# 物理学1A (航空)|PHYS1149 Physics 1A (Aviation)代写 unsw

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This is an introductory level course in physics for students from all disciplines. The course has both a laboratory and theoretical component. Topics covered include the description of motion; forces and momentum; the dynamics of particles; kinetic and potential energy; the conservation of energy; oscillations and simple harmonic motion

An aircraft is equipped with a wing of symmetrical airfoils. The lift curve slope of the total aircraft is estimated to be $\frac{\partial C_{L}}{\partial \alpha}=0.8 \cdot 2 \pi \frac{1}{\mathrm{rad}}$. The stall angle of attack (AOA) is $12^{\circ}$. Wing area is $16 \mathrm{~m}^{2}$. Use $g=g_{0}$ and $\rho=\rho_{0}$.
What is the aircraft’s mass during a flight on which a stall speed of $50 \mathrm{kt}$ was observed.

$$\begin{array}{ll} \text { Given: } \quad & \frac{\partial C_{L}}{\partial \alpha}=5.0265 \frac{1}{\mathrm{rad}} \quad \alpha_{\max }=12^{\circ}=0.20944 \mathrm{rad} \ & V=50 \mathrm{kt}=92.6 \mathrm{~km} / \mathrm{h}=25.72 \mathrm{~m} / \mathrm{s} \ & S_{W}=16 \mathrm{~m}^{2} \end{array}$$
$$C_{L, \max }=\frac{\partial C_{L}}{\partial \alpha} \cdot \alpha_{\max }=5.0265 \frac{1}{\mathrm{rad}} \cdot 0.20944 \mathrm{rad}=1.05275$$
$$L=\frac{1}{2} \rho V^{2} \cdot C_{L, \max } \cdot S_{W}=\frac{1}{2} \rho_{0} V^{2} \cdot C_{L, \max } \cdot S_{W}=W=m \cdot g=m \cdot g_{0}$$
$$m=\frac{\rho_{0} \cdot V^{2}}{2 g_{0}} \cdot C_{L, \max } \cdot S_{W}=\frac{1.225 \mathrm{~kg} \cdot \mathrm{s}^{2} \cdot 25.72^{2} \cdot \mathrm{m}^{2}}{\mathrm{~m}^{3} \cdot 2 \cdot 9.80665 \cdot \mathrm{m} \cdot \mathrm{s}^{2}} \cdot 1.05275 \cdot 16 \mathrm{~m}=696 \mathrm{~kg}$$
Answer: The aircraft’s mass is $696 \mathrm{~kg}$.

## PHYS1131 COURSE NOTES ：

The category of effect of a failure is judged to be hazardous. Following $A C J N o$. 1 . to $J A R$ $25.1309$
a) What is the largest permissible failure probability?
b) What is the mean time to failure MTTF?
Solution
$\begin{array}{ll}F(t) & \text { probability of failure, } \ \lambda & \text { failure rate } \ \text { MTTF } & \text { mean time to failure } \ \text { FH } & \text { flight hour }\end{array}$
a) hazardous: $F(t=1 \mathrm{FH}) \leq 10^{-7}$
b) For small probabilities of failure: $\lambda \approx F / t=10^{-7} \cdot \frac{1}{\mathrm{FH}}$
$$\mathrm{MTTF}=1 / \lambda=\frac{1}{10^{-7}} \mathrm{FH}=10000000 \mathrm{FH}$$
Answer: If a failure has a hazardous effect, the mean time to this failure may not be less than $10000000 \mathrm{FH}$

# 高级物理学1B|PHYS1231 Higher Physics 1B代写 unsw

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This is the second of the two introductory courses in Physics. It is a calculus based course. The course is examined at two levels, with Higher Physics 1B being the higher of the two levels. While the same content is covered as Physics 1B, Higher Physics 1B features more advanced assessment, including separate tutorial and laboratory

\begin{prob}

A car started moving from rest with a constant acceleration. At some moment of time, it covered the distance $x$ and reached the speed $v$. Find the acceleration and the time.

Solution. The formulas for the motion with constant acceleration read
$$v=a t, \quad x=\frac{1}{2} a t^{2},$$
where we have taken into account that the motion starts from rest (all initial values are zero). If $v$ and $x$ are given, this is a system of two equations with the unknowns $a$ and $t$. This system of equations can be solved in different ways.

For instance, one can express the time from the first equation, $t=v / a$, and substitute it to the second equation,
$$x=\frac{1}{2} a\left(\frac{v}{a}\right)^{2}=\frac{v^{2}}{2 a} .$$
From this single equation for $a$ one finds
$$a=\frac{v^{2}}{2 x}$$
Also, one can relate $x$ to $v$ as follows
$$x=\frac{1}{2} a t \times t=\frac{1}{2} v t .$$
After that one finds
$$t=\frac{2 x}{v}$$
and, further,
$$a=\frac{v}{t}=\frac{v}{2 x / v}=\frac{v^{2}}{2 x}$$

## PHYS1131 COURSE NOTES ：

A missile launched from a cannon with the initial speed $v_{0}$ targets an object at the linear distance $d$ from the cannon and at the height $h$ with respect to the cannon. Investigate the possibility of hitting the object and the targeting angles.
Solution. The formula for the motion of the missile has the form (motion with constant acceleration)
$$z=v_{0 z} t-\frac{1}{2} g t^{2}, \quad x=v_{0 x} t .$$
The instance of these general formulas corresponding to hitting the target is
$$h=v_{0 z} t_{f}-\frac{1}{2} g t_{f}^{2}, \quad d=v_{0 x} t_{f} .$$
From the first equation one finds $t_{f}$ as in the preceding problem,
$$t_{f}=\frac{1}{g}\left(v_{0 z} \pm \sqrt{v_{0 z}^{2}-2 g h}\right) .$$