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6

I suspect this will be closed as "opinion based". I don't believe there is a canonical answer. Usually microscopic scale relates to phenomena that occur on a level much smaller than the system under consideration (atoms in a crystal when you are thinking about the crystal, for example). There is an analogy with micro- and macro-economics. Micro-economics ...


5

Let bottomleft point be $(0,0)$ 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, ...


4

The #1 cause of low-earth-orbit decay is atmospheric drag. There is just enough air up there to cause a tiny amount of drag, slowing the satellite down just like an aircraft without engines. End result: everything in LEO will return to the surface rather soon. Things that are higher up, like geosynchronous satellites or things in high inclination orbits ...


3

You are partially correct. If you have two objects with moment of inertia $I_1$ and $I_2$ then if one applies a torque to the other, they will start rotating in opposite directions. So if one object is the reaction wheel and the other is the satellite, the satellite will indeed rotate (while the reaction wheel, internally, is rotating in the opposite ...


3

Assuming the wheel moves at the same speed as the water (ie: neglecting and 'slip' of the wheel past the water), angular velocity is given by: $$\omega = v / r$$ where $r$ is the 'radial' distance from the centre of the wheel to the top of the water. In practice, there will be some water sliding past the wheel, depending upon the hydrodynamics of the ...


3

According to Newton's second law, $\overrightarrow F=m\overrightarrow a$ which means an object accelerates in the direction of the applied force. Acceleration means change in velocity is in the direction of the applied force. The final velocity will not necessarily be in the direction of the applied force, except when the initial velocity is zero.


3

The parallel axis theorem $I_B = I_A + md^2$ only applies when $A$ is the center of mass.


2

Because this radius $a$ is the semi-major axis of the elliptical orbit. It is not the instantaneous distance.


2

Classical mechanics are very important for everyday physics. For the energy scales, relative velocity differences, and mass scales that we experience is our everyday lives, Newtonian physics provide us with an extremely valuable tool of predicting outcomes of events. In other words Newtonian physics are an accurate enough approximation to the more precise ...


2

Newton's describes his notion of absolute time and space his Scholium on Time, Space, Place and Motion. In Newton's time, civil time was still measured by the motion of the Sun. Newton needed to distinguish time as measured by a sundial from the time as measured by a clock (or by the motions of the planets, or of Jupiter's moons). Scientists in Newton's day ...


2

Newton's first law says that unless an object is applied a force it will continue its motion with the same velocity (or linear momentum). This law expresses the homogeneity of the Euclidian space: nothing changes from one place in the space to another. Newton's third law says that if a body A acts on a body B by a force $\vec F$, then the body B acts also ...


2

I believe you want to replace mass with charge and angular velocity with the magnetic induction. The Coriolis effect is an apparent force due to the fact that the observer is measuring with respect to a rotating frame of reference. There is no actual force acting on the body, so this can be made to disappear by changing the frame of reference. Classical ...


2

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, ...


2

When considering motion in a fluid, or of a fluid, there are two types of forces to consider. Everyone immediately thinks of viscous forces, which arise from the viscosity of the fluid, and as you say these disappear in a superfluid. However there are also inertial forces that arise because the fluid has a mass. Accelerating the fluid requires a force just ...


2

Everything in the equation is treated as a function of time. $p(t)$ is defined to be the momentum of the rocket at time $t$. $m(t)$ is defined to be the mass of the rocket at time $t$. $v(t)$ is defined to be the speed of the rocket at time $t$. ("Rocket" here is the system that is actually losing mass. This is where the whole discussion originates from; ...


2

Reaction wheels, momentum wheels, and control moment gyros are three somewhat distinct ways of controlling the rotation and orientation of a spacecraft. Reaction wheels are the easiest to understand, at least in their simplest form. Consider a spacecraft such as a space telescope that is nominally not rotating with respect to inertial space. The reaction ...


2

The frictional force would be in the forward direction if the object is a rotating object. Let us say the object we are talking about is a wheel. At a point of time, the frictional force between the wheel and the surface will be only at the point of contact of the wheel with the surface i.e the bottom most point of the wheel, say A. With respect to the ...


1

I see that you already have the solution with you. So, I will try to help you with the tension in the chain zero part. The tension here will be zero even if the chain is acceelerating because the acceleration is caused by the gravitational force which is pulling it down and this force is acting on every single atom and molecule of the chain. In other words, ...


1

Historically, I'm not really sure what prompted Newton to write down his third law. Physically, however, it is just a statement of momentum conservation. Say object 1 pushes on object 2 with force $F_{12}$.Then by the third law object 2 pushes on object 1 with force $F_{21}=-F_{12}$ Rearranging and using Newton's second law: $F_{12}+F_{21} = \frac{d}{dt} ...


1

I can't quite fathom the source of your confusion (I think it might have something to do with a focus on the notion of rotation here---angular momentum does not require rotational motion), so I'm having trouble writing a really clear response. For the moment I would rather offer a program for practicing the right skills rather than reinforcing the mistaken ...


1

https://www.youtube.com/watch?v=dmnmuTv4pGE I think this is what you are looking for.


1

Solar flares etc can heat the highest reaches of the upper atmosphere and make it expand outwards, thereby increasing the drag on satellites. Using an artificial heating via HAARP was explored as a way of altering the trajectory of incoming ballistic missiles across the pole. Allegedly without much effect (so the US govt claims)


1

Alternative - just because it's fun to think about these things. Moving away a bit from the traditional rocket science, in principle you could land with very little fuel and a good set of wheels/brakes: apply a small fuel burn to change your orbit just enough to make a glancing pass at the surface of the moon, then apply the brakes as you touch down. You ...


1

I don't understand the following: Why in the term (m+dm), the parameter dm is positive, while in the term u(−dm) it is treated as negative? Why in the first term, (m+dm), it is thought that the mass m is gaining mass? Shouldn't it be (m−dm) ? Because that it loses mass. The convention used in that article is that $m(t)$ denotes the mass ...


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 ...


1

It is there, it's just hidden by the change of coordinates. Written in Cartesian coordinates, the kinetic energy is $$ T=\frac12m_1\dot{x}_1^2+\frac12m_1\dot{y}_1^2+\frac12m_2\dot{x}_2^2+\frac12m_2\dot{y}_2^2+\frac12I_1\dot\theta_1^2+\frac12I_2\dot\theta_2^2\tag{1} $$ where the last term is the rotational kinetic enregy. If you let \begin{align} ...


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

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


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

The dynamic equations of classical mechanics are locally time-reversal invariant. You can replace $t$ with $-t$ in them and they won't change their form. A system with friction is NOT described by these equations, and that kind of system is not covered by the reversibility statement. Statistical mechanics and chaos theory give you the real arguments for the ...



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