高级物理学1A|PHYS1131 Higher Physics 1A代写 unsw

This course provides an introduction to Physics. It is a calculus based
course. The course is examined at two levels, with Higher Physics 1A
being the higher of the two levels. While the same content is covered as
Physics 1A, Higher Physics 1A features more advanced assessment.

这是一份unsw新南威尔斯大学PHYS1131的成功案例

高级物理学1A|PHYS1131 Higher Physics 1A代写 unsw代写


\begin{prob}

As an oil well is drilled, each new section of drill pipe supports its own weight and that
of the pipe and drill bit beneath it. Calculate the stretch in a new 6.00 m length of steel
pipe that supports 3.00 km of pipe having a mass of 20.0 kg/m and a 100-kg drill bit.
The pipe is equivalent in strength to a solid cylinder 5.00 cm in diameter.

证明 .

Use the equation $\Delta L=\frac{1}{Y} \frac{F}{A} L_{0}$, where $L_{0}=6.00 \mathrm{~m}, Y=1.6 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$. To calculate the mass supported by the pipe, we need to add the mass of the new pipe to the mass of the $3.00 \mathrm{~km}$ piece of pipe and the mass of the drill bit:
$$
\begin{aligned}
&m=m_{\mathrm{p}}+m_{3 \mathrm{~km}}+m_{\text {bit }} \
&=(6.00 \mathrm{~m})(20.0 \mathrm{~kg} / \mathrm{m})+\left(3.00 \times 10^{3} \mathrm{~m}\right)(20.0 \mathrm{~kg} / \mathrm{m})+100 \mathrm{~kg}=6.022 \times 10^{4} \mathrm{~kg}
\end{aligned}
$$
So that the force on the pipe is:
$$
F=w=m g=\left(6.022 \times 10^{4} \mathrm{~kg}\right)\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)=5.902 \times 10^{5} \mathrm{~N}
$$
Finally the cross sectional area is given by: $A=\pi r^{2}=\pi\left(\frac{0.0500 \mathrm{~m}}{2}\right)^{2}=1.963 \times 10^{-3} \mathrm{~m}^{2}$





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PHYS1131 COURSE NOTES :

Derive the equation for the vertical acceleration of a rocket

The force needed to give a small mass $\Delta m$ an acceleration $a_{\Delta m}$ is $F=\Delta m a_{\Delta m}$. To accelerate this mass in the small time interval $\Delta t$ at a speed $v_{\mathrm{e}}$ requires $v_{\mathrm{e}}=a_{\Delta m} \Delta t$, so $F=v_{e} \frac{\Delta m}{\Delta t}$. By Newton’s third law, this force is equal in magnitude to the thrust force acting on the rocket, so $F_{\text {thnust }}=v_{e} \frac{\Delta m}{\Delta t}$, where all quantities are positive. Applying Newton’s second law to the rocket gives $F_{\text {thrus }}-m g=m a \Rightarrow a=\frac{v_{e}}{m} \frac{\Delta m}{\Delta t}-g$, where $m$ is the mass of the rocket and unburnt fuel.




















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