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1

The height $h$ is probably the vertical displacement pointing downwards. Therefore: $$ h = \left(-\mathbf{\hat j}\right)\cdot\mathbf s = -|\mathbf{\hat j}||\mathbf s|\cos\alpha = -s\cos\alpha $$ Now we can derive: $$ \frac{dh}{ds} = -\frac{d}{ds}\left(s\cos\alpha\right) = -\cos\alpha \quad\Longrightarrow\quad \frac{dh}{ds} = -\cos\alpha $$ Therefore, ...


1

It is there, it's just hidden by the change of coordinates. Written in Cartesian coordinates, the kinetic energy is $$ T=\frac12m\dot{x}_1^2+\frac12m\dot{y}_1^2+\frac12m\dot{x}_2^2+\frac12\dot{y}_2^2+\frac12I\left(\dot\theta_1^2+\dot\theta_2^2\right)\tag{1} $$ where the last term is the rotational kinetic enregy. If you let \begin{align} ...


1

At any instant in time the action and reaction forces are equal - but since as Sofia pointed out they act on different bodies, that doesn't mean they "cancel out": each of the bodies experiences one of the forces and accelerates accordingly. However, in a real world system it takes time for the "information" about the force to make its way through the entire ...


0

When we decompose vector into it's rectangular components, we get two vectors along x-axis and y-axis. So the vector in the second diagram may have its x-component as a vector also whose magnitude is $F\cos\theta$ and direction is same as $\vec{S}$ i.e. along positive x-axis. And hence $\vec{F}$ is the resultant vector of $\vec{Fx}$ and $\vec{Fy}$. When ...


5

Let bottomleft point is $(0,0)$ and assuming each small segment is a uniform square of side 1 unit. The $y$ coordinate of the center of mass will be $y_{cm}=\dfrac{28\cdot4.5+25\cdot3.5+27\cdot2.5+27\cdot1.5+26\cdot.5}{143}\approx2.34$ Similarly the x coordinate will be, ...


0

Case A) Moving ball hits stationary man: $F_{1A}=\frac{m_{1}(v_{1A}-u_{1A})}{t}=-\frac{m_{2}v_{2A}}{t}$ Case B) Moving man hits stationary ball: $F_{2B}=\frac{m_{2}(v_{2B}-u_{2B})}{t}=-\frac{m_{1}v_{1B}}{t}$ You point out that $u_{1A}=u_{2B}$, let's just call it $u$. I don't see what is wrong with that. That forces in both cases must be equal just means ...


1

The center of mass (or gravity) is given by the formula (see Wikipedia, http://en.wikipedia.org/wiki/Center_of_mass) $$ (1) \ \Sigma_{i=0}^N \ m_i (\vec r_i - \vec R) = 0$$ If someone has a problem with the uniformity of the shapes, then we can replace the sum by an integral, and inside each shape consider the mass as a function of $\vec r$, i.e. $m(\vec ...


1

If $\theta$ is obtuse then $\cos\theta$ is negative and thus, $W=\vec{F} \cdot \vec{s}=Fs \cos\theta$ is negative.


0

Yes. Work $W$ is found as: $$W=\vec{F} \cdot \vec{s}=Fs \cos(\theta)$$ The angle will from the dot product take care of the sign for you. A perpendicular force e.g. is not doing any work, since $\cos(90^\mathrm{o})=0$. When $\theta < 90^\mathrm{o}$, $\cos(\theta)>0$, and when $\theta > 90^\mathrm{o}$ you change sign, $\cos(\theta)<0$. ...


1

Assuming that hands remain completely static, and the object do not breaks, then all of its energy can be considered to be converted into heat and sound (as you have already described).


-1

Energy is accepted by molecules of both objects and turns into heat form.


2

You're making a mistake in assuming that there is any left. Heat and sound account for all of it. There is one exception, though. If it crashes into something, and that breaks or bends the object, then the potential energy of the molecules is higher. That's why cars end up smashed after a collision. The molecules of the metal or plastic or whatever have ...


0

Let's assume there is no change in weight due to fuel consumption. That's a rather unrealistic assumption. If you get rid of that assumption, landing requires considerably more energy than does ascent. Some realistic assumptions: The lander Starts docked to some orbiting spacecraft, Separates from the orbiting spacecraft, Performs a de-orbit maneuver ...


1

It is a bit more complicated. You should be able to land without the use of energy but you have to pass your momentum to something - and this passing on of momentum will be the same in both cases. Let us consider a "spherically-symmetric-chicken-in-a-vacuum" model to see this explicitly. Momentum conservation and propelled mass The spacecraft of mass ...


-1

Now it occurred to me that the escape velocity/breaking may be symmetrical - that means you need the same amount of energy to counter the gravity on the way up as you need on the way down - but what also matters is how long you stay "hoovering" in the gravitation filed. This is what consumes fuel no matter what way you go and in reality both descend and ...


1

If I throw a small rock (m = 1kg) at a big rock (100kg) the small rock rebounds. Let's say my weight is 80kg, if I would jump into a big rock instead of bouncing back I would move in the same direction as a big rock. The big rock is heavier but it is not reflecting me. Why is that? There are two conditions which are to be met if a body A ...


0

according to what i have got about the law (maybe wrong or right): the breaking of the glass is the reaction of the glass to the excess force you have exerted on it. and you fill as much as the glass can react when it is not broken and is in touch with your hand.


-1

Landing is gravity-assisted, so requires less energy. A spacecraft on the moon has used up fuel resulting in its mass being less. If it is significantly less, then it takes less energy to lift a less massive craft back into orbit, than landing a heavier craft. Escape velocity applies only to non-powered projectiles, such as shooting a canon ball. There is ...


1

The difference in force to stop the trains you are talking about here is the difference in force is needed to bring the train to a stop within a particular distance Let me tell you what I mean. When you try to stop the train, you'll obviously be dragged in front of the train. Say the dragging causes an uniform force (due to the friction from the ground) ...


3

To first order, air resistance falls into two regimes at subsonic speeds. At very low speeds it can be modeled with a linear response to velocity, while at any higher speed you generally observe a quadratic response. The ratio of Reynold's number approximates the relative contribution for both components and has derivable values depending on the geometric ...


2

Yes, there is. Usually air resistance or other kinds of resistant forces can be considered as $bv^2$ or $bv$ where $b$ is a constant that depends on many things. For example, pressure, density, and so on. These functions are just an approximation and derived experimentally. You know that friction is an actual complicated force! They are usually neglected for ...


1

First of all you should note that Newton's law says when $F$ acts on a mass $m$, then that mass will move with acceleration $a$. Here, we should apply the laws of collision and by using the conservation of momentum, find out what your velocity will be after the collision. Before collision we have: $p_{tot}=mv$ and after collision $p_{tot}'=mv'+MV$ where $M$ ...


1

Two laws govern collisions: conservation of momentum and conservation of energy. Momentum is the product of mass and velocity, so we can write $$\sum m_i\cdot \vec{v_i} = const$$ Conservation or energy is a little bit trickier, since energy can be converted from one type to another. In an elastic collision, the kinetic energy is conserved, so $$\sum ...


1

Your claim that nothing will happen because they initially have no velocity is where things start to go wrong. In fact everything in relativity has the same speed (the speed of light). A particle that looks to be "at rest" in some reference frame simply has all of its velocity pointing in the "time direction". This is an intuitive reason why you would see ...


2

How accurate do you need to be? The problem with these calculations is that massless, frictionless pulleys are usually out of stock at Acme Mail Order. High school physics will give you the tension in the rope, a different set of high school equations will tell you how much extra tension is needed for a certain acceleration. A rough metric is add 10% for ...


2

Third law: analysis We have seen that in '63 Huygens exposed his theory to the Royal Society and in '68 he submitted his paper to the PT of RS in which he presented his law of momentum: "Quantitas motum duorum Corporum augeri minuive potest per eorum occursum; at semper ibi remanet eadem quantitas versus eandem partem, ablata inde quantitate motus ...


1

I have helped some school students through a course which was taught with Matter and Interactions which focussed on using physics principles to program computer simulations. Because of the programming, my students were comfortable with vectors from the first week. The computer made the vectors visual and as simple to manipulate as variables containing ...


3

It seems to me that it's not 'vectors' or 'vector algebra' which these students aren't grasping. Its the connection between a given 'physical phenomenon' and a corresponding 'mathematical representation'. I suspect this has something to do with conceptualising the physics rather than the mathematics. To put is simply: Physics $\neq$ Mathematics When ...


2

I can't comment on the up-till-now, but here's something to try for the "now-and-henceforth". I suspect that many student difficulties have in the past been left unhelped by the fact that lecture courses were "linear" (no pun meant here): there was a set coursework and a set way of thinking about concepts that the lecturer or teacher chose that students ...


0

When vectors are introduced, it should be empathized that a vector has 2 components, a magnitude and a direction. You can make a list of non-vectors and vectors (mass, speed velocity, acceleration, etc) which you can put on a quiz. Also, any equation that utilizes vectors is incorrect if the vector symbol is not included.


1

The other answers are OK, but if I'm correct they are missing information. Firstly, to be completely thorough, a general approach to force questions is to split the forces into components as shown here. If you do that and add the vertical force components and the horizontal force components, you will get a net force. This net force is the direction of ...


1

You can be sure that at least one reasearch team may help answer your question. Try to mail them. It seems that one can make such model. But still, as we can see it's a bit probabilistic field. http://trb.metapress.com/content/v4t5712601175275/ ...


0

Put the arrows after one another. Then draw a new arrow to the point they reach. The net force vector is the new arrow. The direction its' angle. The maginute is its' lenght (just as the magnetude of the two original forces were the lenghts of each).


0

I'll leave the problem for you to solve, but here's a hint. Remember that forces are vectors, and "net" means "sum of components". You can certainly use angles when summing vectors - just be careful with the signs so that you cancel what chould get canceled and add what should get added.


0

UPDATE (Solution): The center of mass of $B_1$ has a displacement of $$\vec x_1 = \vec x_{1,1} + \vec x_{1,2}$$ One due to RWS and the other due to its contact with the other body. Differentiation (twice) leads to $$\vec a_1 = \vec a_{1,1} + \vec a_{1,2} \xrightarrow{algebraic} a_1 = a_{1,2} - a_{1,1}$$. But $$T = m a_1$$ and $$\sum \tau = I ...


0

In one dimensional motion, sign really does mean direction for vectors. You can say positive is upwards and negative is downwards or vice versa. Now, come to your case. You seem to have mathematically understood. Then start with making your intuition. As stated by other answer, suppose you are very,very far away where gravitational force of the earth ...


0

The answer to the above question is the definition of work i.e. how we define it. When we apply force on a body, it is displaced with respect to its position. Work is said to be on the body. The definition of work says, "Work is the product of the component of force in the direction of displacement or vice versa". So the case I stated above has it's force ...


1

You can calculate the work done by gravitational force as the product of its weight and y-displacement. If I have got your question right, the body is freely falling after the force tips it off the table. So the work done by your force will not be as you have written. It would've been correct if the force had been acting on the body throughout its ...


0

Here is a simple model as explanation: Imagine yourself far away from any gravitational field - your potential energy is zero. As soon as you are entering into a gravitational field, you are accelerating and winning kinetic energy. The origin of this energy is that you "borrowed" some energy which is the potential energy you are losing. When you want to ...


2

The center of mass of an object is the point where the first moment of mass is zero. Put differently, when you support the object at that point, it will be balanced. Assume that point is $x_0$, then $$\int_0^\ell (x-x_0) \lambda(x) dx = 0$$ Substitute $\lambda$ and some simple manipulation will give you an expression for $x_0$. Let us know how far you ...


-1

The reason that the Bucket knows that it is spinning is that the Universe has a horizon in much the same way that the Earth has a horizon. The Earth's Horizon is curved line that is there not only because of Perspective Geometry, but also because the Earth is curved. Because the Earth has a two dimensional surface the height tilts back away from the ...


0

While considering 3rd law, forces act on different bodies , and not on same bodies. So the body which is hit is under the influence of applied external force only. The force which the hit body applies back to the hitting object is acting on the hitting object, so no point of cancelling of forces as they are acting on different objects. I too used to think ...


0

The net force on raindrop plus wagon is zero. Consider a single rain drop. Let the momentum in the direction of travel of the combined wagon/raindrop system be p. Now p = p_wagon_before + p_raindrop_before where p_wagon_before & p_raindrop_before are the momentum of the wagon and the raindrop before the drop hits the wagon. We have then: ...


0

This is the opposite of the rocket problem. In a rocket, acceleration occurs becaue mass is thrown out the back end. F = d/dt(mv) = m.dv/dt +v.dm/dt. If F= 0, then m.(-dv/dt) = v.dm/dt <-- note the negative term with the acceration. In a "typical" problem mass does not tend to change significantly, but in the rocket this mass term is highly ...


1

From the definition of Centre of Mass, the entire mass of the system is assumed to acts at the centre of mass. Consider a frame in which the position vector of $i^{th}$ particle is $R_i$ and its mass is $M_i$. Total mass of the system is $$M=\sum_i M_i$$ Let $R_{cm}$ be the location of the centre of mass measured in this frame. Then the placing the body ...


1

Without providing a numerical answer to your question, to do so would help nobody: It would help to first draw a 'free-body-diagram', doing so will help you visualise the forces acting on the pole. You need to calculate the moment at the hand 1m from the end, this is your pivot, from the force exerted by the mass of the pole. Calculate the force required ...


0

But if the velocity of the wagon changes, the net force can't be zero, right? Only true if the mass is constant (it's not if a wagon is filling up with water). If mass and velocity both change, you can't say anything about the force. Record the experiment and play the video backwards. You will see a wagon moving backwards. The wagon is spraying water ...


0

Note that in your suggested motion after the collision, momentum would not be conserved. The Law of Conservation of Momentum is kind of a big deal in Physics, and especially useful when analyzing collisions. It states that for a closed system, momentum is conserved. It's also sometimes restated that for a closed system experiencing a collision, the momentum ...


1

Since the surface is frictionless there is only vertical force. The torque is given by the normal force of the surface multiplied by the horizontal distance to the center of mass (c.o.m.). Now the normal force depends on the vertical acceleration of the c.o.m. - you know that the acceleration of the c.o.m. is a result of all the forces acting on the object, ...


0

Because in your example the action is not in the same direction than the velocity. The action and reaction are normal to the wall of the table, so the billiard ball only feels a reaction force (and thus a change in its velocity) in that direction. The component of the velocity of the ball parallel to the wall is not affected.



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