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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 ...


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( ...


3

Your approach is correct, now simply add indices to everything, i.e. $$y_i = v_{0,i}t_i + \frac12 a_it_i^2\quad\text{where } i\in\{1,2\}$$ and note that $t_2 = t_1 - 2\,\text{s}$. Then solve $15\,\text{m}\stackrel!=y_1 - y_2$.


2

The drift velocity is the average velocity due to an applied electric field. In a conductor, electrons scatter around at the Fermi velocity but have a net zero average (i.e., equal scattering in all directions). When the electric field is applied, the electrons are given a small velocity in one direction. Thus, we can say, $$ v_{drift}=\eta E $$ where $\eta$ ...


2

There is no "speed of gravity". You might be referring to acceleration due to gravity: gravity speeds up a falling object. However, that acceleration is strictly applicable only if there are no other forces. Air drag is another force. It causes the acceleration of the object to reach zero at the object's terminal velocity.


1

I'll assume you're talking about Earth where the average acceleration due to gravity is about $g=9.8m/s^2$. There is simply no 'speed' of gravity. This is simply an acceleration, not a speed. Terminal velocity occurs (from henceforth, I'll be using classical mechanics without relativity since it's unencessary) when the force of drag due to air resistance is ...


1

A bullet's path is affected by gravity. This means that if you shoot a bullet off into the distance as assuming there are no obstructions, the horizontal velocity could still be high enough (after a mile for example) to cause damage by the time it hits the ground. Of course, the ballistic path of the bullet can vary quite a bit and change the speed upon ...


1

If you take the time of throwing of the first object as $0$ then the second object will start falling at $0+2$ second.Now for the second one take the time of its start of fall as 0 and so the ending time will be 2 sec less i.e$0+2 \longrightarrow 0$ and$t \longrightarrow t-2$ So for the second one $$ \frac {dx} {dt} =u+at$$ $$ dx=udt +atdt$$ $$\int^y_0 ...


1

If your position is in 3D space (which means your position vector must be defined), then there is no distinction between displacement and change in position. $s=\boldsymbol{R_f-R_i}=\Delta\boldsymbol{R}$ , where $s$ is displacement and $R$ is position. However, in $v = ds/dt$, $ds$ does not mean change in displacement but rather an infinitesimally small ...


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 ...


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.



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