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The term in equation is: $$\frac{\partial u_i}{\partial x_j}\frac{\partial u_j}{\partial x_i}$$ So let's take a step back and think about what kinds of terms can appear in conservation equations. There can be a production term, a transport term, and a dissipation term. The transport term is the $\vec{u}\cdot\nabla q$ term that you noted. When you look at ...

6

The answer is right there in your own math. You derived that the delta V that results from using the rocket as a single stage rocket is $$\Delta V_{\text{single stage}} = v_{ex}\ln\Bigl(\frac{M}{M-(m_{fa}+m_{fb})}\Bigr)$$ while the delta V from using the rocket as a two stage rocket is $$\Delta V_{\text{two stage}} = v_{ex} \Biggl( ... 10 Put your math aside for a minute, and take a lesson from Robert H. Goddard, in one of my all-time favorite papers. Basically your rocket consists of a payload H, and the rest of the rocket consisting of fuel mass P, plus non-fuel mass (i.e. tank) K. The secret is, as you shed P through combustion, you must also shed K. Otherwise as P gets smaller and ... 0 Since the force is constant, then the acceleration is constant: \overrightarrow {a}=\frac {\overrightarrow {F}}{m} and the velocity would be:$$\overrightarrow {v}(t)=\frac {\overrightarrow {F}}{m}\times t +\overrightarrow {v_0}$$1 I would say,$$\sum \vec{F} = m\,\vec{a}_C$$where the left hand side are the net forces applied, and \vec{a}_C is the acceleration of the center of mass. 1 The x and y velocities should not add to V_0. To understand why, imagine something moving with V_x = 1 \frac{m}{s} and V_y = -1 \frac{m}{s}. This is something going horizontally and down; there's no reason why its velocity should be zero. The answer is that V_0 is the length of the velocity vector \vec{V}, and so it's calculated using ... 0 In an elastic collision the masses of both objects, the total kinetic energy, and the total linear momentum are conserved. The kinetic energy has contributions from the motions of the objects as well as their rotations. If we assume that no exchange between these two forms of kinetic energy occurs, i.e. that both forms are separately conserved, we have$$ ...

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You are probably thinking about the moments of a velocity distribution function. Suppose you had some probability distribution that was a function of space, time, and velocity or $f$ = $f\left( \mathbf{x}, \mathbf{v}, t \right)$, where $f$ has the units of # per unit volume per cubic velocity. From this we can define things like the number density: $$... 0 If you want to go from a to v you have to integrate over time:$$a=\dot{v}=\frac{d v}{dt}\int a dt =\int \dot{v}dt=\int \frac{d v}{d t}dt=\int dv = v$$0 You are almost done. If you simplify the express you found for v_{avg} and compare it with the initial equation for v_{avg}, in which you have to substitute v_i for v(t_i) and v_f for v(t_f), you will see that they are equivalent. Another way of deriving this is by expressing v(t) in terms of v_i and v_f:$$ v(t) = v_i + \frac{v_f - ...

1

There is one error in the derivation, if you want to have $v(t_i)=v_0$, you must have $$v(t) = v_0 + a(t-t_i)$$ You also have to use the fact that $v_f = v(t_f)$. Once you use all this, you should be able to divide out $t_f-t_i$ in the corrected version of your last line and get the result you seek.

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The formula you have written is correct; but they are functions of time. Hence, by inserting the particular instant , say $t$ on the function ,you get the instantaneous components of velocity. Then using phythagoras theorem you will get the total instantaneous velocity. Taking your example, at time $T$ s , the X-comp. is $30$ unit and Y-comp. is $(20 - ... 1 If the drag force is being modeled as a linear function of velocity$(\vec{F}_D=-b\vec{v})$, then the problem is straightforward. The vertical force balance for a falling droplet is $$\Sigma F_y=mg-bv=m\dot{v},$$ which gives the following differential equation for the velocity: $$\boxed{\dot{v}+\frac{b}{m}v=g}.$$ In the limiting case of the maximum ... 0 The acceleration is in the opposite direction from the initial velocity. If you are measuring positive upwards (by taking the initial velocity to be$+16.1$) the acceleration of gravity is negative. Your second answer is correct, but has one place too much accuracy as your data only goes to one place behind the decimal point. 1 Here's a point of view from thermodynamics that might be useful. Typically, the intensive quantities (in the form they're usually defined) arise as derivatives of the total (internal) energy$U$by some particular extensive quantity. Thus: Temperature$T=\frac{\partial U}{\partial S}$, the derivative with respect to the entropy Pressure$P=-\frac{\partial ...

0

This is a very interesting question. Velocity can be considered as either an intensive or an extensive property, depending on whether we are inquiring about the parts of a single system, or considering relations among separate systems. Velocity must be an intensive property, for consider: If I and my passenger and my books are traveling in my car, and if ...

1

If I understand your question correctly you are asking why you should use $t_2-t_1$ in the denominator instead of $t_2+t_1$. The reason is that both measurements are taken with respect to some initial time. This initial time is arbitrary; think about it as the time when you started your stopwatch. The total time elapsed between the events occurring at ...

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You're thinking of space with an external time, not spacetime. In spacetime, all objects move, because they trace out timelike curves through spacetime. The statement that an object is "stationary" in a coordinate system adapted to a reference frame is merely a statement that it's 4-velocity is of the form $(a,0,0,0)$, where $|a|^{2} = ... 2 What a layman calls "air resistance", a physicist would call drag. Drag is affected by the area and shape of the solid object, the speed and orientation of the object relative to the fluid, and various properties of the fluid such as its density and kinematic viscosity. The drag on a solid, rigid object isn't affected by the object's mass. However, drag ... 0 Final answer found here: http://digitalcommons.mtu.edu/cgi/viewcontent.cgi?article=1697&context=etds "DEVELOPMENT OF THE ECOCAR 3 PROPOSAL AND GUIDELINES FOR MODELING AND DESIGN IN YEAR ONE OF ECOCAR 3 - Tyler B. Daavettila - Michigan Technological University"$P = \frac{m_e}{2\epsilon_t \underline {t_a}} \left( v_f^2+v_b^2 \right)+\frac 2 3 m g ...

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It can be said that the tangential speed of the moon in its orbit is represented by a vector that is constant in magnitude, but not so his direction. This variation of the vector direction (always remains tangent to lunar orbit), is actually a change in velocity, and therefore acceleration. Why the moon does not fall on the ground? Simplifying to a ...

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Acceleration can change velocity in two ways - by changing its magnitude, and by changing its direction. Essentially, Earth's gravity is constantly steering the moon around the Earth. Your initial premise - "If moon travels with constant speed in one direction" - is incorrect. The moon's direction is constantly being changed by the gravitational ...

2

If you truly have a vector equation, then you really have three quadratic equations - one each for the X, Y and Z component. Let's write them: $$s_x = v_x \Delta t + \frac12 a_x (\Delta t)^2\\ s_y = v_y \Delta t + \frac12 a_y (\Delta t)^2\\ s_z = v_z \Delta t + \frac12 a_z (\Delta t)^2$$ If there is only one value of $\Delta t$, then this is an ...

1

You've written a vector equation, but any solution involving numbers has to involve one coordinate at a time, or what amounts to the same thing, three simultaneous equations. For simplicity, let's assume a one-dimensional version. All of the displacements, velocities, and the acceleration point in the same direction: $$s = v_i\Delta t + 1/2\, a\,(\Delta ... 6 The behaviour you are describing is a consequence of the virial theorem. Without going into the gory details this tells us that if some interacting system of many objects has an average total potential energy of <U> then its average total kinetic energy <T> is related to <U> by:$$ <T> = \tfrac{1}{2} <U> $$The proof of ... 3 Because of the homework policy, i'll just remind you of the definitions.$$Average \hspace{1mm}Speed= \frac{Distance \hspace{1mm}Traveled}{Time\hspace{1mm} of\hspace{1mm} Travel}Average \hspace{1mm}Velocity= \frac{Displacement}{Time} With this, you should be able to answer your own question.

3

Actually given that the first postulate says that all physical laws are the same in all inertial frames, you could replace the second postulate by the postulate: "Maxwell's equations are the physical laws for electromagnetism". From Maxwell's laws you can derive that the speed of light in vacuum has a specific, constant value, in SI units ...

2

Maxwell's theory had predicted that the speed of light varies with the speed of the observer. Initially (prior to Fitzgerald and Lorentz advancing the ad hoc length contraction hypothesis) the Michelson-Morley experiment was compatible with the assumption that the speed of light varies with the speed of the light source (as predicted by Newton's emission ...

1

Einstein did not prove this postulate ; he simply asked "what if it is true?". He had very good reasons for asking that question. His efforts to answer the question challenged a whole raft of "beliefs" about time and space, none of which were based on proof either ; they were (up until then) assumed true by so-called "common sense" alone. He made ...

1

The way I think of it, as a non physicist who quite likely has a few things wrong, is as follows: time actually passes slower for Alfred, and thus there is no difference in the speed of light. If you say, "Why does time pass slower for Alfred and not Bernard, after all, motion is relative, right!?" Well, this stumped me for a long time as a non-physicist, ...

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The answer is simple: Maxwell's equations. Maxwell published his electromagnetic theory in the 1860s. This generated a huge schism in physics. Maxwell's electromagnetism was in direct conflict with Newtonian mechanics. There is no allowance in Maxwell's electrodynamics for the speed of the emitter or the speed of the receiver. The speed of light is constant ...

2

If we are going for the maximum range, we could treat the collisions with things on the way down as elastic, and in the worst case, the collision would preserve the kinetic energy of the object, but change its direction so that it might move in a different direction. The first problem to solve is that the range is for a projectile launched from a given ...

2

Why? Because you did not answer the question as stated... The question explicitly says that the time spent at each speed is the same. You specifically started by saying the distance at each speed is the same, and along the way you find that the time at each speed is different Different question... Different answer... The best way to find the average ...

2

Your approach, though original, does not yield the average. There are a few important factors here. The car travels each speed for the same amount of time Those speeds are constant (there is no speeding up/slowing down) Due to factor #1 being true, we don't have to worry about weighing the average. We can take this average in the simplest way: average ...

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