# Tag Info

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 $$where \eta ... 1 If you take the time of throwing of the first object as 0 then the second object will start falling at 0+2 second.Now for the second one take the time of its start of fall as 0 and so the ending time will be 2 sec less i.e0+2 \longrightarrow 0 andt \longrightarrow t-2 So for the second one$$ \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 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 $(20 - ... 1 There is one error in the derivation, if you want to have$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. 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.

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