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In introductory problems about work you're normally taught that it's force times distance: $$W = F \times x$$ and you treat the force as constant. If you look at the problem this way then you're quite correct that if the force is $F = mg$ then the box can't accelerate so it can't move. However a more complete way to define the work is: $$W = ... 19 Chaotic is not the same as random. A chaotic system is entirely deterministic, while a random system is entirely non-deterministic. Chaotic means that infinitesimally close initial conditions lead to arbitrarily large divergences as the system evolves. But it's impossible, practically speaking, to reproduce the same initial conditions twice. Given ... 3 Newton's first law states that: "An object at rest will remain at rest unless acted on by an unbalanced force. An object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force." It is often called the law of inertia. So if You want to move an object with zero velocity, at first moment You have ... 3 Speaking as an ex-hockey player, it's not so much that a dull blade doesn't glide as well: it's that the sharp edge allows us to get a solid grip in the direction perpendicular to our motion (i.e. "pushing off"). Same for turns. Granted, a dull blade with scrapes and divots won't glide well for obvious reasons. But there's little-to-no difference in the ... 3 I) OP is using the period formula$$\tag{1} T~=~2\pi\sqrt{\frac{I}{MgR}} $$for a compound/physical pendulum (in the small amplitude limit) to estimate the gravitational acceleration constant$$\tag{2} g~=~\left(\frac{2\pi}{T}\right)^2 \frac{I}{MR}. $$Here I is the moment of inertia around the pivot point; R is the distance from CM to the pivot ... 3 The buoyant block does exert a force on the water, it's force is equal to the mass of the displaced water, so the pressure of the water immediately beneath the block is exactly the same as the pressure of the water at that height in the rest of the container. Indeed, the mass of the system is just the mass of container with water + mass of block 3 Perhaps a better question to ask is: why is a single pendulum non-chaotic? Almost all real systems are chaotic at least to some extent; the fact that we can write out the solution for a single pendulum for all points in time is really quite peculiar, and only true because it is a highly simplified system. The reason these non-chaotic systems are so prevalent ... 2 A rather elegant approximation of what you describe may be set up by tying a string between two horizontally-separated mounting points so that the center hangs down a slight distance, and then tying a length of string from the center of that string to a weight. If one sets the weight in motion at a diagonal to the first string, one will observe that the ... 2 The force due to gravity balances the buoyant force exerted on the block The buoyant force is there because of gravity (There is a difference in pressure as we go deeper in to the ocean). There's an easy way to think it through. Imagine a beaker with some water, and it is standing still. There is no internal motion. Consider a small portion of this ... 2 Let's suppose I have some system and I know M, the system's total mass, \vec{r}_{cm}, the system's center of mass position and \vec{L}_{cm}, the systems angular velocity in the frame where the center of mass is the origin. How do I find \vec{L}', the angular momentum with respect to some other origin, say \vec{r}_{cm} + \Delta \vec{r}, which is ... 2 You posed two questions: First, Earth exerts a gravitational force on the car. The car, in turn, exerts a gravitational force on Earth. These are equal and opposite. This can be seen from the fact that the formula for magnitude of the (Newtonian) gravitational force:$$F_g=\frac{GM_1M_2}{r^2}$$remains of exactly the same form if the masses are switched. The ... 1 This is a question everyone asks at first because it intuitively seems like a contradiction. However, it is not. Conceptual Examples I think you are not far off but perhaps the third law is the one tripping you up, not the 1st... But anyway, here are some conceptual examples, which might help... Example 1. Consider the particle in the frame for a ... 1 You have a couple of mistakes. First, if you say that G = 6\times 10^{-11} k, then k should be \frac{\text{m}^3}{\text{kg}\cdot \text{s}^2}. If we instead define k as you did, then it is a dimensionless number: the conversion factor between the two sets of units. You should rather have said that 6\times 10^{-11} / k is the value of G in the ... 1 This turned out to be more complicated than I expected, here is my attempt at making sense of the problem using some physics and I think that Ryan M has the right idea. The force of friction will always oppose motion since it dissipates energy, if the ball is slipping there will be friction which opposes this motion, and if it is rolling there will be ... 1 I believe that the direction of the friction would actually be different depending where you hit below or above the center of mass. If you hit the ball below the centre the ball still rolls forward but at first the spin is in the opposite direction but changes direction due to the friction and overall forward inertia of the ball. The friction would have to ... 1 What you need is to match mechanical impedance between what the motor produces, and the tuned mass damper. Read more here http://www.bksv.com/doc/17-179.pdf and here http://arkansas.s.jniosh.go.jp/en/indu_hel/pdf/43-3-3.pdf 1 The value of initial velocity will be different for different angles θ with the horizontal.. So I got this result.$$ u=(gR/(sinθcosθ-cos^{2}θ))^{1/2} $$or$$ u=(2gR/(sin2θ-2cos^{2}θ))^{1/2} $$or$$ u=(39.2/(sin2θ-2cos^{2}θ))^{1/2}  This is my attempt for the solution(i have attached image): From A to B displacement is FB From C to B displacement is ...