 30Sin30.t + 1/2(-10).t^2\nt = 3s.\n.\n.\nHorizontal distance = horizontal velocity x time\ndistance = 30Cos30 x 3= 77.9 metres
654062
I think you can use a simpler equation than the one for distance.\nWe don't need to find distance, just time.\n\nv = u + a.t\nv= 0, u= 64 f/s, a= -32f/s^2\nsolve for t. \nt = 2 s\nThats the time it takes to go up just before it starts to fall again.\nThe time taken to fall is going to be the same: another 2 s\n\ntotal time: 4 s
360496
Hmmm, I get 0.033 Joules, not 0.33 Joules:\n\nW = F * r\nW = m * a * r\nW = (3.35e-5 kg) * (9.8 m/s^2) * (100m) = 0.033 Joules\n\nAs for the air resistance, well, the equation we did above was just the work done by gravity.  The total force is actually zero because the velocity is constant.  Therefore, there must an equal and opposite force preventing the  raindrop from accelerating...the air resistance.  \n\nF(total) = 0\nF(total) = F(grav) - F(air resistance)\ntherefore,\nF(grav) = F(air resistance)\n\n...note that the forces are the same over the whole length of the fall, but opposite in direction.  Since work is the dot product of force and distance (i.e. direction matters), the work done by air resistance will be equal to the work done by gravity but of opposite sign...-0.033 Joules.\n\nLet's be clear on this: The total work done on the raindrop is zero because it's not accelerating...but the gravitational force *is* doing work, and the air resistance is doing an equal amount of negative work.
504563
If the feather floats at a constant velocity of 2.0m/s, it means that it is in equilibrium, i.e. its weight must be equal to the air resistance. Therefore, \n\nWeight=0.0005*9.8\n           =0.0049N\nAir resistance=0.0049N\n\nAgain since the feather is floating at a constant velocity of 2m/s, you can use the formula:\n\nTime=Distance/velocity\n       =10/2\n       =5s\n\nNote: this formula is not applicable if the feather were accelerating downwards.
439734
The climber falls through a height h (= 0.75 m) \n\nh = 0.5*g*t^2\n\nSolve for t, the fall time before the rope affects the motion. At the end of that time the climber's speed is\n\nv = g*t\n\nAt the point that the rope runs out of slack, the climber's total energy is\n\nm*g*y + 0.5*m*v^2\n\nWhen the rope runs out of slack it stretches, behaving like a spring, so that when the climber stops falling (briefly) his total energy is now\n\nm*g*(y - x) + 0.5*k*x^2\n\nwhere x is the distance the rope stretched, and (in both the above expressions) y is the height above the ground. Consevration of energy says\n\nm*g*y + 0.5*m*v^2 = m*g*(y-x) + 0.5*k*x^2\n\nSimplifying this leads to\n\n0.5*m*v^2 = -m*g*x + 0.5*k*x^2\n\nand rearranging gives\n\nx^2 - (2*m*g/k)*x - m*v^2/k = 0\n\nThis quadratic equation can be solved for x, the distance through which the rope stretched. Note: be sure to take the physically consistent root when solving the quadratic equation.
550855
You can do this largely without equations if you wish.  The two sisters are approaching each other at a rate of 140 m/hr (just the sum of the two speeds).  When they meet, the total distance traveled will be 1680 meters (the whole perimeter).  Dividing the 1680 meters by 140 m/hr gives a time of 12 hours.  12 hours is 12x60 = 720 minutes.
501319
force of air is less than it's weight therefore it rises and it drops.\n\na=b
460234
1) The gravitational potential energy of water at the top of a waterfall is first converted to kinetic energy (falling) and then converted to both acoustic (sound) and thermal (temperature) energy when it hits the pool of water at the bottom. The temperature of the water in the pool will be slightly higher than the temperature of the water at the top of the waterfall.\n\n2) The thing to remember for this problem is that the energy in a system before some action = the energy in that system after some action:\n\nP0 + K0 + Th0 = P1 + K1 + Th1\nWhere P is gravitational potential energy (mgh), K is kinetic energy (0.5mv^2), and Th is thermal energy (leftover).\n\n22*9.81*4.5 + 0 + 0 = 0 + 0.5*22*2.5^2 + Th1\nSolve for Th1:\nTh1 = 22*9.81*4.5 - 0.5*22*2.5^2 =  902.44 Joules of thermal energy were generated due to friction.\n\n3) Split this problem into two cases, the first 10m and the second 10m. Find the velocity of the crate after the first 10m:\na = F/m = 200/70 = 2.86 m/s^2\nv = sqrt(2*a*d) = sqrt(2*2.86*10) = 7.56 m/s\n\nFor the second 10m, the force of friction on the crate is:\nFf = 0.3*70*9.81 = 206 N\n\nThe net force on the crate is 6N in the direction opposite of movement. The acceleration for the second 10m is:\na = F/m = 6/70 = 0.086 m/s^2\nvf = sqrt(vi^2 + 2*a*d) = sqrt(7.56^2 + 2*-0.086*10) = 7.45 m/s\n\n5) Since watts (W) are measured in J/s and joules (J) are measured in N*m, a watt is the same as N*m/s. Convert lb to N and ft to m to go from horsepower to watts:\n\n550 ft*lb/s * 0.3048 m/ft * 4.45 N/lb = 746 N*m/x = 746 W\n\nPS - You have in your problem description that horsepower is equal to 764 W but that should be 746 W.\n\nI hope this helps!
217429
It depends on the object.  Tests with coins have measured velocities too low to cause signifcant damage to the average human head.\n\nTerminal velocity, in this case the maximum speed a falling object will reach, depends on the shape of the object, the accelleration, and the qualities of the atmosphere.\n\nWhen an object is dropped near earth, it accellerates towards the earth at about 32 ft/sec/sec.  It goes faster and faster until the frictional drag from the air, has a slowing effect equal to the speeding effect from gravity.  From that point on it will stay approximately the same speed.  A feather has a very low terminal velocity.  A crowbar has a much higher one.  Its all about friction in air.\n\nA penny dropped near a planet without an atmosphere would accellerate toward the planet, getting faster and faster until it hit.  This scenario could be deadly for the space-suited person in the way.\n\nWater falling has an additional complication in that the air drag changes the shape of the water droplet and that in turn affect the drag.  A water droplet's changing shape on impact also affects the damage it might do a person.
375138
Galileo postulated  the "feather drop" which states that when a feather and a stone are dropped from the same height, both objects will hit the ground at the same time.  \n\nWe see that the feather and a stone fall at different rates when dropped from a certain height because there is air resistance.  The feather will be affected by the air when falling and reaching terminal velocity before the stone will.  Therefore, the feather will fall more slowly than the stone.  \n\nHowever, if somehow we were to remove the air and then drop the feather and the stone, both the feather and the stone will fall and hit the ground at exactly the same time.  Since gravity will cause the same acceleration in both objects (feather and the stone) of 9.80 m/s^2 downwards, both objects will accelerate uniformly at the same rate towards the ground.  Hence, both objects will hit the ground at the same time, as long as we are able to ignore air resistance.