Tag Info

New answers tagged

-1

Wind will affect rainfall measurement. It is a serious problem in meteorological data called wind-induced undercatch, and it is why NOAA advises protecting gauges from the wind saying: "As a general rule, the windier the gauge location is, the greater the precipitation error will be." Take a gauge that is a cube $1\ m$ on each edge with an open top and ...


-2

Zeph we cant ignore that Dark Energy is a form of Energy yet to be understood but one thing for sure the name it self has a form of energy that governs 73% of the Universe while the rest that we see "pitch black" we called it Dark Matter, in our planet we interact with molecules like air carbon and much others depending where your at and when we speed we ...


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 ...


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. ...


0

You can do either. If the chocolate bar is not moving relative to the cart and you don't care how much force is between the cart and the chocolate bar, then you can treat them as a single system. If you do care about the forces or the chocolate bar is sliding around within the cart, then you'd have to model the system with two separate independent masses. ...


0

Can we rigorously define force? Presumably this question is asked in the context of Newton's mechanics (not to be confused with Newtonian mechanics). I'll be unconventional and say the answer is "no". Newton's three laws and his first few corollaries do not define force. They instead describe what forces do and how they relate to mass (also undefined), ...


2

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 ...


0

As the car turns around, since the car is a rigid body(ideally), not only will the rear wheel experience a centripetal force, but also other parts of the car. You can illustrate this by recalling your experience in a car when the car turns around. You will feel a force pushing you in the direction of rotation. So every part of the car is experiencing a ...


0

When the body is falling down without the affect of air resistance,it will experience only one force,I.e.the gravitational force which pulls out downwards towards the earth..and the object will fall to the surface of the earth will an acceleration 'g'. Coming to the first case...when the object is falling down in the presence of a drag force (friction due ...


0

Kinetic energy is the work required to accelerate a mass from rest to a velocity (KE = 1/2 mv^2). Momentum is a measure of the amount of movement a mass has at a velocity (p = mv). Kinetic energy may be considered a process, and momentum may be considered the result of a process. Momentum is conserved, but kinetic energy seems to come and go, as it ...


0

That's easy. Think in a simple example that this happens. Imagine two particles of equal masses moving at $\vec v_1 = \vec v$ and $\vec v_2 = -\vec v$. Their momentum: $\vec p_1 = m\vec v_1$ and $\vec p_2 = m\vec v_2$. The momentum of the system is therefore: $$ \vec p = \vec p_1 + \vec p_2 = m\vec v_1 + m\vec v_2 = m\vec v - m\vec v = \vec 0 $$ The ...


0

When we use F = change in momentum / change in time, is that the average force applied? No. $F=\frac{dp}{dt}$ is the force in a negligibly small period of time. That is, the instantaneous force in this very moment. $F$ is equal to the change of momentum $p$ in this very moment. There is no timespan, so there is nothing to take average of. ...


0

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.


0

Actually this seems pretty simple, if you have 45N stored in the spring - this is the force that will be applied to the block when it is launched (assuming perfect perfectness perfection) from there you simply need your Second Law of motion, which states that the force is equal to the mass times the acceleration. From here you have your acceleration, at this ...


0

You can actually use the conservation of energy principle to solve this.Account for the initial energy of the system...initially the potential energy of the system is 45 J + the potential energy of the block with respect to the earth and equate it to the final energy I.e. The kinetic energy and the potential energy.


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 ...


0

You can treat it as one system. Since the weight of the bar and the car are downwards, you can add them.


0

Since momentum is a vector, the quantity being measured did indeed change. As you state, while the magnitude is a constant $10\ \mathtt{kg\cdot m/s}$, the direction has altered. But what does that mean? The meaning of this change lies in understanding the distinction between momentum and impulse. The change of momentum is called an impulse. The common ...


1

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 ...


0

The answer by RGJ is not exact: the ball will always travel back up eh. The only difference is that:if the collision is inelastic $e<1$, if the collision is elastic $e=1$, But as correctly said by angel, it is almost impossible that this may happen and the coefficient of restitution will be at most 0.99...... This because e is the ...


0

If force F is horizontal and the blocks do not move, the friction force is the only force acting against gravity, so it should be equal to the sum of the weights, and its value should not depend on F or friction coefficients.


1

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 ...


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

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) = ...


1

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 ...


6

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


0

The (strong) equivalence principle states that physics is the same as that of special relativity, in a local inetial frame. This means that conservation of 4-momemtum (ie. Newton 3rd law ) holds locally (point particles). In general notice that GR does not tell you anything new about non gravitational physics that you did not know already in SR.You obtain ...


0

The period for normal pendulum is proportional to (L/g)^1/2 so you replace g with gcos(30) the new restoring force is cos(30)=(root(3)/2) so period will increase by factor root(2/root(3))


0

All correct except the total force should be divided by $m_1 + m_2$ (not just by $m_1$) since both masses are being accelerated.


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. ...


5

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 ...


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.


0

Imagine sitting on a swivel chair; hold your arm in front of you with your elbow to the side, then flex it quickly. Your chair will start to rotate (I know, I am sitting on a swivel chair as I write this and just did the experiment). The question you can ask yourself in a situation like you describe is this: if I am floating in space, would something move? ...


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 ...


6

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 ...


-4

In GR everything become geometric, gravity becomes curvature in spacetime. Actually, this isn't true. Spacetime curvature relates to the tidal force as opposed to the force of gravity. You will not find Einstein referring to a gravitational field as curved spacetime. How do we think of Newton's third law in the context of GR? What corresponds to ...


0

Suppose a box is pushed in a vacuum.Due to this push the box at rest starts to move.But in a vacuum there are no other forces which oppose the motion of the box.So it will continue to move.Or to think in energy terms,once we gave the box energy there is no mechanism for the energy to transfer from the box.The energy has been "trapped" in it.


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 ...


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 ...


0

First look at the "loop" made up of the two small pulleys and the upper horizontal bar ( the $\frac13$-$\frac23$ one) and the connecting rope that is being pulled on. If you pull $s$ meters of rope out of this loop, it gets shorter (top to bottom) by $ \frac s2$ This loop is part of a bigger loop, made up of the two large pulleys, the previous loop, the ...


0

The net force on the 5kg block will be 15N in the direction of the hand pushing. the subsequent acceleration or the 5kg block will be 3m per second per second. the net force on the 10kg block will be 30N giving an acceleration 3m per second per second. The accelleration will continue until the the hand is removed resulting in the blocks continuing at a ...


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: $$F=ma+Ma$$ $$a=F/(m+M)$$ Putting the values you may get your result and your resultant force.


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.


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 ...


1

You don't change the mass of the object by fragmenting it, you just create many more smaller pieces. Unless you do this at a sufficiently large distance (so that you create a lot of additional angular momentum and the particles will miss the Earth), what you are doing instead is turning a rifle into a shot gun. While a rifle bullet is quite deadly if it ...


1

Ablation of the surface inherently is less risky than fragmentation of the entire object, as the trajectories of fragments are not predictable with any great degree of accuracy in the absence of knowing the exact composition, internal stresses, and fault lines of the object. However, a possibly greater danger than the unforeseen trajectories of fragments ...


1

Initially, I agreed with Olaf Chujko's answer to this question; however, on further reflection, I think the most accurate answer is 'it depends': Firstly, from the schematic that is given, when switch A is pressed, the cylinder volumes above the two cylinders will be vented to a reservoir at ambient pressure (trust me, I work in Oil & Gas and I look at ...


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 ...


0

When two surfaces are at rest, the only friction between them is static friction. Static friction is a force, and takes a value between 0 and $F_{max} = \mu_\mathrm{s} F_{n}$ When two surfaces are in relative motion, the only friction between them is kinetic friction. Kinetic friction is a force and has a constant value of $F_{k} = \mu_\mathrm{k} F_{n}$ ...


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) - ...



Top 50 recent answers are included