# Tag Info

0

There are two mistakes in your thinking. Considering static friction first, $\alpha$ is the minimum angle at which the body first starts to slide; if $\tan\alpha$ is less than $\mu_s$, the body won't slide at all. Considering kinetic friction, when $\tan\alpha=\mu_k$, the body can be moving at constant velocity; at larger angles, it will be accelerating.

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You can find the position (as a function of time) for the balloon with simple kinematics: $y_B=200-\dfrac{1}{2}gt_B^2$ (with $g=+9.8\:m/s^2$). After five seconds, the archer shoots the arrow upward, the position of which can also be represented with kinematics: $y_A=40t-\dfrac{1}{2}gt_A^2$ Now, they're on separate time scales, where $t_i=0$ corresponds ...

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I think your confusion is between quasi-static extension of spring and extension of spring by a constant force. Here are some pointers: The potential energy of the spring $U_s$ is always going to be $\frac{1}{2} k\Delta x^2$ for a displacement $\Delta x$ from its "relaxed" state which we can set as the zero potential energy state. The work done in getting ...

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If the net hitting force passes through the center of mass of the object being hit, it will have linear movement, otherwise, it will have some rotation as well as movement.

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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 ...

1

You have to approach this from an energy standpoint. You start with a certain amount of energy, in this case gravitational potential energy, which is dependent on the initial height ($h=4\ \text{ft}$). At the end, you have a final energy, in the form of kinetic energy, which is given by $\frac{1}{2}mv^2$. With no other energy terms involved (meaning you ...

0

It is actually an example of continuity equation, which might help you to understand the process. The continuity equation of course implies mass conservation, in a given case. I find this example very useful as it contrasts typical consideration of continuity equation in which there is a flow of liquid through a pipe of variable diameter and speed of liquid ...

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In the limiting case when there is no dissipation (loss of energy), this is called an elastic collision, and answers are well known (see other answers). In the real world, this is never exactly the case, and depending on how much energy is dissipated (in the vacuum, by deformation of the bodies) you will have a result more or less close to this ideal limit. ...

0

In Elastic Collision Velocity of $Ball_1$ is Given by $v_1=u_1\frac{m_1-m_2}{m_1+m_2}+\frac{2u_2m_2}{(m_1+m_2)}$ $Ball_2$ $V_2=\frac{2m_1u_1}{(m_1+m_2)}+\frac{u_2(m_2-m_1)}{m_1+m_2}$ Where u is initial Velocity(Velocity before Collision) In Inelastic Collision Kinetic Energy after Impact is Less than kinetic energy before impact. The Loss in Kinetic ...

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As per your question events 2 & 3 (i.e. more mass and less mass are relatively the same events). Here I am considering that the events occur in one dimension. Event 1: When two bodies of equal mass collide elastically, their velocities get mutually interchanged. Events 2 & 3 are same: When a very lighter mass collides with heavy mass ...

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If the line of action of the force is is not through the centre of mass you can transform the original force $\vec a$ by adding two forces $\vec a$ and $-\vec a$ (net force zero) acting at the centre of mass as shown below: You can now consider these three forces as follows: a force equal in magnitude and direction to the original force but passing ...

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By "it is a case of conservation of mass. " Book means , that it uses Conservation of mass and is not an example of it. Or more clearly, when water falls down from a tap, it is a case which uses conservation of mass.

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Any linear force not going through the centre of mass will create torque, which I hope you know, is related to how far from the centre of mass the line of force is. So, if you manage to hit the object exactly at its centre of mass, i.e. the line of force is directly passing through the centre of mass, then it will show NO ROTATION. It will go straight ahead ...

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If not influenced by any other forces then after pushing It will move in a straight line and most likely rotating as it goes. It would be real hard to push it without giving it some kind of rotation but it will always move in a straight line.

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If you give a tangential force it would rotate. If you give a force at centroid, it will move in straight line. Along anyother point, between tangent and centroid , it will show joint motion.

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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

We have to consider the contribution of each little piece of mass $\mathrm{d}M = \rho \mathrm{d}V$ and how far it is from the other mass, $m$ resulting in a small force $\mathrm{d}F$. If you have uniform distribution of mass, $\rho=M/(LA)$, where $A$ is the cross sectional area of the rod, and we probably don't integrate over that because it's a small ...

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My feeling is that I have at some point (without realizing) assumed that this is my basis. I think this was likely the step in which I assumed that $|\psi\rangle$ is an eigenvector of $H$. Is that correct? That was not the step where this happened. There were two crucial steps: in the first you invoked "resolution of the identity" which implicitly ...

0

In order to understand the manipulations you made, you should first of all understand that the column vector notation is a relative one, in the sense that it is defined with respect to some (arbitrary) fixed basis (at least in the context of formal vector spaces). Every finite-dimensional vector space admits a basis; so you can always decompose one of its ...

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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 ? 0 If the pulse is moving to the right relative to the string (and you), then string (and you) are moving to the left relative to the pulse. It's like when you are in a car; if you are moving forward relative to the landscape, the landscape is moving backward relative to you. The next time you are in a car traveling at a constant speed along a straight road, ... 1 In the first instance, we are stationary w.r.t. the string, and the pulse propagates from left to right as follows: In the second instance, we are stationary w.r.t the pulse, and so the pulse appears in the same place while the string appears to move from right to left (look at the end points of the string): 0 You have a net force $$F = F_G - F_D$$ $$ma = -mg + C_0 v^2$$ Dividing both sides by m, letting$C_1 = C_0 / m$, $$a = -g + C_1 v^2$$ $$\frac{dv}{dt} = -g + C_1 v^2$$ $$\frac{dv}{-g+C_1 v^2} = dt$$ Now to integrate both sides, we need to use the limits$[0,t]$for$dt$,$[v_0,v]$for$dv$(where$v_0$is our initial velocity. $$\int_{v_0}^{v} ... 2 If it's falling only then you have F_d=+Cv^2, where up is the positive direction. You said there is a gravitational force F_g=-mg. Write a Newton's 2nd Law equation, set a=\frac{dv}{dt}, rearrange, to get dv/g(v) = dt, (I'll let you find g(v)) and integrate away. The v integral is not trivial. Look it up in an integral table, if your teacher will ... 2 The basic answer was given here: In a fluid, why are the shear stresses \tau_{xy} and \tau_{yx} equal?. Angular momentum conservation follows from linear momentum conservation (expressed by the Euler/Navier-Stokes equation) combined with the symmetry of the stress tensor. Momentum conservation is the equation$$ \frac{\partial}{\partial t}\pi_i + ... 1 Hint : the block starts from rest 11 cm from equilibrium. You don't give it a push. That means that after one oscillation, it'll come back at the same place and never go further away than 11 cm. That makes the rest of the work become trivial. 2 When the net force is zero, all you can tell is that the body will remain in the same state of motion (ie. same velocity). If the object is at rest and the force applied is equal to the resisting force, the object will remain at rest : the velocity remains zero. An example of this : someone pushing on a parked car. The static friction exerted by the asphalt ... 2 Kinetic energy tells us how much work is required to stop a body. The total work done on a body equals its change in kinetic energy. That's what the kinetic energy theorem says. The "work of a body" makes no sense. Work is something done ON a body BY an agent. It can be seen as an exchange of energy from that agent to the body. For instance, if you make a ... 0 It says "no slip". That means more than enough friction, so you can treat the string as if were a chain draped over a sprocket. They're trying to simplify the problem for you. 0 Heat energy and thermal energy are pretty much the same thing. My science teacher taught me, that because potential energy is related to height, it would be the kinetic energy that is converted into th 1 It seems like people watch too much TV around here. Who says there is no injurious recoil? Watch this, or just go to Youtube and search for "gun recoil". Mostly you'll find rifle recoil videos, but there are some featuring handguns, such as the second vignette in this one (The fourth vignette in that last video is quite troubling.) 0 There are several reasons for this: The momentum of the bullet and the momentum of the gun are distributed over different area. The recoil distributed over all the area of the gun's butt, while the bullet's momentum is applied to a much smaller area. Therefore momentum per unit square is much less in the case of the gun. You need to include also the ... 1 Think of force as the change in momentum with time Mathematically formalising this: $$F=\frac{dp}{dt}$$ $$p=mv$$ By conservation of momentum if the bullet leaves the barrel at some velocity (albeit quite high) comparatively the mass is incredibly small with respect to your hand So although the momentum imparted to your hand would be in the opposite ... 1 Because the bullet is very small compared to the handle of the gun. The bullet exerts a high pressure (Force per Area) and is able to pierce through skin. The backreaction of the handle is distributed onto the whole hand. 2 There is actually a back reaction on the shooters arm. But since the gun is much heavier than the bullet, the velocity of the gun is much smaller. This follows from momentum conservation, which states that the velocity of the gun is$v_g=v_b\frac{m_b}{m_g}$. Next the energy of the gun is also absorbed by a larger impact area, than the bullet, lessening the ... 1 It's a mistake to think of a body in orbit having its forces 'balanced' in some way - the forces are not balanced, because the body is accelerating towards the centre of the orbit! See my BBC article here, which explains this and weightlessness in general terms. I hope it helps. http://news.bbc.co.uk/1/hi/magazine/4625150.stm 2 Presumably you're okay with the intermediate step $$\int 2\vec h\cdot \vec r_\text{cm} dm = 2\int \vec h\cdot \vec r_\text{cm} dm \tag1$$ since the number two is a constant, and integration is linear over a constant. The authors assert that$\vec h\$ is also constant, so it comes out as well. If it's motion of the dot product across the integral sign ...

1

When the person starts moving, the boat will start moving backward due to reaction. backward momentum = forward momentum. So, when the person stops, the process will reverse so that center of mass of the boat and of the person combined remains where it was before he started moving. The relative position of person will be now different inside the boat but the ...

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Inertia is the measure of an object's ability to resist CHANGES in motion (acceleration). Mass is directly related to Inertia.

-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

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 ...

1

I am giving the solutions of original task (to get the speed at point B). I am not sure if the questions are necessary to perform the task. If you are sure the path taken does not matter (and I will assume that per your statement). So, let us consider a straight line path. Vertical component of F overcomes gravity and causes vertical move. Only horizontal ...

2

If we assume all other things being equal other than the downward force due to gravity, the vehicle on Earth would be capable of greater acceleration. The ability of the tires to grip the surface on which they are resting depends on the downward force keeping them in contact (Coefficient of friction). That will pretty much relate the gravitational ...

3

It is independent of the mass of the planet if you assume the bearings are frictionless. Also assume that the tyres do not make dents in the ground. However, in reality, the bearings have friction. Additionally, there is rolling resistance as the tyre makes small deformations in the ground as it is rolling. This is why you see a characteristic "W" shape in ...

1

the kick will have to be less than the kinetic friction (which is less than the static friction threshold...) Not quite. The kick will have to apply less force than the static friction. There is no need to consider kinetic friction at all. Only when the kick applies a larger force than the static friction limit, will the foot start to slide and kinetic ...

2

Always, always, always start problems like this by drawing a diagram: This make it obvious why cos and sin are used as they are.

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.

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