Tag Info

Hot answers tagged

7

No, you don't need to account for the mass of humans. We're a part of the total mass. Children who grow and gain weight are merely transferring mass from other parts of the Earth to them. The mass of humanity versus the mass of everything else on the Earth is a zero sum game. To see whether the Earth is changing mass, you need to look outside the box (or in ...


6

Your position is well-taken. But note that the discussion does not mention any rotation of the turntable... Since the turntable is frictionless, there is no way that the rotary motion, if any, of the turntable can influence the block. The only force between the turntable and the block is the normal force balancing gravity. Any motion of the block, caused ...


6

Yes you need to include the mass of the 7 billion or so humans in the total mass of the Earth, though since their total mass is something like $5 \times 10^{11}$kg and the mass of the Earth is around $6 \times 10^{24}$kg humans make up only about 0.00000000001% of the total mass. We don't make any great contribution to the escape velocity. The point of your ...


5

There is approximately 7 billion people on earth. If we take some average of about 75 kg each, that means $$ m_{\rm population}=7\cdot10^9\times75\,{\rm kg}=5.25\cdot10^{11}\,{\rm kg} $$ The mass of the earth is roughly $$ m_{\rm earth}=5.97\cdot10^{24}\,{\rm kg} $$ which is about 13 orders of magnitude bigger than the total population of the world. So no, ...


3

If you have another model, then use it to generate a set of predictions, and let's compare those to the predictions we get in the world we see. If you think there's some ether filling the universe, then write down some properties it has, and make some predictions with those properties. Talking in vague generalities will only lead you down a rabbit hole of ...


3

Your wife could be right about the rocket. Whenever we say something is small physically, we need to be sure what it is small with respect to. In the case of a thread, the drag force exerted by air beats the centripetal force keeping it taut. Because centripetal forces scale with mass, a denser thread will work. The issue facing a space elevator is ...


3

You can use the conservation of momentum to check their solution. Since no external forces are added, the initial momentum of the spaceship must equal the sum of the momentum of A,B,C in X, Y, and Z directions. The initial x momentum is 30000Ns and y and z momentum are 0. In the X-Dir: $ p_x = m_Av_{xA} + m_Bv_{xB} + m_Cv_{xC} = ...


3

In principle there is an effect, but firstly it's tiny and secondly it averages to zero. The mass of the ISS is about 420 tonnes, or about 5000 times the mass of an astronaut. That means if an astronaut pushes themselves off a wall at 1 m/sec the ISS moves in the other direction at about 0.0002 m/sec. But the ISS isn't very large so after only a couple of ...


3

i do not understand the part of clamping the center of the coil or cutting it off The solution method appeals to symmetry. Essentially, the system is symmetric (or even about the center of the spring) and this symmetry is exploited to arrive at an answer. If you have taken a first semester in electrostatics, you may be aware of the method of images. ...


2

If the particle is moving along some constrained curve then there must be some external force, $\mathbf{F_e}$, acting to constrain it to the curve. So the net force on the particle is: $$ \mathbf{F_{net}} = \mathbf{F} - \mathbf{F_e} $$ and the work done on the particle is: $$ W_{net} = \int (\mathbf{F} - \mathbf{F_e}).d\mathbf{s} = \int ...


2

I agree with CuriousOne that the example is more confusing than helpful, but this is the way I would explain it. Suppose you take a spring, place it on the ground then compress it. If you now suddenly let go of the spring it will rebound and bounce upwards off the ground: The spring clearly has work done on it because its kinetic energy increases and ...


2

Since the box is attached to the center of the table by a string, if you set it in motion with a tangential velocity, it will keep moving in a circle around the center, with the string supplying the required centripetal force. The movement of the box does not need to match how the table itself rotates -- as User58220 explains the turning of the table cannot ...


2

The concept of inertia is indeed useful in two ways. I think your notion of it as a technical promotion of the everyday word "sloth" (without the baggage given it by the Roman Catholic translation of the "deadly sin" Ἀκηδία) as extremely close to the mark. In physics the notion of "inertia" has two, very alike uses: The first is practical, through a weak ...


2

Your maths is almost correct. I think you should treat the lever as two separate rods which each have a pivot at one end. If the mass per unit length is $\sigma = M/L$, then the following applies to the two rods of length $r_1$ and $r_2$ (with $r_2=L-r_1$), where I split the rod up into infinitesimal chunks of length $dr$, mass $dm=\sigma dr$ and moment of ...


2

Let's make a concrete example with numbers: Suppose that $v_a = 6m/s$ and $v_b = 0 \rightarrow E_k = 0.5 * 6^2 = 18, p_a = 1 * 6 = 6, v_{cm} = p/M = 2$ . According to the conservation of energy and momentum: Kinetic energy and momentum are conserved only in a perfect elastic collision, if the bodies stick together the collision is inelastic an ...


2

Yes, the insect does actually travel at 0mph when it reverses direction. The critical aspect that you are missing is that the time that it is not moving is infinitely short. Actually, I would think that the head stops, then the thorax and finally the tail as it is squished up against the train.


2

Both the insect and the train window are deformable. Microscopically so, but deformable nonetheless. Because of that fact, the insect slows continuously to zero, reverses direction, and then speeds up in the direction the train is going. To our human perception this happens imperceptibly fast.


2

This is a fairly common misconception among schoolkids. The "action" force and "reaction" force act upon different bodies, and hence there's no cancellation. Take for example, the simple process of walking. You push the ground backwards ("action" force), the ground pushes you forward ("reaction" force). It is clear that they are not acting on the same ...


2

The first one is correct. The problem is that in general the force acting on the particle will no longer be conservative in the moving reference frame and we can no longer associate the potential energy $U$ to it. To see why this is so, realize that a force $\vec{F}$ is conservative if and only if: $$\oint_C\vec{F}\cdot d\vec{r}=0$$ to any closed curve $C$, ...


1

Newton's third law of motion is not applied on a single body. Suppose, there are two bodies $A$ & $B$ .Now $A$ exerts force $\vec{F}$ on $B$. Then Sir Newton said that the object $B$ will exert same force $\vec{F}$ on $A$ ie. in the opposite direction. So, they act on different bodies. There is no case of neutralization or cancelling out. A practical ...


1

Think of it like this and you will never be confused: Friction always tries to keep the two objects together. This regards both static and kinetic friction ($f_s$ and $f_k$) . If something is sliding to the right over asphalt (kinetic friction), friction will try to stop this relative motion, and hold the object and the asphalt together. So $f_k$ will ...


1

But in the real world,instead of storing that work, the agent of the conservative force returns it to the body itself which will be stored as potential energy. I don't believe this is the case. For example, consider the simple mass-spring system. When the spring is compressed and the mass is motionless, all of the energy of the mass-spring system ...


1

A conservative force only returns the energy back when the object moves in a closed path, that is, it returns to the initial position (it doesn't matter if he returns due to other forces). This can be demonstrated as a theorem, but the intuitive explanation is that a conservative force depends only on the spatial coordinates, and not in the direction of ...


1

B will move faster, the reason is that the acceleration, $a$, of A is smaller for two reasons (remember that $F_{applied}-F_{drag}=ma$) : 1) the same force forward is applied so the contribution to the acceleration on the smaller ball will be larger 2)the drag force on the larger ball will be larger (see Rennie's comment) on A because the cross sectional ...


1

I do not think the question ask what happens if the bobs collide, rather, it is a small amplitude so the pendulums interact through the horizontal cord until they reach the same average kinetic energy at equilibrium (by virial's theorem). If they have the same kinetic energy they each will rise the same relative height from their vertical position, so the ...


1

This is a side-effect of treating angles as dimension-less. For translational systems, we have \begin{align*} [\text{linear momentum}] &= [\text{action}][\text{length}]^{-1} \\ [\text{force}] &= [\text{linear momentum}][\text{time}]^{-1} \\&= [\text{energy}][\text{length}]^{-1} \end{align*} Correspondingly, for rotational systems, we have ...


1

The problem with your solution is that the inelastic collision and assumption that kinetic energy is conserved are mutually exclusive. You can see that in your math when you try to solve for $v_2$. Rewriting equation $(1)$ gives $v_1=\left(m+M\right)v_2/m$ which inserted into $(2)$ yields $$ m \left(\frac{m+M}mv_2\right)^2 = \left(m+M\right)v_2^2.$$ This ...


1

I reminded you elsewhere that you know very well when 3rd law is applicable, and that you even teach it to others for example here Newton's third law of motion is not applied on a single body. - user36790 but then, inexplicably, you forget to use it properly in your own posts: Let a body move and a conservative force oppose its motion. ...


1

I think this is a good example, and should be studied and understood carefully. But I don't know where the book goes after making it. Whether or not it is poorly stated depends on the surrounding context. It focuses attention on two things: 1.) the skater is a deformable body. The center of mass is not fixed in the body. 2.) work is defined as a ...


1

An initially ice-skater pushes away from a railing and then slides over the ice. Her kinetic energy increases because of an external force F⃗ on her from the rail . However, that force does not transfer energy from the rail to her. Thus, the force does no work on her. This is simply confused: This example is nothing but an elastic collision ...



Only top voted, non community-wiki answers of a minimum length are eligible