If you drop an apple from a height of 500 meters on Earth and on the Moon, it will fall faster on the Moon because it keeps accelerating.  Could someone please show the math for proving this, assuming that the apple reaches a terminal velocity of 42 m/s, Earth's gravity=10m/s/s and Moon's gravity=1.7m/s/s.  It would be preferred if you could use relatively simple equations like v=at, d=at^2/2, etc.
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The correct formula to use is one of the simple equations of uniform linear motion.\n\nv² = u² + 2as\n\nu = initial velocity = 0\na = acceleration = 1.7 m/s²\ns = height = 500\n\nWhen rearranged\n\nv = sqrt(2as) = 41.2 m/s on the moon\n\nThe apple on earth will approach terminal velocity asymptotically due to air resistance. To prove the apple reached a faster velocity from this height on the moon is difficult without doing calculations involving air resistance. For a height higher than 500m it is trivial. The apple on the moon will reach a higher velocity before impact, whereas the apple on earth will still be limited to falling at its terminal velocity of 42m/s.
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Plug the numbers into the Newtons Equation :\n\nF =  (GMm) / r^2\n\nF = Gravitational Force / weight  \nG = Universal gravitational constant   6.673*10^-11 (note the - sign on the exponent)\nM = mass of earth (kg)                 5.98*10^24 kg\nm = mass of satellite (kg)             150kg\nR = radius of earth    (m)             6,378,137 m\nh = height above earth (m)             12,800,000m\nr   = h+R\n\nThe answer will come out as the weight being @ 162.67N\n\nNot weightless, although someone standing on the satellite would feel weightless as its in continuous free fall.\n\nEver been in an elevator and felt lighter as it falls or been weighless on the big dipper as it plummets?\n\nThe gravitational force (weight) of 162.67N is what keeps it in orbit and prevents it from flying off at a tangent. Basically it pulls the satellite towards the earth but the satellite is travelling so fast it misses and continues round in its orbit.\n\nAdded:\n\nWeight is measured in Newtons and is the attractive force between the two masses. (M & m)\n\nIn this case the earth and the satellite. A 1 Kg mass on the surface of the earth will be pulled towards the earth's centre with a force of 9.8N, therefore its weight at ground level is 9.8N. If you plug take m as 1kg, h as 0 and plug them into Newton's formula should get the answer 9.8N, which is where we get the constant g from. \n\nAs you get further away from the centre of the mass the gravity field decreases in inverse proportion to the distance squared. Hence the divide by r^2 on Newtons fornula. Therefore the weight of a 1Kg mass at 12,800km is 1.08N.\n\nBecause the satellite is in a stable orbit the opposing forces of the satellite trying to escape orbit and the gravitational pull towards the earth are equal so yes the weight is 162.7N and the centrefugal force is 162.7N. This is the correct answer to the problem. \n\nAs an aside; for simplicicities case in this scenario the mass of the satellite can be regarded as constant due to its relatively low velocity compared to the speed of light. \n\nHowever as you come closer to the speed of light mass increases according to Einstein's theory of relativity, eventually reaching infinity at light speed.
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It seems the 16 m/s is not relative to this question. That is velocity in the horizontal direction. Gravity only effects in the verticle in this case. \n\ntime = sqrt(2 x height/ g) = sqrt(2 x 45 x 9.8) = 3.03 seconds\n\nvelocity = time x acceleration of gravity = 3.03 x 9.8 = 29.694m/s
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All objects are pulled by earth with constant acceleration\n\nAcceleration is change in velocity per unit time = change in velocity / time\n\nIn one second the velocity changes by g unit.\nIn t seconds it velocity reduces by gt.\nIf u is the initial velocity and if in t seconds it is reduced to zero then u = gt.\n\nOr t = u/g.\n\nThe time to go up is u/g.\n\nThe height it has gone up is given by average speed x time. = u/2 x t = u/2 x u/g = uu/ 2g.\n\nWhen it falls down it starts with zero speed, but falls with the same acceleration g.\n\nWhen it reaches the ground let its speed is v and time of fall is t2.\n\nv= gt2.\n\nSince the distance is = average velocity x time = v/2 x t2 = gt2/2 x t2= gt2t2/2  \n\ngt1t2/2 = uu/ 2g. or t2t2 = uu/ gg  ot t2 = u/g .\n\nThus time of fall is equal to time of up.
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k ill try to help u  even though i dont condone cheating. i just know how it is to not understand something and its due in a few hours and it sucks. just remember, nobody will be able to help u on the exam.\n\nso 1. \na. acceleration is change in velocity divided by time. \nsince it started from rest, initial velocity is 0 and then final velocity is given as 25.3 m/s. and the time is given as 0.59s. so to get acceleration   you would do  \na =  Vf-Vi / t\n  =  25.3 - 0 / 0.59\n  = 42.88......\n\nb. the force is equal to mass x acceleration. since you know the mass and acceleration, you just multiply them and you get F = 38421.69... N. \n\nc. you have to find the normal force in this case, which is the force that opposes gravity. like upward force. so in this case, gravity is acting on the driver and force of gravity is equal to mass x 9.8 = 666.4 N. and that means the normal force is 666.4N (since he is not moving up or down the total vertical force = 0)\n\n2. ok well in this case, you know the initial velocity is 0, the distance is 404m and the time is 4.944s. if its constant acceleration, you could assume that the average velocity is just Vi+Vf/2. so if thatz average velocity, then distance = average velocity x time\n     404 = 4.944s  x  (0+Vf)/2\n 404 x 2 / 4.944 = Vf\nVf = 163.43........m/s\n\nthe acceleration would just be Vf-Vi/t. and since you know Vf now  you just do   a = 163.43-0/4.944\na = 33.056....m/s^2\n\n3. ok well acceleration is 9.8m/s^2 (gravity)     distance is 9.7m. initial velocity is 0m/s. so this time, you have to solve 2 equations. the first one is the acceleration equation. a = Vf-Vi / t = Vf / t = 9.8\n9.8 = Vf/t\nVf = 9.8t\nand the other one is the distance = t  x  Vi+Vf/2   average velocity one. \n9.7 = t x Vf / 2\n19.4 = Vf x t\n19.4 = (9.8t) x t\n19.4 = 9.8t^2\nt^2 = 1.979...\nt = 1.406....s\n\nnow that you know the time, you find the Vf by putting it in any of those 2 equations. Vf = 9.8t = 9.8 (1.406)\nVf = 13.788... m/s\n\npart b. so in this case, you use the initial velocity as the one u just found (the velocity of the swimmer at the surface of the water)   so Vi = 13.788m/s\nand Vf = 0    since they come to a stop. and d = 2m\nso you have to find the acceleration. first you need to find the time by using the average velocity equation.\nd = t x Vi+Vf/2\n2m = t x (0+13.788)/2\n4 = 13.788t\nt = 0.29s\nnow that you know the time, you have to find acceleration\na = Vf-Vi/t\n  = 0-13.788 / 0.29\na = -47.5286m/s^2    its negative because its downwards and up is positive in our case. so the force = mass x acceleration and you just got acceleration and her mass is 65kg so you just multiply by 65kg to get F = 3089.36N\n\n\n4. well F = ma and you have the force. when they say it weighs this much newtons, then it means  use gravitational acceleration which is 9.8. so 2350N = m x 9.8m/s^2\nm = 2350/9.8\nm = 239.79kg\n\ni hope you actually understand this stuff. try to do your homework by yourself   dont take advantage of this site. youll just end up screwing yourself over. and if u dont get anythin, let me know and ill try to help
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the moon does not rotate.  The same side faces the earth all the time.  The acceleration = 0.  Speed = 0. Period = infinite.
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The answer involves a bit of algebra and is easy to lose track of where you are.\n\nConservation of energy says that\n\n0.5*m*v^2*sin(q)^2 + m*g*h = m*g*H\n\nwhere H is the height at the peak of the object's travel, q is the initial angle, h is the starting height, v is the magnitude of the velocity, and g is the gravitation acceleration. Also, we know that\n\n(1): H - h = 0.5*g*T1^2\n\nwhere T1 is the time is takes to reach the top of its travel. This expression was obtained based on the fall from the peak to the starting height h. Using this expression and the conservation of energy expression\n\n(2): T1 = v*sin(q)/g\n\nThe remaining distance h is covered in a time T2 given by\n\n(3): h = v*sin(q)*T2 + 0.5*g*T2^2\n\nGiven the range of travel (L) we know that\n\n(4): v*cos(q)*T = L\n\nand the relation among the individual times is\n\n(5): T = 2*T1 + T2\n\nT1 is counted twice for the upward and downward motion of the arc. That is the time to reach the top of its travel from its initial height and to return to that height..\n\nUsing expression (5) and replacing T2 in expression (3) we have\n\n(6): 0.5*g*(T^2 + 4*T1^2 - 4*T1*T) + v*sin(q)*(T - 2*T1) - h = 0\n\nFrom equation (2) we can substitute for T1\n\n(7): 0.5*g*(T^2 + 4*v^2*sin(q)^2/g^2 - 4*v*sin(q)/g * T) + v*sin(q)*(T -2*v*sin(q)/g) - h = 0\n\nSimplifying this becomes\n\n(8): 0.5*g*(T^2 - 4*v*sin(q)/g*T) + v*sin(q)*T - h = 0\n\nUsing Equation (4) to replace T\n\n(9): 0.5*g*(L^2/(v^2*cos(q)^2) - 4*sin(q)*L/(cos(q)*g)) + L*sin(q)/cos(q) - h = 0\n\nwhich, after rearranging, becomes\n\n(10): g*L^2/2 - (h*cos(q)^2 + L*sin(q)*cos(q))*v^2 = 0\n\nor\n\nv^2 = g*L^2/(2*(h*cos(q)^2 + L*sin(q)*cos(q)))\n\nGiven the values you mentioned above, v = 9.125 m/s.
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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.
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force=mass x acceleration(a)\nweight=mass x acceleration due to gravity(g)\n\nweight = force\nSo unit of weight is newton(N).\n\nValue of acceleration due to gravity(g) on the earth is 9.8 m/s^2\n\nValue of acceleration due to gravity(g) on the moon is 1/6th value acceleration due to gravity(g) on the earth and that is 9.8/6=1.6 m/s^2\n\nMass of a material remains same every where.\nWeight of meteorite on moon is 65/6=10.83N
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"Add 9.81m/s² for gravity = about 36.48 m/s² (since this is accelerating straight up against the force of gravity, not sideways)"\n\n"so the effective acceleration caused by your rockets engine is 26.67m/s/s +9.81 =36.47"\n\nTwo problems:\n1) you are assuming it has constant acceleration.  It does not because it is being slowed down as gravity pulls DOWN.  Now assuming this is just an ideal estimate, you should be ok using the average acceleration which is what you guys found.\n2) Noticed that "DOWN" part.  You are accelerating up, gravity is down, acceleration is a vector quantity.  Even if you used 26.67, it would be 26.67 - 9.81 not 26.67 + 9.81\n3) As stated in part 1, you guys found the average acceleration, this takes into account the down pull of gravity already.  Just plug in the 26.67 for acceleration.
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OK a little homework help;\n\n1. Gravity 'pulls' at 32ft/s^2 at the surface. The higher you go, the less gravity has an effect. To find the rate of acceleration from very high altitudes, you would need to integrate 32ft/s^2 from max altitude to zero.\n\nOddly enough, since gravity is proportional to mass, the gravity is at maximum at the surface. As you leave the surface of earth, gravitational pull gets less. AND, as you go to the center of the earth, gravitational pull becomes less again. In fact, when you are at the center of the earth, you would weigh zero pounds.\n\n2, mass x acceleration = force.\nso, mass = force / acceleration\nmass = 1225 N / 9.8 m/s\n\n5. momentum = mass x velocity so;\n500 x 64, 250 x 128, 1000 x 32 - which is larger\n\n4. force = mass x acceleration = 1357 x 8\n\n3.\nLaw I. Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.\n\nLaw II. The relationship between an object's mass m, its acceleration a, and the applied force F is F = ma.
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How about using it in a sentence and discussing it?\n\nIf an object is dropped from a great height, so that it will fall for a significant period of time, the force of gravity will accelerate the object downward. But as it accelerates, the air through which it passes will resist its motion (wind resistance) with a force that increases as its downward velocity increases. \n\nEventually the object will have been accelerated to a velocity at which the force of wind resistance equals the force of gravity. At that point, the net force on the object is 0 (wind resistance offsets gravity), and its acceleration will also be 0 (F = ma, so in this case 0 = m (0)). The velocity at which this occurs is called the "terminal velocity." It could be called the equilibrium velocity, or the ultimate velocity, or any other descriptive name, but we have come to call it "terminal velocity."\n\nNote that terminal velocity is not a fixed number. Obviously, terminal velocity is rather slow in the case of a balloon (and negative in the case of a helium balloon), since its wind resistance is great relative to its mass. And for a dense, streamlined object (think of a steel spear), terminal velocity would be extremely high, perhaps approaching the speed of sound (Mach 1). \n\nBesides the density and shape of the object, other variables that affect terminal velocity include air density and the force of gravity, both of which are significantly reduced at very high altitudes.
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I think you have got it wrong. Its mis spelled by you.Its Terminal Velocity.\nThe terminal velocity of an object falling towards the ground, in non-vacuum, is the speed at which the gravitational force pulling it downwards is equal and opposite to the atmospheric drag (also called air resistance) pushing it upwards. At this speed, the object ceases to accelerate downwards and falls at constant speed. An object moving downwards without power at greater than the terminal velocity (for example because it previously used power to descend, it fell from a thinner part of the atmosphere or it changed shape) will slow down until it reaches terminal velocity.\n\nFor example, the terminal velocity of a skydiver in a normal free-fall position with a closed parachute is about 195 km/h (120 Mph). It would take about 5.5 seconds to reach that speed. This speed increases to about 320 km/h (200 Mph) if the skydiver pulls in his limbs—see also freeflying. This is also the terminal velocity of the Peregrine Falcon diving down on its prey.\n\nThe reason an object reaches a terminal velocity is because the drag force resisting motion is directly proportional to the square of its speed. At low speeds the drag is much less than the gravitational force and so the object accelerates. As it speeds up the drag increases, until eventually it equals the weight. Drag also depends on the cross-sectional area. This is why things with a large surface area such as parachutes and feathers have a lower terminal velocity than small objects like bricks and cannon balls.\n\nMathematically, terminal velocity is described by the equation\n\n    V_t= &#92;sqrt{&#92;frac{2mg}{&#92;rho A C_d }}\n\nwhere\n\n    Vt is the terminal velocity,\n    m is the mass of the falling object,\n    g is gravitational acceleration,\n    Cd is the drag coefficient,\n    ρ is the density of the fluid the object is falling through, and\n    A is the object's cross-sectional area.\n\nThis equation is derived from the drag equation by setting drag equal to mg, the gravitational force on the object.\n\nNote that the density increases with decreasing altitude, ca. 1% per 80 m (see barometric formula). Therefore, for every 160 m of falling, the "terminal" velocity decreases 1%. After reaching the local terminal velocity, while continuing the fall, speed decreases to change with the local terminal velocity
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Two factors at least determine the spin rate of the Earth. The orbital distance from the Sun where the solid material condensed to form the Earth's body had a fundamental orbital velocity around the sun. As these materials coalesced into the Earth, they had to maintain the energy of their momentum during accretion. This resulted in the basic rotational speed of the solid Earth. Second, the Moon has an effect on the Earth. Since the Earth is about 80 times heavier than the moon, most of this effect is upon the Moon, which now turns at a rotational speed which places the heaviest side always facing the earth. On Earth the moon provides a slight attraction which is most easily seen in tides. This gravitational force is slowing the rotation of the Earth slightly. A possible 3rd factor is if the moon is the result of an impact of a large object on the Earth which caused ejected material to coalesce into the Moon. Such a large impact would effect the rotational speed of the Earth. Or the Moon may be a sister condensation that formed at the same time as the Earth and spun off into its current orbit. This would carry off material and momentum and set the Earth spinning as well.\nIn any case, momentum must be preserved. The final result is the current rotational speeds of both the moon and the Earth.
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Air drag would not be negligible. However if it was, It wouldn't necessarily take 7 seconds to fall. This would only be the case if the firework accelerated up constantly at 9.81m/s2. It could be more or less than this.\nThe acceleration vertically (downwards) would be 9.81m/s2 like everything else that falls\nthe path would be a parabola effectively it is a y=x^2 as the horizontal velocity is linear (proportional to t) and the vertical velocity is a function of t^2\n\nAgain air drag has a huge effect. Acceleration down would stop in about 0.5 second.
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The Moon's mass is less but you're closer to the centre.   Gravity goes up as the square of your distance from the centre.
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Mathematically, on the moon you could now throw the ball 240m.\n\nScientifically, on the moon you could throw the ball 240m and farther. On the moon, their is only gravity bringing the ball to a stop. On Earth, gravity and friction of the air slows the ball down