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In general relativity, rather than a two objects exerting a gravitational force on each other, the two objects are both part of the stress-energy tensor. This tensor determines the shape of spacetime (via the spacetime metric), and the spacetime metric determines what the geodesics are (roughly speaking, the metric determines how an object will move when no ...

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Integrate the jerk 3 times then using starting conditions to work out the integration constants.

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The website to which you linked doesn't seem to understand the purpose and results of Prof. Schwab's experiment. In fact, it didn't really describe the experiment at all. It just rehashed a lot of quantum mumbo-jumbo to make it look as though some power of "mind" causes quantum effects, rather than the more mundane cause-and-effect of having to use tools ...

5

There are two parts to this question. Part 1: will the card slide?if I have a card at an angle, is there a limiting vertical force that will make it slide sideways? The force diagram looks something like this: This is a bit like the "climbing a sliding ladder" problem, in which case there is going to be a limiting force F - once you exceed that force, ...

5

When you inhale you create an area of low pressure immediately in front of your mouth, like the Venturi of a carburetor. You would be drawn toward the low pressure area as the incoming stream of air accelerates down your throat, maintaining the low pressure in front of your mouth. Until your lungs are full. When you turn 180 degrees and exhale, you reverse ...

4

Okay, the system is in equilibrium so all the forces must be balanced. First consider the weights We have $W$ downwards from the centre of each of the 3 balls (assuming their centres of mass are at their geometric centres). We also have $N_A = N_B = \frac{3}{2}W$ upwards from the points of contact between the plane and balls $A$ and $B$ respectively. ...

4

Fundamentally, this is no different from computing the friction in a fluid (shear viscosity). The theory of viscosity goes back to Maxwell and Boltzmann, and microscopic calculations are possible for many fluids. Solid friction is more complicated, because the exact preparation of the surface obviously matters. First principles theories therefore concentrate ...

3

If you look at the instantaneous motion of your arm at any moment while it is flexing it will have a single direction; your arm goes up and then toward your center. During this time the opposite reaction is on the rest of your body, your torso is pulled slightly downward while you lift your arm and then slightly outward as you pull your arm toward your ...

3

I cannot see the image for some reason, but I think $\gamma$ is rather small there. The term $\gamma^2\omega^2$ shifts the maximum position, as a matter of fact. You took a rather strong "friction" ($\gamma=1$), which makes the resonance "frequency" smaller (longer period T). It is physically comprehensible.

3

We consider the integral: $$\sum_{i\lt j}\int_{t_0}^{t_f} \mathbf{F}_{ij}(\mathbf{r}_j(t))\cdot(\mathbf{r}'_j(t) - \mathbf{r}'_i(t))dt$$ For a rigid body, the distance between any two masses is always held constant, a fact that we can express as: $$\vert\mathbf{r}_i(t) - \mathbf{r}_j(t)\vert^2 = \Delta_{ij}$$ or $$(\mathbf{r}_i(t) - ... 3 Looking very close at the surfaces that touch at friction, this is an illustration Both surfaces are rough. They have ticks, holes, gabs, pits, spikes, and edges on the microscopic level. The smoother, the lower the coefficient of friction \mu. This constant is thus to be considered as a combined "roughness" between these two surfaces. Intuitively and ... 3 Newton's second law is a generalization of experience. It has no derivation in simpler terms. 3 This is the simplest analogy I could think of. Imagine a long narrow carpet sliding across a huge ice rink at 1kph. On the rear end of the carpet stands a very fat (200kg) man wearing roller skates. You want to bring him to a standstill. You could grab the man and dig your ice skates into the ice until he eventually stops. Alternatively, you could grab the ... 3 You're mistaken. If I visualize , the speed of rotation increases as the person's hands fold inwards, this indicates that the angular velocity increases. The net angular momentum however, remains conserved. (If we consider the system to be isolated, that is. For every isolated system, the angular and linear momentum is always conserved.) Also, for the given ... 2 ...a truck in motion and it has stack of hay (lets suppose) on the back. Now if the truck comes to a sudden stop will it stop faster if the force exerted by the truck on hay had overcome the friction force (another wording: will it be faster if the hay slips forward) or will it stop faster if the hay remains constant. I tried to find a braking ... 2 Nothing in physics can be proven in the mathematical sense. Sometimes mathematical physicists begin with axioms that are thought to model an aspect of the natural world and then work out what follows from these axioms. Their derivations are proofs in the mathematical sense of what follows from the axioms, but ultimately they do not prove anything about the ... 2 First We'll see the FBD (Free Body Diagram)and we get: Where Fc is normal force acting (Force of contact). From FBD of 5 Kg block (By newton's 2nd law)$$F-F_c = ma$$(1) From FBD of 10 Kg block$$F_c=Ma$$Solving above equations we will get:$$F=ma+Maa=F/(m+M)$$Putting the values you may get your result and your resultant force. 2 Why doesn't my hand just produce normal force on the book, cancelling out the force of gravity It does. Since the book doesn't accelerate downwards, another force is compensating for the weight. That is the normal force from your hand. and costing me no effort whatsoever? You are right that the normal force does not require energy to withstand the ... 2 From "what is a kettlebell" website: So just what is a kettlebell? A kettlebell is a cast iron ball with a handle attached to the top of it (picture a cannonball with a handle on the top). This design makes kettlebells different from training with dumbbells because the weight of a kettlebell is not distributed evenly, thus creating the need to counter ... 2 Knowing only "jerk" (third derivative of position), you cannot determine the distance traveled. To get distance traveled (or equivalently, position as a function of time) from jerk, you need to integrate three times. Each integration produces a constant of integration representing an initial value; your final equation looks something like this:$$p(t) = ...

2

The issue here is that your front wheels are turned/steered by the same angle. When you try to find the instantaneous centre of curvature, you may first want to assume the wheels won't slip from side to side, like you may get if you drive around a corner on a slippy road. As there is no slip, the velocity of each wheel must occur in the direction the ...

2

I looked up leap second in Wikipedia. It is a second added (usually) to clocks to keep them in sync with the atomic clock. Civilian clocks use Coordinated Universal Time (UTC), sometimes erroneously called Greenwich Mean Time (which no longer exists). Atomic clocks use International Atomic Time (TAI). UTC and TAI are in sync. Civilian clocks tick at the ...

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The explanation comes from earlier in that paragraph: If all the co-ordinates and velocities are simultaneously specified, it is known from experience that the state of the system is completely determined and that its subsequent motion can, in principle, be calculated. This is just saying the familiar thing that if you know the laws of physics for the ...

1

Change in direction won't impact the amount collected per unit area. Amount collected per unit area can be affected only if there is some lensing/dispersion. Since, mass of water that falls, and the area on which it falls (any increase in area in direction of wind is compensated by corresponding decrease in opposite direction) are both constant w.r.t. ...

1

For sure the object you will put in cart will act as one body until and unless they both have same velocity or I say that they both move relatively and this is possible if both have contact with each other. And if you want to take internal forces then you have to work on that but besides everything of what you ask the answer is simple that you have to take ...

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Look at a free body diagram. With red and the contact normal forces, with pink the friction forces, and with gray the gravity forces. If in the end any of the friction forces come up to being negative, then flip the orientation.

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Consider the total energy of the particle $$E=\frac{mv^2}{2}+mgh$$ Then (assuming $k>0$): $$\dot{E}=mv\dot{v}+mgv=mv[-g-mkv+g]=-m^2kv<0$$ So when the particle is thrown up and returns to a given height it has less energy than when it was first there. Since the potential energies are the same the speed has fallen. That is it comes down slower than ...

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Yes. The manner of which two surfaces in contact interact is highly investigated by the Tribology community.In particular, the field exploring the mechanics of the interaction is called contact mechanics. Tackling problems of contact mechanics analytically/numerically is often done by solving the elasticity equations. By predicting quantitatively the forces ...

1

Well, if the string was pulled such that it was along a radial arm from the centre of the "earth", in other words, in a vertical, then it wouldn't sag. Any deviation such that it is no longer exactly vertical, then the above answers come into play. I know this isn't what you are really asking, but let's admit, it does apply to the question as stated.

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You are making this rather hard for yourself. You correctly solved for the velocity, which is of the form $$v(t) = c_1 e^{-\alpha t} - \frac{g}{a}$$ where $a = mk$ and $c_1$ is found from the initial conditions. Integrating this expression should just give you $$x(t) = -\frac{c_1}{a} e^{-at} - \frac{gt}{a}$$ I think that because you ended up splitting ...

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