# 液体和等离子体|PHYS3202 Fluids and Plasma代写

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“Fluids and Plasmas” is an applied physics course designed to give the students experience in working with, predicting and measuring the behaviour of fluid flows and plasmas. The course begins with an outline of the fluid equations of motion, which lead to solutions for waves in fluids, convection and buoyancy-driven flows.

Here we consider some special cases of $(100)$ obtained by specializing $a, b, c$, and $d$ in $H$ of (77). Our choices for these four functions will determine the structure of the first integrals $x_{0}$ and $y_{0}$ through (99). For the cases we consider, their structure will be easy to discern and will give some insight into the behavior of $\xi$. How $\mathbf{B}_{p}$ will propagate in each case is pointed out to make the discussion more physically concrete. To conclude, a physical interpretation for the terms of $H$ and the role they play in determining how solutions propagate are discussed as well.

The first case we consider is a rather drastic simplification of the general result (99): we take $a, b, c$, and $d$ all to be zero, getting rid of Hentirely. Then we are simply left with
$$x_{0}=x \quad \text { and } \quad y_{0}=y .$$
Thus, in this case, the general solution for $\xi$ is of the form
$$\xi=\xi(x, y, z-\gamma \tau),$$
which corresponds to a structure propagating toroidally with specd $\gamma$.

$$\psi=(1 / \gamma)\left(\xi-\alpha \nabla_{1}^{2} \xi\right)(x, y, z,-\gamma \tau) .$$
The arguments in parentheses stress that $\psi$ moves in exactly the same way as $\xi$ : surfaces of constant poloidal flux simply propagate in the $z$ direction with constant velocity $\gamma$. Applying $\mathbf{B}{p}=-\epsilon B{T} \hat{\mathbf{z}} \times \nabla_{1} \psi$ to (105) shows that the disturbance $\mathbf{B}{p}$ also propagates in the same way: if we follow a point moving along a characteristic curve, $\mathbf{B}{p}$ at the point will be a constant vector. However, from $(105)$ and the arguments given at the end of Sec. III F, the solution is not necessarily an Alfvén-like wave because, in general, $\mathbf{B}{p}$ will not be proportional to $\mathbf{v}{t}$ for this case.

## PHYS3202 COURSE NOTES ：

Having introduced the fluid equations, we next discuss a method for arriving at exact solutions of them.

We denote the partial derivative of a quantity by a subscript, e.g., $\partial U / \partial \tau \equiv U_{\tau}$. Then, after rearranging the terms of $(9)$ and $(10)$ and subtracting (14) from (9), we can write
\begin{aligned} &U_{r}+[\phi, U]+J_{z}+[J, \psi]=0, \ &\psi_{r}+(\phi-\alpha \chi){z}+[\phi-\alpha \chi, \psi]=0, \end{aligned} This is the nonlinear system we will study. Note that we are taking $\hat{\eta}=0$ in (16); the resistivity of the plasma is neglected for all that follows. To satisfy (17) we take $$\chi=g(z)+U,$$ where $g$ is an arbitrary function of $z$. This is by no means the general solution to (17); it is simply a special case that satisfies (17) with little effort. Defining \begin{aligned} &\xi \xi \phi-\alpha g(z), \ &\text { and recasting (15) and (16) in terms of } \xi \text { gives } \ &\qquad U{+}+[\xi, U]+J_{z}+[J, \psi]=0 \ &\text { and } \ &\qquad \psi_{\tau}+(\xi-\alpha U){z}+[\xi-\alpha U, \psi]=0, \end{aligned} where (18) has been used. We note in passing that from (19) and (6), the definition of $U$, we have $$U=\nabla{1}^{2} \xi,$$
a relation that will be used often in what follows.
Now we have to find solutions to (20) and (21). Let us first consider the simpler case of axisymmetric equilibrium.

# 物理基础1A| Physics 1A: Foundations PHYS08016代写

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This is an introductory-level course, covering the classical physics of kinematics, dynamics, oscillations, forces and fields, and touching on aspects of contemporary physics, including relativity and chaos. The course is designed for those with qualifications in physics and mathematics at SCE-H level or equivalent.

During an IFR flight, a passenger looks out the window while relaxing in his seat. He observes a turn and estimates the bank angle to be $30^{\circ}$. At the same time, the passenger observes the free surface of the orange juice in his glass: it is parallel to the tray.
a) The passenger assumes the turn being flown as a rate one turn. Explain the term rate one turn. Why is it correct to assume a rate one turn.
b) The passenger assumes the turn being flown as a coordinated turn. Explain the term coordinated turn. Why is it correct to assume a coordinated turn.
c) Calculate the aircraft’s true airspeed.

a) A rate one turn is a turn with a heading change of $180^{\circ}$ in 60 seconds. During IFR flights turns are performed as rate one turns.
b) In a coordinated turn (correctly banked turn)

• the lift force lies in the aircraft plane of symmetry,
• the ball in the turn and slip indicator is centered,
• there is no acceleration along the $y$-axis of the aircraft.
This phenomenon is also shown by the free surface of the orange juice in the glass which is parallel to the tray.
c) $\tan \Phi=\frac{V \cdot \Omega}{g}$
$$V=\frac{g}{\Omega} \cdot \tan \Phi=\frac{9.81 \mathrm{~m} \cdot 60 \mathrm{~s}}{\mathrm{~s}^{2} \cdot \pi} \cdot \tan 30^{\circ}=108.2 \frac{\mathrm{m}}{\mathrm{s}}=210 \mathrm{kt}$$
Answer: The aircraft’s true airspeed is $210 \mathrm{kt}$.

## FINM2001 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 $M T T F$ ?
Solution
$F(t) \quad$ probability of failure,
$\lambda \quad$ failure rate
MTTF mean time to failure
FH flight hour
a) hazardous : $F(t=1 \mathrm{FH}) \leq 10^{-7}$
b) For small probabilities of failure: $\quad \lambda \approx F / t=10^{-7} \cdot \frac{1}{\mathrm{FH}}$
$$\mathrm{MTTF}=1 / \lambda=\frac{1}{10^{-7}} \mathrm{FH}=10000000 \mathrm{FH}$$
Answer: If a failure has a hazardous effect, the mean time to this failure may not be less than $10000000 \mathrm{FH}$.

# 公司财务|FINM2001 Corporate Finance代写

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This course focuses on tools and techniques used in modern financial management. Material in the course has an applied focus and is designed to provide students with the knowledge and skills required for understanding, exploring and analysing financial management issues. The course draws upon topical material in order to contextualise theoretical discussion, and present students with examples in practice.

Throughout the late 1950 ‘s, Myron J. Gordon (initially working with Ezra Shapiro) formalised the impact of distribution policies and their associated returns on current share price using the derivation of a constant growth formula, the mathematics for which are fully explained in the CVT text.

Required:

1. Present a mathematical summary of the Gordon Growth Model under conditions of certainty.
2. Comment on its hypothetical implications for corporate management seeking to maximise shareholder wealth.

These questions not only provide an opportunity to test your understanding of the companion text, but also to practise your written skills and ability to editorialise source material.

1. The Gordon Model
According to Gordon (1962) movements in ex-div share price $\left(\mathrm{P}_{2}\right.$ ) under conditions of certainty relate to the profitability of corporate investment and not dividend policy.

Using Gordon’s original notation and our Equation numbering from $C V T$ (Chapter Three) where $\mathrm{K}{e}$ represents the equity capitalisation rate; $E{1}$ equals next year’s post-tax earnings; $b$ is the proportion retained; (1-b) $E_{1}$ is next year’s dividend; $r$ is the return on reinvestment and r.b equals the constant annual growth in dividends:
(16) $\quad \mathrm{P}{0}=(1-\mathrm{b}) \mathrm{E}{1} / \mathrm{K}{\varepsilon}-\mathrm{rb} \quad$ subject to the proviso that $\mathrm{K}{\varepsilon}>$ r.b for share price to be finite.
You will also recall that in many Finance texts today, the equation’s notation is simplified with $D_{1}$ and g representing the dividend term and growth rate, subject to the constraint that $\mathrm{K}{\varepsilon}>\mathrm{g}$ (17) $\mathrm{P}{\mathrm{b}}=\mathrm{D}{1} / \mathrm{K}{\mathrm{c}}-\mathrm{g}$

1. The Implications
In a world of certainty, Gordon’s analysis of share price behaviour confirms the importance of Fisher’s relationship between a company’s return on reinvestment $(\mathrm{r})$ and its shareholders’ opportunity cost of capital rate $(\mathrm{K})$ ).

## FINM2001 COURSE NOTES ：

Moving into a world of uncertainty, Gordon (op cit) explains why rational-risk averse investors are no longer indifferent to managerial decisions to pay a dividend or reinvest earnings on their behalf, which therefore impacts on share price.

Required:

1. Present a mathematical summary of the difference between the Gordon Growth Model under conditions of certainty and uncertainty.
2. Comment on its hypothetical implications for corporate management seeking to maximise shareholder wealth.
An Indicative Outline Solution
Again, these questions provide opportunities to test your understanding of the companion text and practise your written and editorial skills.
3. The Gordon Model and Uncertainty
According to Gordon (ibid) movements in share price under conditions of uncertainty relate to dividend policy, rather than investment policy and the profitability of corporate investment. He begins with the basic mathematical growth model:
(16) $\mathrm{P}{0}=(1-\mathrm{b}) \mathrm{E}{1} / \mathrm{K}{c}-\mathrm{rb} \quad$ subject to the proviso that $\mathrm{K}{\varepsilon}>$ r.b for share price to be finite.
This again simplifies to:
(17) $\mathrm{P}{0}=\mathrm{D}{1} / \mathrm{K}{c}-\mathrm{g}$ subject to the constraint that $\mathrm{K}{c}>\mathrm{g}$
But now, the overall shareholder return (equity capitalisation rate) is no longer a constant but a function of the timing and size of the dividend payout. Moreover, an increase in the retention ratio also results in a further rise in the periodic capitalisation rate. Expressed mathematically:
$$\mathrm{K}{\varepsilon}=f\left(\mathrm{~K}{\mathrm{el}}<\mathrm{K}{e 2}<\ldots \mathrm{K}{\mathrm{en}}\right)$$
4. The Implications
According to Gordon’s uncertainty hypothesis, rational, risk averse investors adopt a “bird in the hand” philosophy to compensate for the non-payment of future dividends.

They prefer dividends now, rather than later, even if retentions are more profitable than distributions (i.e. $r>\mathrm{K}{\mathrm{e}}$ ). They prefer high dividends to low dividends period by period. (i.e. $\mathrm{D}{1}>\mathrm{D}{2}$ ). . Near dividends and higher payouts are discounted at a lower rate ( $\mathrm{K}{\mathrm{ct}}$ now dated) ,
Thus, investors require a higher overall average return on equity $\left(\mathrm{K}_{c}\right.$ ) from firms that retain a higher proportion of earnings with obvious implications for share price. It will fall.

# 微积分2|Calculus 2 MAST10006代写

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This subject will extend knowledge of calculus from school. Students are introduced to hyperbolic functions and their inverses, the complex exponential and functions of two variables. Techniques of differentiation and integration will be extended to these cases. Students will be exposed to a wider class of differential equation models, both first and second order, to describe systems such as population models, electrical circuits and mechanical oscillators. The subject also introduces sequences and series including the concepts of convergence and divergence.

Find the mass of a ball 9 of radius $a$ whose density is numerically equal to the distance from a fixed diametral plane.

Let the ball be the inside of the sphere $x^{2}+y^{2}+z^{2}=a^{2}$, and let the fixed diametral plane be $z=0$. Then $M=\iiint|z| d V$. Use the upper hemisphere and double the result. In spherical coordinates.
\begin{aligned} M &=2 \int_{0}^{2 \pi} \int_{0}^{2 \pi} \int_{0}^{a} z \cdot \rho^{2} \sin \phi d \rho d \phi d \theta=2 \int_{0}^{2 \pi} \int_{0}^{\pi / 2} \int_{0}^{a} \rho \cos \phi \cdot \rho^{2} \sin \phi d \rho d \phi d \theta \ &\left.=2 \int_{0}^{2 \pi} \int_{0}^{\pi / 2} \frac{1}{4} \rho^{4} \cos \phi \sin \phi\right]{0}^{a} d \phi d \theta=\frac{1}{2} a^{4} \int{0}^{2 \pi} \frac{1}{2} \sin ^{2} \phi \int_{0}^{\pi / 2} d \theta=\frac{1}{4} a^{4} \int_{0}^{2 \pi} d \theta=\frac{1}{4} a^{4} \cdot 2 \pi=\frac{1}{2} \pi a^{4} \end{aligned}

## MAST10006 COURSE NOTES ：

Find the surface area $S$ of the part of the sphere $x^{2}+y^{2}+z^{2}=4 z$ inside the paraboloid $z=x^{2}+y^{2}$.
$\square$ The region $\mathscr{R}$ under the spherical cap (Fig. 44-33) is obtained by finding the intersection of $x^{2}+y^{2}+z^{2}=4 z$ and $z=x^{2}+y^{2}$. This gives $z(z-3)=0$. Hence, the paraboloid cuts the sphere when $z=3$, and $\mathscr{B}$ is the disk $x^{2}+y^{2} \leq 3 . \quad S=\iint_{a} \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^{2}+\left(\frac{\partial z}{\partial y}\right)^{2}} d A . \quad 2 x+2 z \frac{\partial z}{\partial x}=4 \frac{\partial z}{\partial x}, \quad \frac{\partial z}{\partial x}=-\frac{x}{z-2} . \quad$ Similarly, $\frac{\partial z}{\partial y}=-\frac{y}{z-2}$. Hence,
$$1+\left(\frac{\partial z}{\partial x}\right)^{2}+\left(\frac{\partial z}{\partial y}\right)^{2}=1+\frac{x^{2}}{(z-2)^{2}}+\frac{y^{2}}{(z-2)^{2}}=\frac{(z-2)^{2}+x^{2}+y^{2}}{(z-2)^{2}}=\frac{\left(x^{2}+y^{2}+z^{2}\right)-4 z+4}{(z-2)^{2}}=\frac{4}{(z-2)^{2}}$$
Therefore,
$$\left.S=\iint_{\pi} \frac{2}{z-2} d A=\int_{0}^{2 w} \int_{0}^{\sqrt{3}} \frac{2}{\sqrt{4-r^{2}}} r d r d \theta=-\int_{0}^{2 \theta} 2 \sqrt{4-r^{2}}\right]{0}^{\sqrt{3}} d \theta=-2 \int{0}^{2 \pi}(1-2) d \theta=2 \cdot 2 \pi=4 \pi$$

# 金融学基础|FINM1001 Foundations of Finance代写

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This course is designed to familiarise students with the components of the financial system as well as to introduce them to the three basic ideas underpinning finance, namely the time value of money, diversification and arbitrage. In doing so, the course provides students with introductory exposure to financial transactions, institutions and markets including money markets, stock markets, foreign exchange and derivative markets and the instruments traded therein. It also provides students with a solid foundation for later studies in finance.

If you wish to provide 520,000 for your newborn’s University education, how much should you imvest now, given the interest rate that will accrue on the investment is $10 \%$ p.a compounded monthly?

In order to determine how much you should invest now, calculate the present value of $\$ 20,000received 18 years from now, bearing in mind that interest is compounded monthly. \begin{aligned} P V &=\frac{\ 20,000}{\left(1+\frac{0.10}{12}\right)^{12_{x} 13}} \ &=\ 3,330.73 \end{aligned} ## FINM1001 COURSE NOTES ： A company needs\$10,000$ in 5 years to replace a piece of equipment. How much must be invested each year at $8 \%$ p.a compounded semi-annualy in order to provide for this replacement?

To determine the amount the company must invest annually, simply use the future value of an annuity formula, bearing in mind that the interest rate is compounded semi-annually. Therefore, as investments will be made on an annual basis, we must calculate an annual effective interest rate to use in the annuity calculation.
\begin{aligned} r &=\left(1+\frac{0.08}{2}\right)^{2}-1 \ &=0.0816 \ &=8.16 \% \end{aligned}
\begin{aligned} \ 10,000 &=F\left[\frac{(1.0816)^{3}-1}{0.0816}\right] \ F &=\frac{\ 10,000}{\left[\frac{(1.0816)^{5}-1}{0.0816}\right]} \ &=\ 1,699.14 \end{aligned}

# 物理1|Physics 1 PHYC10003代写

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This subject is designed for students with a sound background in physics, and aims to provide a strong understanding of a broad range of physics principles.

An atomic clock is placed in a jet airplane. The clock measures a time interval of $3600 \mathrm{~s}$ when the jet moves with speed $400 \mathrm{~m} / \mathrm{s}$.

• How much larger a time interval does an identical clock held by an observer at rest on the ground measure?

We take the $S$ frame to be attached to the Earth and the $S^{\prime}$ frame to be the rest frame of the atomic clock. It follows from that
and from that
$$\gamma \simeq 1+\beta^{2} / 2$$
It follows that $\delta t=3.2 \mathrm{~ns}$ when $v=400 \mathrm{~m} / \mathrm{s}$ and $\Delta t^{\prime}=3600 \mathrm{~s}$.

## PHYC10003 COURSE NOTES ：

Two spaceships approach each other, each moving with the same speed as measured by a stationary observer on the Earth. Their relative speed is $0.70 c$,

• Determine the velocities of each spaceship as measured by the stationary observer on Earth.
Solution
Text Eq. (1.32) gives the Lorentz velocity transformation:
$$u_{x}^{\prime}=\frac{u_{x}-v}{1-u_{x} v / c^{2}}$$
where $u_{x}$ is the velocity of an object measured in the $S$ frame, $u_{x}^{\prime}$ is the velocity of the object measured in the $S^{\prime}$ frame and $v$ is the velocity of the $S^{\prime}$ frame along the $x$ axis of $S$.

We take the $S$ frame to be attached to the Earth and the $S^{\prime}$ frame to be attached to the spaceship moving to the right with velocity $v$. The other spaceship has velocity $u_{x}=-v$ in $S$ and velocity $u_{x}^{\prime}=-0.70 c$ in $S^{\prime}$.
It follows fromthat
$$0.70=\frac{2 \beta}{1+\beta^{2}}$$
solving which yields $\beta=0.41$. As measured by the stationary observer on Earth, the spaceships are moving with velocities $\pm 0.41 c$.

# 天文学入门|PHYS1160 Introduction to Astronomy代写 unsw

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Is there life beyond the Earth? How common might life be? Millions of dollars are spent by the Australian government and other countries each year on space exploration, so what do we actually learn from modern day telescopes and satellites? In this course, students will explore an introduction to the Universe and the study of astronomy.

Earth has an orbital period of 365 days and its mean distance from the Sun is $1.495 \times 10^{8} \mathrm{~km}$. The planet Pluto’s mean distance from the Sun is $5.896 \times 10^{9} \mathrm{~km}$. Using Kepler’s third law, calculate Pluto’s orbital period in Earth days.

\begin{aligned}
&T_{E}=365 \text { days }\
&r_{E}=1.495 \times 10^{8} \mathrm{~km}\
&r_{P}=5.896 \times 10^{9} \mathrm{~km}\
&T_{P}=\text { ? }\
&\left(\frac{T_{E}}{T_{P}}\right)^{2}=\left(\frac{r_{E}}{r_{P}}\right)^{3}\
&\left(\frac{365 \text { days }}{T_{P}}\right)^{2}=\left(\frac{1.495 \times 10^{8} \mathrm{~km}}{5.896 \times 10^{9} \mathrm{~km}}\right)^{3}\
&\left(\frac{365 \text { days }}{T_{P}}\right)^{2}=\left(2.54 \times 10^{-2}\right)^{3}\
&\left(\frac{1.32 \times 10^{5} \text { days }^{2}}{T_{P}{ }^{2}}\right)=1.63 \times 10^{-5}\
&T_{P}=\sqrt{\frac{1.32 \times 10^{5} \text { days }^{2}}{1.63 \times 10^{-5}}}\
&T_{P}=9.00 \times 10^{4} \text { days }
\end{aligned}

## PHYS1160 COURSE NOTES ：

A spherical planet orbits an F-v star with luminosity $1.5 \times 10^{34}$ erg $^{-1}$. The orbit is circular with a radius from the star of $2 \mathrm{AU}$. Assume that the planet is rapidly rotating, has an atmosphere and reflects $20 \%$ of the light that falls on it, but absorbs the other $80 \%$ and, in the assumed steady state, radiates it as a blackbody.
(a) Ignoring any greenhouse effect, what is the temperature of the planet?
Solution:
The solution to the planetary tempearture comes from assuming a state of “balanced power”. The energy received per second by the planet from its host star, $\dot{E}{\text {in }}$ is balanced by the energy it radiates per second as a blackbody, $\dot{E}{\text {out }}$. If this were not so, then the temperature of the planet would rise or fall until the balance was achieved. This would take something like the heat capacity of the atmosphere divided by $E_{\text {in }}$ or in the case of the earth a few weeks.

The energy in is the cross sectional area of the planet as viewed from the star, a disk with radius $R_{p}$ intersects the radiation that would have passed througha disk of area $\pi R_{p}^{2}$ (not $2 \pi R_{p}^{2}$, then one would have to integrate $\cos \theta$ for the incident radiation, which adds work). The flux passing through each square $\mathrm{cm}$ of the disk is $L_{} /\left(4 \pi d^{2}\right)$ where $d$ is the distance from the star to the planet. Additionally it was specified that only $80 \%$ of the flux gets through to the planets surface and contributes to its warming, so $$\dot{E}{\text {in }}=0.8 \frac{L{}}{4 \pi d^{2}}\left(\pi R_{p}^{2}\right)$$

# 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,
$$t_{f}=\frac{1}{g}\left(v_{0 z} \pm \sqrt{v_{0 z}^{2}-2 g h}\right) .$$

# 高级物理学1A|PHYS1131 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 Physics 1A being the lower of the two levels.Electricity and Magnetism: electrostatics, Gauss’s law, electric potential, capacitance and dielectrics, magnetic fields and magnetism, Ampere’s and Biot-Savart law, Faraday’s law, induction and inductance. Physical Optics: light, interference, diffraction, gratings and spectra, polarization. Introductory quantum theory and the wave nature of matter. Introductory solid state and semiconductor physics: simple energy band picture.

\begin{prob}

(a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where $E=0$ ) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.
(b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.

(a) Let the equilibrium of the test charge be stable. If test charge is in equilibrium and it is displaced from its position in any direction, then it experiences a force towards a null point, where the electric field is zero. All the field lines near the null point are directed inwards and towards the null point. There is a net inward flux through a closed surface around the null point. According to Gauss’s law, the flux of electric field through a surface, which is not enclosing any charge, is zero. Hence, the equilibrium of the test charge can be stable.
(b) Two charges of same magnitude and same sign are placed at some distance. The mid-point of the joining line of the two charges is the null point. When a test charged is displaced along the line joining charges, it experiences a restoring force. If it is displaced normal to the joining line of charges, then the net force takes it away from the null point. Hence, the charge is not stable because stability of equilibrium requires restoring force in all directions.

## PHYS1131 COURSE NOTES ：

What is the force between two small charged sphere having charges of $2 \times$ $10^{-7} \mathrm{C}$ and $3 \times 10^{-7} \mathrm{C}$ placed $30 \mathrm{~cm}$ apart in air?
Solution:
Given:
Charge on the first sphere, $\mathrm{Q}{1}=2 \times 10^{-7} \mathrm{C}$ Charge on the first sphere, $Q{2}=3 \times 10^{-7} \mathrm{C}$
Distance between the spheres is, $r=30 \mathrm{~cm}, \mathrm{r}=0.3 \mathrm{~m}$
The electrostatic force between two charges is given by the formula, $\mathrm{F}=\frac{\mathrm{kQ}{1} \mathrm{Q}{2}}{\mathrm{r}^{2}}$
Where $\mathrm{k}=\frac{1}{4 \pi \varepsilon}=9 \times 10^{9}$
$$\mathrm{F}=\frac{9 \times 10^{9} \times 2 \times 10^{-7} \times 3 \times 10^{-7}}{(0.3)^{2}}=6 \times 10^{-3} \mathrm{~N}$$
Hence, the force between the spheres is $6 \times 10^{-3} \mathrm{~N}$. Charges are of the same polarity so force between them will be repulsive in nature.