New answers tagged

-1

(1) is right. (2) is just wrong because it's wrong. The net electric force between two charged particles acts along a line joining them and its magnitude is given by the coulomb formula. You can take this as an experimental fact. You need to first calculate that force, and then you can calculate its components using trigonometry. There's no reason to think ...


-1

Ignoring friction, final speed at the bottom will be same for both because they loose same potential energy (assuming same mass). Therefore, acceleration down the flatter path will be less, and it will take longer time to attain final speed. In other words, steeper the path, faster is the downward acceleration. Longer path, (irrespective of shape of the ...


-1

@ BowlOfRed. The question states would both objects come down at the same time with ideal environment, that is, not friction, same mass, and g=9.8m/s2. So think about it this way, here the only work done is going to be by gravity. W=Fdcostheta, the force would be gravity, therefore W=mgh. Since gravity is a conservative force, it does not matter what path ...


0

It's the same reason a knife cuts through an object so easily. Use a blunt object and you'll have a harder time. When a person is standing facing the waves or away from them, the force exerted on them by the wave covers a greater area all across them and therefore produces a greater net force on the object that would be their body. Standing on your side, ...


-2

This is actually really simple. Yes! Both people will get to the end at the same time. In a perfect environment (say mass is the same, no friction force, and earth gravity), time is not a factor, and therefore both will come at the same time. The only forces acting on the people are $F_g$ and $F_n$, where $F_g = mg$ and $F_n$ really doesn't play a role so we ...


0

The answer, is the mechanism called a 'crank'. The pedals of a bicycle take linear force from your feet, and produce torque (twisting force) in a gear, and this is transmitted to another gear on the back wheel, where it is still a torque. The wheel, then, acts like the crank, producing (at the small footprint where it reaches the ground) a forward force to ...


5

So this depends very strongly on the shape of the slide. The easiest way to see this is to push it to its extreme: suppose one slide is purely vertical and has a length of 100 meters (i.e. $H = L$, then in the absence of friction getting to the bottom requires a free-fall time, which is gotten by solving $H - \frac12 g t_1^2 = 0$ to get a time $t_1 = ...


-2

Friction less environment - do you intend to say vacuum. If vacuum and if both went down the two slides of equal length, they reach the bottom at same time irrespective of their mass as the only accelaration is due to GRAVITY. Considering they both are at different lengths, the time taken will be greater if greater the height or length. Reason being the ...


0

That's a good question, and the answer is no, it depends on the shape on the slide. Then the question you can ask is which shape gives the shortest travelling time on the slide ?


1

TL:DR; There is no "classical" explanation... The speed of light is given by: $$ c = {1 \over {\sqrt {\mu_0\epsilon_0}}} $$ $\mu_0 = $ permeability of free space, $\epsilon_0 = $ permettivity of free space So this simple equation shows that the speed of light depends on the ability of free space (i.e., the vaccuum) to support electric and magnetic ...


0

Speed of any wave is property of the medium through which it travels. So, it is property of empty space that electromagnetic waves travel at a certain speed (no more, no less). The vacuum is not a medium. With a medium the propagation speed is related to the bulk and/or Young's modulus depending on the wave type. That's why it's a property of the ...


1

Suppose given an equation of the form $$y = Cx^n.$$ If you plot $y$ vs $x$ then it would be hard to find the value of $C$ or $n$ by looking at it. But if we take log on both sides we get $$ log(y) = n ~log(x) + log(C).$$ Now plotting $log(y)$ vs $log(x)$ would give us a straight line with slope $n$ and intercept $log(C)$. In your question the relation is ...


2

You apparently expect the plot of force vs acceleration to be linear, but not everything you will plot (e.g. position versus time) will be linear. Another example: plot the mean orbital radius of the planets vs. the period of revolution the planets. It's not linear. But if you plot the log of the orbital radius versus the log of the period it will be ...


3

Modern high-end bowling balls are extremely non-homogeneous in construction. If you look around at vendor sites, you'll see that there's an off-center "dumbell" of different density from the body of the sphere. All sorts of eccentric motion can occur as a result.


1

I would answer that: My mass is 80 kg. My weight is 784.1 N


-1

When you are walking you are doing work against gravity and friction. Consider this - when you walk on a flat surface, you shift your body weight on to say right leg. Lift the left leg and move it by a step. For the next step, you shift the weight on the left leg, lift the right leg and move forward. Thus you move. What is the work done? The leg consists ...


2

This is harder than it sounds. There are three things to consider. For a very narrow tube, for example $\lt 1~\rm{mm}$ diameter, the surface tension will be able to support a column of liquid - even if the other end of the tube is open. The pressure difference that can be sustained across a curved liquid surface with radius of curvature $R$ is calculated ...


0

I am not exactly sure what you wish to show. You can set up a water barometer in the same way as a mercury barometer is set up. As it will be about 10 metres high is has to be made of thick wall plastic/rubber tubing with the final metre or so a piece of thick walled glass tubing sealed at the end. This is so you can see the water meniscus at the top of ...


2

The net effect of the charging process is the movement of electron from one plate which then has a net positive charge to the other plate which then has a net negative charge. The battery facilitates this by creating an electric field in the wires and it is this electric field which applies forces on the electrons which makes them move. The movement of ...


10

In electromagnetism there are both positive and negative charges. Hence the force due to electric charges can be attractive or repulsive. Gravity, when treated as a classical force field, can only be attractive, there are not two types of "gravitational charge". What this means is that in electromagnetism, a given medium, may contain both positive and ...


17

First remember that $k = \dfrac {1}{4 \pi \epsilon_r\epsilon_o}$ where $\epsilon_r \ge 1$ It is because a medium can be polarised by an external E-field. The dipoles so set up produce the external E-field produce an E-field in the opposite direction so the net E-field (the sum of the external and dipole produced E-fields) is smaller. Thus the force a given ...


0

If A would gain kinetic energy, it would move far from B. As A would move more far, Potential Energy of B won't increase as distance had increased proportionally.


1

There is no conflict here. Let the two charged particles ($M,Q$) be the system with no external forces acting. Momentum is conserved and so for all time $M_BV = M_A V_{Af} + M_BV_{Bf}$ Energy is also conserved and so for all time $\frac 12 M_B v_B^2 + \dfrac{kQ^2}{R_i} = \frac 12 M_B v_{Bf}^2 + \frac 12 M_A v_{Af}^2 + \dfrac{kQ^2}{R_f}$ The ...


6

Potential and potential energy are defined for pairs of objects, not individual objects. It's meaningless to say "the potential energy of A". One must say "the potential energy of the system consisting of A and B". There is only one potential and potential energy in your problem. Perhaps the confusion comes from the way potential is introduced in ...


1

Note that pendulum motion is only sinusoidal for small angular displacements; as you increase the amount of swing the harmonic approximation fails. Lagrangian mechanics gives you a handle on all of the cases.


0

Your force will transmit through the cube to the bottom and it will spread like a projection from your pressure point. having sand underneath it will give away a pression mark the size of your cube showing how your force transmitted. the height will no matter, being a solid cube and your force will not deformed it and lose itself in the process.


1

The expression for the drag coefficient ($F_D = C_D \tfrac{\rho}{2} U_\infty^2 A$) appears from dimensional analysis, not by integrating the momentum equations over a control volume. Applying the Buckingham's $\Pi$ theorem you get different ways to express your dimensionless parameter $C_D$, but the dynamic pressure ($\tfrac{\rho}{2} U^2$) is more suited. ...


1

The force balance is real - but you may have to take into account acceleration of mass along the way. Acceleration gives rise to an additional force - if I push onto a 1 kg mass with a force of 10 N, it will accelerate at 10 m/s/s. But if I put a second mass in front of the first, their total mass is 2 kg and when I push with a force of 10 N, they will ...


0

Because your system does not only contain your hands, if the weaker arm is moved then you most probable experience a rotation with your upper body. Additionally you are also connected to the ground, which means some of the force will be hand off to the earth.


0

This is due to an imbalance in the forces exerted by your right and your left hand. While it is true that together they feel the same force, the force that is pressing them together, both your hands feel a net force as one. Similarly, you can push a box in one direction, even though, the reaction forces between you and the box are equal. The next ...


3

Gravity acts on all matter, not just water (it just so happens that water flows with less resistance than rock) which is why we get noticeable water tides but not very noticeable earth tides. However, if you were to bring a very large gravitating body too close to earth, you would find that the earth isn't quite as solid as it feels. The answer to your ...


0

No, the minus sign is not used to show direction. A vector is just a direction and magnitude and cannot be negative. But it may be subtracted from another vector. This minus sign does not show direction but just a mathematical procedure. But it turns out that subtracting a vector $\vec A$ form another vector $\vec B$: $\vec A-\vec B$ is the same as adding ...


0

After thinking over the answers and discussion, I believe that the problem is just that using sign to denote direction only really holds up in one dimensional problems (unless you're using complex numbers). Therefore, the solution is to use the formula $F_f=\mu F_f$ to deal with magnitude only, and if using sign notation, you'll just have to apply the sign ...


1

TL;DR: $R \neq mg\cos(\alpha)$, because there is a component of $\frac{mv^2}{r}$ in the same direction. Relative to horizontal and vertical The horizontal force must be equal to resultant force towards the center of the circle (for circular motion). $$ R \sin(\alpha) = \frac{mv^2}{r} \\ $$ The particle is in equilibrium in the vertical direction, so $$ ...


1

I'll answer since you've given it the old college try. Start by defining a coordinate system in which x is towards the center of the circle and y is upwards. The components of the normal force are $$R_x=R\sin\alpha,\qquad R_y=R\cos\alpha$$ Then Newton's second law gives you $$\sum F_x=R\sin\alpha=ma_c=m\frac{v^2}{r}$$ $$\sum F_y=R\cos\alpha-mg=0$$ The ...


0

I would normally solve this problem with the following diagram: The reaction force is normal to the surface, and it's bigger than just the component of gravity because the force provided to the object adds another component. This is the dreaded "centrifugal force" that doesn't exist - except in a rotating frame of reference, where it is a very real ...


0

Clearly, if the tension in one part of a rope is different than in another part, there will be a gradient in tension - which in turn means that if you look at a particular part of the rope at location $x$ where there is a gradient in tension $\frac{dT}{dx}$, then there is a net force on an element of length $\Delta x$: $$F = \frac{dT}{dx}\Delta x$$ For a ...


0

It also depends on whether or not the rope is inextensible. Assume it isn't. You tie one end to a rigid wall and pull on the other end with a constant tension $T$. Then at some time $t_0$, you start to pull with a slightly higher tension $T_1 > T$. It will take some time $\tau$ of order $\frac{L}{c}$, $c$ being the typical celerity of mechanical waves ...


-1

Since the LINE, a section of ROPE, has no attached mass or restraint and it seems it is accelerated as a whole, there is no tension. If one end is pulled to accelerate it then F=ma=0.0 N.


1

The small object exerts a force in the opposite direction to the normal force on the cart


0

You are using the word "resultant" in place of a more precise mathematical term - "net", or "sum". Add up all the forces on an object, and if they equal zero, the object experiences no acceleration. For your box, since the forces are in opposite directions, $$\sum F_i=F-F=0$$ For the car, the forces are actually quite complicated, but start with the fact ...


0

Depends on how you grab the block. If you press down or up on the block to balance, it would be a normal force. If you grabbed the side of the block, it would be a frictional force. It really depends on how much detail you want to involve in determining the force by your hand. You could also just call it $F_{Hand}$ if you really don't care how the force ...


1

No, you're not right. The container with more mass on the right will weight more on the scale. Pressure always acts normal to a surface, and the pressure forces acting on the slanted walls of the container will have downward components, which exert net downward forces on the container. You must include these downward forces in the force balance on the ...


1

As Nogueira stated, the air is made up of more than one particle. If the idea you're using is that a force occurs between two objects, then you'd have to treat each air molecule as an individual object, and you'd have many, many forces, each between the skydiver (who you could treat as a single body) and an individual air molecule. Of course, this gets messy ...


0

You can't add forces on different objects to get an acceleration of one object. All the forces should act on the center of the sphere to have effect on breaking the sphere. All the three forces: the tidal force $F_{T}$, the gravitational force $F_{g}$ and the Tension force because of the rope pulling the sphere $T$ act at the center of the sphere for the ...


0

Yes. The air is a gas made by a lot of particles. The air resistence takes the energy from the skydiver and transform into heat. In other words, the molecules that made up the gas hits the skydiver, getting with some kinetic energy of the skydiver.


1

It would probably be wiser to state the friction law as: $$|F_F|=\mu |F_N|$$ where $|F_N|$ denotes the modulus of the Normal force. Now consider the following diagram: Both blocks and slopes are identical. Left: some net force on the block causes an acceleration $a$ (left and up). The friction force $F_F$ points in the opposite direction: it opposes ...


0

In the relation $F_f = \mu F_{N}$, $F_f$ and $F_{N}$ are magnitudes of the frictional force and the normal force, and they are both positive. (I assume you are considering kinetic friction.) The direction of the frictional force is in such a way that it opposes the relative motion between the two surfaces. This cannot be inferred from $F_f = \mu F_{N}$.


0

Inertia is the term we use to say that an object can't change its velocity without a force acting on it. This does mean that a force always changes the velocity of an object (if net force is not zero). The wall scenario is a bit different because the wall will feel the force and will try to change its motion accordingly but in this case, inertia has less to ...


1

(Ignoring friction) if the connections are welded, then the bottom rod is irrelevant to balancing the forces (also ignoring deflection effects). If you removed the bottom rod entirely the diagram would remain the same. With pinned connections the second figure is unstable and will collapse, and there will be tension in the bottom rod in the first figure. ...



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