# 牛顿力学代写|NEWTONIAN MECHANICS MATH122 University of Liverpool Assignment

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## Instructions:

Newtonian mechanics, also known as classical mechanics, is a branch of physics that deals with the study of motion and its causes. It was founded by Sir Isaac Newton in the 17th century and is based on three laws known as Newton’s laws of motion.

Newton’s first law of motion states that an object at rest will remain at rest, and an object in motion will continue in motion with a constant velocity unless acted upon by an external force. This law is also known as the law of inertia.

Newton’s second law of motion states that the force acting on an object is equal to the mass of the object times its acceleration. Mathematically, F=ma, where F is the force, m is the mass, and a is the acceleration.

Newton’s third law of motion states that for every action, there is an equal and opposite reaction. This law implies that forces always occur in pairs.

In addition to these laws, Newtonian mechanics also includes the concepts of energy, work, and momentum. The laws of conservation of energy, work-energy theorem, and conservation of momentum are also important principles in this branch of physics.

Newtonian mechanics has been instrumental in the development of modern physics, and its principles are still widely used in many fields of science and engineering.

A rocket in zero gravitational field has a mass of $m_{r, i}=2.81 \times 10^7 \mathrm{~kg}$, which is the sum of the mass of the fuel $m_{f, i}=2.46 \times 10^7 \mathrm{~kg}$ and the dry mass of the rocket (empty of fuel) $m_{r, d} \equiv m_{r, i}-m_{f, i}=0.35 \times 10^7 \mathrm{~kg}$. The fuel is ejected at a speed $u=3000$ $\mathrm{m} / \mathrm{s}$ relative to the rocket. The total burn time is $510 \mathrm{~s}$ and the fuel is burned at a constant rate. (a) What is the final speed $v_f$ of the rocket in meters/second after all the fuel is burned assuming it starts from rest?

The total mass of the rocket decreases during the burn due to the ejection of fuel. The change in velocity of the rocket is given by the rocket equation:

$\Delta v=v_{f}-v_{i}=u \ln \left(\frac{m_{r, i}}{m_{r, f}}\right)$

where $v_i = 0$ is the initial velocity of the rocket, $v_f$ is the final velocity of the rocket, $m_{r,i}$ is the initial mass of the rocket, $m_{r,f}$ is the final mass of the rocket, and $u$ is the relative velocity of the ejected fuel. At the end of the burn, all the fuel has been ejected, so the final mass of the rocket is the dry mass $m_{r,d}$:

$m_{r, f}=m_{r, d}=0.35 \times 10^7 \mathrm{~kg}$

Thus,

$\Delta v=u \ln \left(\frac{m_{r, i}}{m_{r, d}}\right)=3000 \mathrm{~m} / \mathrm{s} \ln \left(\frac{2.81 \times 10^7 \mathrm{~kg}}{0.35 \times 10^7 \mathrm{~kg}}\right) \approx 7974.4 \mathrm{~m} / \mathrm{s}$

Therefore, the final speed of the rocket is $v_f = \Delta v = 7974.4$ m/s.

(b) Now suppose that the same rocket burns the fuel in two stages, expelling the fuel in each stage at the same relative speed $u=3000 \mathrm{~m} / \mathrm{s}$. In stage one, the available fuel to burn is $m_{f, 1, i}=2.03 \times 10^7 \mathrm{~kg}$ with burn time $150 \mathrm{~s}$. The total mass of the rocket after all the fuel in stage 1 is burned is $m_{r, 1, d}=m_{r, i}-m_{f, 1, i}=0.78 \times 10^7$ kg. What is the change in speed after stage one is complete?

In stage one, the initial mass of the rocket is $m_{r,1,i}=m_{r,i}=2.81 \times 10^7 \mathrm{~kg}$ and the initial mass of fuel is $m_{f,1,i}=2.03 \times 10^7 \mathrm{~kg}$. The mass of the rocket after all the fuel in stage 1 is burned is $m_{r,1,f}=m_{r,1,i}-m_{f,1,i}=0.78 \times 10^7$ kg. By the principle of conservation of momentum, the change in velocity of the rocket after stage one is complete is given by:

$\Delta v_1=u\ln \left(\frac{m_{r,1,i}}{m_{r,1,f}}\right)$

where $u=3000 \mathrm{~m}/\mathrm{s}$ is the relative speed of the ejected fuel and $m_{r,1,i}$ and $m_{r,1,f}$ are the initial and final masses of the rocket in stage one, respectively. Substituting the values:

$\Delta v_1=3000 \mathrm{~m}/\mathrm{s} \ln \left(\frac{2.81 \times 10^7 \mathrm{~kg}}{0.78 \times 10^7 \mathrm{~kg}}\right) \approx 3066 \mathrm{~m}/\mathrm{s}$

Therefore, the change in speed after stage one is complete is approximately $3066 \mathrm{~m}/\mathrm{s}$.

(c) Next, the empty fuel tank and accessories from stage one are disconnected from the rest of the rocket. These disconnected parts had a mass $m=1.4 \times 10^6 \mathrm{~kg}$, hence the remaining dry mass of the rocket is $m_{r, 2, d}=2.1 \times 10^6 \mathrm{~kg}$. All the remaining fuel with mass $m_{f, 2, i}=4.3 \times 10^6 \mathrm{~kg}$ is burned during stage 2 with burn time of $360 \mathrm{~s}$. What is the change in speed in meters/second after stage two is complete?

To calculate the change in speed after stage two, we can use the rocket equation, which relates the change in velocity of a rocket to the mass ratio of the rocket before and after the burn, and the effective exhaust velocity of the propellant.

The mass ratio of the rocket before and after stage two is:

$$\frac{m_{i}}{m_{f, 2}}=\frac{m_{r, i}+m_{f, i}+m}{m_{r, 2, d}}=\frac{2.81 \times 10^7 \mathrm{~kg}+2.46 \times 10^7 \mathrm{~kg}+1.4 \times 10^6 \mathrm{~kg}}{2.1 \times 10^6 \mathrm{~kg}}=22.95$$

where $m_{i}$ is the initial mass of the rocket (including all the fuel and the dry mass) and $m_{f,2}$ is the final mass of the rocket after stage two (including the dry mass and all the fuel burned in stage two).

The effective exhaust velocity $v_{e}$ is given by:

$$v_{e}=u \ln\frac{m_{i}}{m_{f,2}}=3000 \mathrm{~m/s} \ln\frac{2.81 \times 10^7 \mathrm{~kg}}{2.1 \times 10^6 \mathrm{~kg}+4.3 \times 10^6 \mathrm{~kg}}=2445.5 \mathrm{~m/s}$$

where $u$ is the speed of the ejected fuel relative to the rocket.

Now we can use the rocket equation:

$$\Delta v=v_{e} \ln\frac{m_{i}}{m_{f,2}}=2445.5 \mathrm{~m/s} \ln\frac{2.81 \times 10^7 \mathrm{~kg}}{2.1 \times 10^6 \mathrm{~kg}+4.3 \times 10^6 \mathrm{~kg}}=2434.4 \mathrm{~m/s}$$

Therefore, the change in speed after stage two is $\Delta v = 2434.4$ $\mathrm{m/s}$.

# 计算物理学代写|INTRODUCTION TO COMPUTATIONAL PHYSICS PHYS105 University of Liverpool Assignment

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Assignment-daixieTM为您提供利物浦大学University of Liverpool INTRODUCTION TO COMPUTATIONAL PHYSICS PHYS105计算物理学代写代考辅导服务！

## Instructions:

Computational physics is a subfield of physics that involves the use of computers to solve problems in physics that are too complex to be solved analytically. The use of computational techniques has become increasingly important in physics research and has led to many breakthroughs in our understanding of the physical world.

To get started with computational physics, it’s important to have a strong foundation in mathematics and programming. In terms of programming languages, Python is a popular choice for computational physics because of its ease of use and large user community.

There are many resources available online to help you learn computational physics. Some good places to start include:

• The Open Source Physics Project: This project provides a collection of open-source software tools and simulations for teaching and researching physics. They have a range of resources and materials available for learning computational physics, including tutorials and sample code.
• Coursera: Coursera offers a range of online courses in computational physics, including courses on Python programming and numerical methods for physics.
• GitHub: GitHub is a platform for collaborative software development, and there are many open-source repositories available on GitHub that contain code for solving physics problems. You can use these repositories to learn from existing code and contribute to the development of new code.
• Textbooks: There are many textbooks available on computational physics, including “Computational Physics” by Mark Newman and “An Introduction to Computational Physics” by Tao Pang.

I hope this information is helpful! Let me know if you have any other questions.

(a) Write a procedure that solves quadratic equations using the quadratic formula: It should take as arguments three numbers a, b, and c. It should print error messages if a is zero, or if the roots are complex. Otherwise it should print the two roots.

# a.) code for roots function
def roots(a, b, c):
#form of equation is a*x**2 + b*x + c = 0
#quadratic formula is   x = (-b + sqrt(b**2 - 4 * a * c))/(2 * a)
#                   or  x = (-b - sqrt(b**2 - 4 * a * c))/(2 * a)
#roots are complex when the discriminant (b**2 - 4*a*c) is negative
discriminant = b**2 - (4 * a * c)
if discriminant < 0:
return "Roots are complex"
return "x = "+str((-b + math.sqrt(discriminant)) / (2 * a))+" or x = "+\
str((-b - math.sqrt(discriminant)) / (2 * a))

# Test Cases (assertions are just to automate)
##print '\nTesting roots'
##
##print roots(1, 2, 1) #(x+1)^2: double root at x = -1
##assert roots(1, 2, 1) == 'x = -1.0 or x = -1.0'
##
##print roots(1, -2, -3) #(x+1)(x-3): roots at x = 3 or x = -1
##assert roots(1, -2, -3) == 'x = 3.0 or x = -1.0'
##
##print roots(2, 2, 2) #2x^2 + 2x + 2: complex roots
##assert roots(2, 2, 2) == 'Roots are complex'

(b) Modify your procedure to handle the case of complex roots.

# b.) same code, modified to handle complex roots
def roots(a, b, c):
discriminant = b**2 - (4 * a * c)
if discriminant < 0:
discriminant = discriminant + 0j
return "x = "+str((-b + discriminant**0.5) / (2 * a))+ " or x = " +\
str((-b - discriminant**0.5) / (2 * a)) # math.sqrt does not work

# Test Cases (assertions are just to automate)
##print '\nTesting roots'
##
##print roots(1, 2, 1) #(x+1)^2: double root at x = -1
##assert roots(1, 2, 1) == 'x = -1.0 or x = -1.0'
##
##print roots(1, -2, -3) #(x+1)(x-3): roots at x = 3 or x = -1
##assert roots(1, -2, -3) == 'x = 3.0 or x = -1.0'
##
##print roots(2, 2, 2) #2x^2 + 2x + 2: complex roots
##assert roots(2, 2, 2) == 'x = (-0.5+0.866025403784j) or x = (-0.5-0.866025403784j)'

###########################################
## 2.) procedure for evaluating polynomials
###########################################
def eval_poly(x, coeffs):
total=0 #will keep a running total of the sum
coeffs.reverse() # to put the low order coeffs first
for i in range(len(coeffs)):
total+=coeffs[i]*(x**i) #add the curent term to the total

# Test cases
##print '/nTesting Polynomial'
##
##print eval_poly(1,[1,2,3])
##assert eval_poly(1,[1,2,3]) == 6
##
##print eval_poly(2,[1,2,3,4])
##assert eval_poly(2,[1,2,3,4]) == 26


Write a procedure that evaluates polynomials. It should take two arguments. The first is a number $x$. The second is a list of of coefficients ordered from highest to lowest:
$$a_n, a_{n-1}, \ldots, a_2, a_1, a_0$$
Your procedure should return the value of the polynomial evaluated at $x$ :
$$a_n x^n+a_{n-1} x^{n-1}+\ldots+a_2 x+a_1 x+a_0$$

) procedure for evaluating polynomials
###########################################
def eval_poly(x, coeffs):
total=0 #will keep a running total of the sum
coeffs.reverse() # to put the low order coeffs first
for i in range(len(coeffs)):
total+=coeffs[i]*(x**i) #add the curent term to the total

# Test cases
##print '/nTesting Polynomial'
##
##print eval_poly(1,[1,2,3])
##assert eval_poly(1,[1,2,3]) == 6
##
##print eval_poly(2,[1,2,3,4])
##assert eval_poly(2,[1,2,3,4]) == 26

###########################################
## 3.) procedure to make change
###########################################
def make_change(cost, paid):
change=paid-cost #calculate the change due
bills={20:0,10:0,5:0,2:0,1:0} #dictionary that maps each bill to the amount of it
bill_list=bills.keys() #bill_list is a list of the available bills
bill_list.sort() #sort it in ascending order
bill_list.reverse() #now reverse it to be in descending order
#this is to make sure you get the smallest number of bills
for bill in bill_list:
while change >=bill:
bills[bill]+=1 #increment the amount of that bill
change-=bill #decrease the change by that bill

print "Change is:"
for bill in bill_list:
if bills[bill] == 1:
print bills[bill], bill, "dollar bill"
elif bills[bill] > 1:
print bills[bill], bill, "dollar bills"

#return bills #return the dictionary with amounts of each bill needed

make_change(1, 6)
make_change(4, 109)

# 天体物理学代写|ASTRONOMY 1 ASTRO1001 University of Glasgow AssignmentONLINE TUTOR

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Assignment-daixieTM为您提供格拉斯哥大学University of Glasgow ASTRONOMY 1 ASTRO1001天体物理学代写代考辅导服务！

## Instructions:

Astronomy is the scientific study of celestial objects and phenomena beyond the Earth’s atmosphere, including planets, stars, galaxies, and other celestial bodies. It is a fascinating field that has captured the attention and imagination of people for thousands of years.

There are several subfields within astronomy, including solar system physics, positional astronomy, and dynamical astronomy.

Solar system physics is the study of the physical properties of the planets, moons, asteroids, and comets in our solar system. This includes their composition, structure, atmospheres, and orbits. Solar system physics also investigates the formation and evolution of the solar system, as well as the interactions between the various bodies.

Positional astronomy is concerned with measuring the positions, distances, and motions of celestial objects. This involves the use of telescopes, cameras, and other instruments to gather data on the positions of stars, galaxies, and other objects. Positional astronomy also includes the study of celestial coordinate systems, which are used to locate objects in the sky.

Dynamical astronomy is the study of the motions of celestial objects, including their orbits and gravitational interactions. This field includes the study of celestial mechanics, which deals with the mathematical description of the motions of objects in space. Dynamical astronomy also investigates the formation and evolution of galaxies, and the interactions between galaxies and other celestial bodies.

In conclusion, astronomy is a vast and complex field that encompasses many different areas of study. Solar system physics, positional astronomy, and dynamical astronomy are just a few of the many subfields within this fascinating discipline. The study of astronomy has helped us to understand the universe around us and has inspired us to ask even more questions about the nature of the cosmos.

a. A globular cluster has $10^6$ stars each of apparent magnitude +8 . What is the combined apparent magnitude of the entire cluster?

$\begin{gathered}+8=-2.5 \log \left(F / F_0\right) \ F=6.3 \times 10^{-4} F_0 \ F_{\text {cluster }}=10^6 \times 6.3 \times 10^{-4} F_0=630 F_0 \ m_{\text {cluster }}=-2.5 \log (630)=-7\end{gathered}$

b. Find the distance modulus to the Andromeda galaxy (M31). Take the distance to Andromeda to be $750 \mathrm{kpc}$.

$\mathrm{DM}=5 \log \left(\frac{d}{10 \mathrm{pc}}\right)=5 \log (75,000)=24.4$

c. An eclipsing binary consists of two stars of different radii and effective temperatures. Star 1 has radius $R_1$ and $T_1$, and Star 2 has $R_2=0.5 R_1$ and $T_2=2 T_1$. Find the change in bolometric magnitude of the binary, $\Delta m_{\text {bol }}$, when the smaller star is behind the larger star. (Consider only bolometric magnitudes so you don’t have to worry about color differences.)

$$\begin{gathered} \mathcal{F}{1 \& 2}=4 \pi \sigma\left(T_1^4 R_1^2+T_2^4 R_2^2\right) \ \mathcal{F}{\text {eclipse }}=4 \pi \sigma T_1^4 R_1^2 \ \Delta m=-2.5 \log \left(\frac{\mathcal{F}{1 \& 2}}{\mathcal{F}{\text {eclipse }}}\right) \ \Delta m=-2.5 \log \left(1+\frac{T_2^4 R_2^2}{T_1^4 R_1^2}\right) \ \Delta m=-2.5 \log \left(1+\frac{16}{4}\right)=-1.75 \end{gathered}$$
So, the binary is 1.75 magnitudes brighter out of eclipse than when star 2 is behind star 1 .

# 物理代写|PHYSICS 1 PHYS1001 University of Glasgow Assignment

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Assignment-daixieTM为您提供格拉斯哥大学University of Glasgow PHYSICS 1 PHYS1001物理代写代考辅导服务！

## Instructions:

Physics is a branch of science that deals with the study of matter, energy, and their interactions. It seeks to understand the fundamental laws of nature and the behavior of the universe at the most basic level. Physicists study everything from the tiniest subatomic particles to the largest structures in the universe, such as galaxies and black holes. They use mathematical models and experimental methods to develop theories and test them against observations. Physics plays a critical role in many areas of modern technology, including electronics, communications, and medicine. It has also contributed significantly to our understanding of the natural world and the universe we live in.

The line integral of a scalar function $f(x, y, z)$ along a path $C$ is defined as
$$\int_C f(x, y, z) d s=\lim {\substack{N \rightarrow \infty \ \Delta s_i \rightarrow 0}} \sum{i=1}^N f\left(x_i, y_i, z_i\right) \Delta s_i$$
where $C$ has been subdivided into $N$ segments, each with a length $\Delta s_i$. To evaluate the line integral, it is convenient to parameterize $C$ in terms of the arc length parameter $s$. With $x=x(s)$, $y=y(s)$ and $z=z(s)$, the above line integral can be rewritten as an ordinary definite integral:
$$\int_C f(x, y, z) d s=\int_{s_1}^{s_2} f[x(s), y(s), z(s)] d s$$

Yes, that is correct.

When we parameterize the curve $C$ using the arc length parameter $s$, we can express the coordinates of any point on the curve as functions of $s$ as $x(s)$, $y(s)$, and $z(s)$.

Using this parameterization, we can express the line integral as an ordinary definite integral over $s$ as shown above. The integrand $f[x(s), y(s), z(s)]$ represents the value of the scalar function $f$ evaluated at the point on the curve corresponding to the arc length parameter $s$.

The limit as $N\to\infty$ and $\Delta s_i\to 0$ in the definition of the line integral is necessary because we want to calculate the exact value of the integral, which involves summing up infinitely many infinitesimal contributions along the path $C$.

A solid cylinder of length $L$ and radius $R$, with $L \gg R$, is uniformly filled with a total charge $Q$. a. What is the volume charge density $\rho$ ? Check units!

a. The volume charge density $\rho$ is defined as the charge $Q$ per unit volume. In this case, the total charge $Q$ is uniformly distributed throughout the volume of the cylinder. The volume of a cylinder is given by $V=\pi R^2L$. Therefore, the volume charge density is:

$$\rho = \frac{Q}{V} = \frac{Q}{\pi R^2 L}$$

Checking units: $[Q]=C$, $[R]=m$, and $[L]=m$, so $[\rho]=\frac{C}{m^3}$.

b. Suppose you go very far away from the cylinder to a distance much greater than $R$. The cylinder now looks like a line of charge. What is the linear charge density $\lambda$ of that apparent line of charge? Check units!

b. When we are far away from the cylinder, it appears as a line of charge with linear charge density $\lambda$. The linear charge density is defined as the charge per unit length. We can find $\lambda$ by considering an element of length $dl$ on the cylinder. The charge on this element is $dq=\rho Adl$, where $A$ is the cross-sectional area of the cylinder (which is just a circle of radius $R$). The element of length $dl$ subtends an angle of $d\theta=\frac{dl}{R}$ at the observation point, and the distance from the element to the observation point is $r\gg R$. Therefore, the contribution to the electric field from this element is:

$$dE=\frac{1}{4\pi\epsilon_0}\frac{dq}{r^2}=\frac{1}{4\pi\epsilon_0}\frac{\rho Adl}{r^2}$$

The total electric field at the observation point is obtained by integrating over the entire length of the cylinder:

$$E=\int_{-L/2}^{L/2} dE = \int_{-L/2}^{L/2} \frac{1}{4\pi\epsilon_0}\frac{\rho A}{r^2}dl=\frac{\rho A}{4\pi\epsilon_0 r^2}\int_{-L/2}^{L/2}dl$$

The integral gives the total length of the cylinder, which is just $L$. Therefore, we have:

$$E=\frac{\rho AL}{4\pi\epsilon_0 r^2}$$

On the other hand, we know that the electric field due to a line of charge with linear charge density $\lambda$ is given by:

$$E=\frac{\lambda}{2\pi\epsilon_0 r}$$

Comparing the two expressions for $E$, we get:

$$\frac{\lambda}{2\pi\epsilon_0 r}=\frac{\rho AL}{4\pi\epsilon_0 r^2}$$

Solving for $\lambda$, we obtain:

$$\lambda=\frac{\rho AL}{2r}=\frac{Q}{2\pi R L}\qquad\text{(since }A=\pi R^2\text{)}$$

Checking units: $[Q]=C$, $[R]=m$, and $[L]=m$, so $[\lambda]=\frac{C}{m}$.

# 机械学和物质代写|Mechanics and Matter PX1121 Cardiff University Assignment

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Assignment-daixieTM为您提供卡迪夫大学Cardiff University PX1121 Mechanics and Matter代写代考辅导服务！

## Instructions:

Mechanics is the branch of physics that deals with the motion of objects under the influence of forces. It is concerned with describing and predicting the behavior of physical systems, including both macroscopic objects like cars and airplanes, as well as microscopic particles like atoms and molecules.

Matter, on the other hand, refers to anything that has mass and takes up space. It is the basic building block of the universe, and includes everything from tiny subatomic particles to massive planets and stars.

Mechanics and matter are intimately connected, as the behavior of matter is governed by the laws of mechanics. For example, the motion of a ball thrown into the air can be described using the principles of mechanics, as can the behavior of atoms and molecules in a gas or liquid.

Understanding the relationship between mechanics and matter is crucial for many areas of science and technology, including engineering, materials science, and many others.

Surface tension: Thermodynamic properties of the interface between two phases are described by a state function called the surface tension $\mathcal{S}$. It is defined in terms of the work required to increase the surface area by an amount $d A$ through $d W=\mathcal{S} d A$.
(a) By considering the work done against surface tension in an infinitesimal change in radius, show that the pressure inside a spherical drop of water of radius $R$ is larger than outside pressure by $2 \mathcal{S} / R$. What is the air pressure inside a soap bubble of radius $R$ ?

The work done by a water droplet on the outside world, needed to increase the radius from $R$ to $R+\Delta R$ is
$$\Delta W=\left(P-P_o\right) \cdot 4 \pi R^2 \cdot \Delta R$$
where $P$ is the pressure inside the drop and $P_o$ is the atmospheric pressure. In equilibrium, this should be equal to the increase in the surface energy $\mathcal{S} \Delta A=\mathcal{S} \cdot 8 \pi R \cdot \Delta R$, where $\mathcal{S}$ is the surface tension, and
$$\Delta W_{\text {total }}=0, \Longrightarrow \Delta W_{\text {pressure }}=-\Delta W_{\text {surface }},$$
resulting in
$$\left(P-P_o\right) \cdot 4 \pi R^2 \cdot \Delta R=\mathcal{S} \cdot 8 \pi R \cdot \Delta R, \quad \Longrightarrow \quad\left(P-P_o\right)=\frac{2 \mathcal{S}}{R}$$
In a soap bubble, there are two air-soap surfaces with almost equal radii of curvatures, and
$$P_{\text {film }}-P_o=P_{\text {interior }}-P_{\text {film }}=\frac{2 \mathcal{S}}{R}$$
$$P_{\text {interior }}-P_o=\frac{4 \mathcal{S}}{R}$$
Hence, the air pressure inside the bubble is larger than atmospheric pressure by $4 \mathcal{S} / R$.

(b) A water droplet condenses on a solid surface. There are three surface tensions involved $\mathcal{S}{a w}, \mathcal{S}{s w}$, and $\mathcal{S}_{s a}$, where $a, s$, and $w$ refer to air, solid and water respectively. Calculate the angle of contact, and find the condition for the appearance of a water film (complete wetting).

When steam condenses on a solid surface, water either forms a droplet, or spreads on the surface. There are two ways to consider this problem:
Method 1: Energy associated with the interfaces
In equilibrium, the total energy associated with the three interfaces should be minimum, and therefore
$$d E=S_{a w} d A_{a w}+S_{a s} d A_{a s}+S_{w s} d A_{w s}=0 .$$

Since the total surface area of the solid is constant,
$$d A_{a s}+d A_{w s}=0 .$$
From geometrical considerations (see proof below), we obtain
$$d A_{w s} \cos \theta=d A_{a w} .$$
From these equations, we obtain
$$d E=\left(S_{a w} \cos \theta-S_{a s}+S_{w s}\right) d A_{w s}=0, \quad \Longrightarrow \quad \cos \theta=\frac{S_{a s}-S_{w s}}{S_{a w}} .$$
Proof of $d A_{w s} \cos \theta=d A_{a w}$ : Consider a droplet which is part of a sphere of radius $R$, which is cut by the substrate at an angle $\theta$. The areas of the involved surfaces are
$$A_{w s}=\pi(R \sin \theta)^2, \quad \text { and } \quad A_{a w}=2 \pi R^2(1-\cos \theta) .$$
Let us consider a small change in shape, accompanied by changes in $R$ and $\theta$. These variations should preserve the volume of water, i.e. constrained by
$$V=\frac{\pi R^3}{3}\left(\cos ^3 \theta-3 \cos \theta+2\right)$$
Introducing $x=\cos \theta$, we can re-write the above results as
\left{\begin{aligned} A_{w s} & =\pi R^2\left(1-x^2\right), \ A_{a w} & =2 \pi R^2(1-x), \ V & =\frac{\pi R^3}{3}\left(x^3-3 x+2\right) . \end{aligned}\right.

The variations of these quantities are then obtained from
\left{\begin{aligned} d A_{w s} & =2 \pi R\left[\frac{d R}{d x}\left(1-x^2\right)-R x\right] d x \ d A_{a w} & =2 \pi R\left[2 \frac{d R}{d x}(1-x)-R\right] d x \ d V & =\pi R^2\left[\frac{d R}{d x}\left(x^3-3 x+2\right)+R\left(x^2-x\right)\right] d x=0 . \end{aligned}\right.
From the last equation, we conclude
$$\frac{1}{R} \frac{d R}{d x}=-\frac{x^2-x}{x^3-3 x+2}=-\frac{x+1}{(x-1)(x+2)}$$

Substituting for $d R / d x$ gives,
$$d A_{w s}=2 \pi R^2 \frac{d x}{x+2}, \quad \text { and } \quad d A_{a w}=2 \pi R^2 \frac{x \cdot d x}{x+2},$$
resulting in the required result of
$$d A_{a w}=x \cdot d A_{w s}=d A_{w s} \cos \theta .$$
Method 2: Balancing forces on the contact line
Another way to interpret the result is to consider the force balance of the equilibrium surface tension on the contact line. There are four forces acting on the line: (1) the surface tension at the water-gas interface, (2) the surface tension at the solid-water interface, (3) the surface tension at the gas-solid interface, and (4) the force downward by solid-contact line interaction. The last force ensures that the contact line stays on the solid surface, and is downward since the contact line is allowed to move only horizontally without friction. These forces should cancel along both the $y$-direction $x$-directions. The latter gives the condition for the contact angle known as Young’s equation,
$$\mathcal{S}{a s}=\mathcal{S}{a w} \cdot \cos \theta+\mathcal{S}{w s}, \Longrightarrow \cos \theta=\frac{\mathcal{S}{a s}-\mathcal{S}{w s}}{\mathcal{S}{a w}} .$$
The critical condition for the complete wetting occurs when $\theta=0$, or $\cos \theta=1$, i.e. for
$$\cos \theta_C=\frac{\mathcal{S}{a s}-\mathcal{S}{w s}}{\mathcal{S}{a w}}=1$$ Complete wetting of the substrate thus occurs whenever $$\mathcal{S}{a w} \leq \mathcal{S}{a s}-\mathcal{S}{w s} .$$

(c) In the realm of “large” bodies gravity is the dominant force, while at “small” distances surface tension effects are all important. At room temperature, the surface tension of water is $\mathcal{S}_o \approx 7 \times 10^{-2} \mathrm{Nm}^{-1}$. Estimate the typical length-scale that separates “large” and “small” behaviors. Give a couple of examples for where this length-scale is important.

(c) In the realm of “large” bodies gravity is the dominant force, while at “small” distances surface tension effects are all important. At room temperature, the surface tension of water is $\mathcal{S}_o \approx 7 \times 10^{-2} \mathrm{Nm}^{-1}$. Estimate the typical length-scale that separates “large” and “small” behaviors. Give a couple of examples for where this length-scale is important.

• Typical length scales at which the surface tension effects become significant are given by the condition that the forces exerted by surface tension and relevant pressures become comparable, or by the condition that the surface energy is comparable to the other energy changes of interest.

Example 1: Size of water drops not much deformed on a non-wetting surface. This is given by equalizing the surface energy and the gravitational energy,
$$S \cdot 4 \pi R^2 \approx m g R=\rho V g R=\frac{4 \pi}{3} R^4 g$$

$$R \approx \sqrt{\frac{3 S}{\rho g}} \approx \sqrt{\frac{3 \cdot 7 \times 10^{-2} \mathrm{~N} / \mathrm{m}}{10^3 \mathrm{~kg} / \mathrm{m}^3 \times 10 \mathrm{~m} / \mathrm{s}^2}} \approx 1.5 \times 10^{-3} \mathrm{~m}=1.5 \mathrm{~mm} .$$
Example 2: Swelling of spherical gels in a saturated vapor: Osmotic pressure of the gel (about $1 \mathrm{~atm})=$ surface tension of water, gives
$$\pi_{g e l} \approx \frac{N}{V} k_B T \approx \frac{2 S}{R}$$
where $N$ is the number of counter ions within the gel. Thus,
$$R \approx\left(\frac{2 \times 7 \times 10^{-2} \mathrm{~N} / \mathrm{m}}{10^5 \mathrm{~N} / \mathrm{m}^2}\right) \approx 10^{-6} \mathrm{~m} .$$

# 电、磁和波代写|Electricity, Magnetism and Waves PX1221 Cardiff University Assignment

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Assignment-daixieTM为您提供卡迪夫大学Cardiff University PX1221 Electricity, Magnetism and Waves电、磁和波代写代考辅导服务！

## Instructions:

Electricity, magnetism, and waves are three interconnected branches of physics that deal with the behavior and properties of electromagnetic fields and phenomena.

Electricity refers to the flow of electric charge, usually carried by electrons in a conductor, and is characterized by voltage, current, and resistance. It is essential to many aspects of modern life, from powering our homes and electronic devices to driving industry and transportation.

Magnetism refers to the behavior of magnets and magnetic fields, which are produced by moving electric charges or certain materials called ferromagnetic materials. Magnetic fields are important in many applications, including electric motors, generators, and medical imaging.

Waves refer to the propagation of energy through space, which can be in the form of electromagnetic waves or mechanical waves like sound waves. Electromagnetic waves, such as radio waves, microwaves, light waves, and X-rays, are characterized by their wavelength, frequency, and amplitude and play a vital role in communication, broadcasting, and imaging.

The study of electricity, magnetism, and waves is fundamental to understanding many aspects of the physical world, from the behavior of atoms and molecules to the functioning of modern technology.

(a) Find the total charge $Q$ on the rectangular surface of length $a$ ( $x$ direction from $x=0$ to $x=a$ ) and width $b(y$ direction from $y=0$ to $y=b)$, if the charge density is $\sigma(x, y)=k x y$, where $k$ is a constant.

To find the total charge $Q$ on the rectangular surface, we need to integrate the charge density $\sigma(x, y)$ over the surface area:

$$Q = \int_{0}^{b} \int_{0}^{a} \sigma(x, y) dx dy$$

Substituting $\sigma(x,y)=kxy$ into the above expression and performing the integrals yields:

\begin{align*} Q &= \int_{0}^{b} \int_{0}^{a} k x y dx dy \ &= k \int_{0}^{b} \left[\frac{x^2}{2} y \right]{x=0}^{x=a} dy \ &= k \int{0}^{b} \frac{a^2}{2} y dy \ &= \frac{k a^2}{2} \left[\frac{y^2}{2} \right]_{y=0}^{y=b} \ &= \frac{k a^2 b^2}{4} \end{align*}

Therefore, the total charge on the rectangular surface is $\boxed{\frac{k a^2 b^2}{4}}$.

(b) Find the total charge on a circular plate of radius $R$ if the charge distribution is $\sigma(r, \theta)=k r(1-\sin \theta)$.

To find the total charge on the circular plate of radius $R$ with charge distribution $\sigma(r,\theta) = kr(1-\sin\theta)$, we need to integrate the charge density over the entire area of the plate.

The charge density $\sigma$ is a function of $r$ and $\theta$, where $r$ is the radial distance from the center of the plate and $\theta$ is the angle measured from some reference direction. Since the plate is circular, we can integrate over $r$ and $\theta$ as follows:

\begin{align*} Q &= \int_{0}^{2\pi}\int_{0}^{R} \sigma(r,\theta)r,dr,d\theta \ &= \int_{0}^{2\pi}\int_{0}^{R} k r^2 (1-\sin\theta),dr,d\theta\ &= k \int_{0}^{2\pi}\left[\frac{r^3}{3} – r^2\cos\theta\right]{r=0}^{r=R},d\theta\ &= k \int{0}^{2\pi}\left(\frac{R^3}{3} – R^2\cos\theta\right),d\theta\ &= k\left[\frac{R^3}{3}\theta – R^2\sin\theta\right]_{\theta=0}^{\theta=2\pi}\ &= k\left(\frac{2}{3}\pi R^3 – 0\right)\ &= \frac{2}{3}\pi k R^3 \end{align*}

Therefore, the total charge on the circular plate is $\frac{2}{3}\pi k R^3$.

Use Gauss’s Law to find the direction and magnitude of the electric field in the between the inner and outer cylinders $(a<r<b)$. Express your answer in terms of the total charge $Q$ on the inner cylinder cylinder, the radii $a$ and $b$, the height $l$, and any other constants which you may find necessary.

To apply Gauss’s Law, we need to choose a closed surface that encloses the region of interest, which in this case is the region between the inner and outer cylinders. A convenient choice is a cylindrical Gaussian surface of radius $r$ and height $l$, centered on the axis of the cylinders. The electric field will be perpendicular to the surface at every point, so the flux of the electric field through the surface is simply the product of the magnitude of the field and the area of the curved surface:

$$\Phi_E = \oint_S \vec{E} \cdot d\vec{A} = E(r) 2\pi rl$$

where $E(r)$ is the magnitude of the electric field at radius $r$ and $d\vec{A}$ is a differential area element of the surface.

By Gauss’s Law, the flux through the surface is equal to the enclosed charge divided by the permittivity of free space:

$$\Phi_E = \frac{Q_{enc}}{\epsilon_0}$$

where $Q_{enc}$ is the charge enclosed by the Gaussian surface. Since the inner cylinder is the only source of charge inside the surface, the enclosed charge is simply $Q_{enc}=Q$.

Setting these two equations equal to each other, we have

$$E(r) 2\pi rl = \frac{Q}{\epsilon_0}$$

Solving for $E(r)$, we obtain:

$$E(r) = \frac{Q}{2\pi\epsilon_0 rl}$$

The direction of the electric field is radial, i.e., pointing outward from the axis of the cylinders if $Q$ is positive and inward if $Q$ is negative.

Therefore, the magnitude of the electric field between the cylinders is given by:

$$\boxed{E(r) = \frac{Q}{2\pi\epsilon_0 rl}}$$

# 动态环境代写|World of Dynamic Environments EA1300 Cardiff University Assignment

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Assignment-daixieTM为您提供卡迪夫大学Cardiff University EA1300 World of Dynamic Environments动态环境代写代考辅导服务！

## Instructions:

Dynamic environments refer to environments that are constantly changing and evolving over time. These changes can be driven by various factors, such as the behavior of agents within the environment, external events, or the passage of time.

Examples of dynamic environments include weather systems, traffic patterns, and financial markets. In these environments, the state of the system can change rapidly and unpredictably, requiring agents to adapt and adjust their strategies in real-time.

Dynamic environments are often modeled using dynamic systems theory, which provides a mathematical framework for understanding how systems change over time. This theory can be used to develop models and algorithms that can help agents navigate dynamic environments and make decisions in the face of uncertainty.

Machine learning techniques such as reinforcement learning, which involves training agents to learn optimal actions through trial and error, can be particularly effective in dynamic environments. By continually updating their policies based on new information, these agents can learn to adapt to changing conditions and optimize their behavior over time

Show that
$$\rho \frac{D}{D t}\left(\frac{q_i q_i}{2}\right)=\rho f_i q_i+\frac{\partial\left(\sigma_{i j} q_i\right)}{\partial x_j}-\sigma_{i j} \frac{\partial q_i}{\partial x_j}$$
where $\sigma_{i j}$ is the viscous stress tensor.

Starting from the left-hand side of the equation, we have:

$\rho \frac{D}{D t}\left(\frac{q_i q_i}{2}\right)=\rho \frac{d}{d t}\left(\frac{q_i q_i}{2}\right)+\rho \frac{\partial}{\partial x_j}\left(u_j \frac{q_i q_i}{2}\right)$

where we have used the material derivative and the fact that $\rho u_i$ is the momentum density, which is related to the velocity by $u_i=q_i/\rho$. Now we can expand the first term on the right-hand side using the chain rule:

\begin{aligned} \rho \frac{d}{d t}\left(\frac{q_i q_i}{2}\right) & =\rho \frac{\partial}{\partial t}\left(\frac{q_i q_i}{2}\right)+\rho u_j \frac{\partial}{\partial x_j}\left(\frac{q_i q_i}{2}\right) \ & =\rho \frac{\partial}{\partial t}\left(\frac{q_i q_i}{2}\right)+\rho q_i \frac{\partial u_j}{\partial x_j}+\rho u_j q_i \frac{\partial}{\partial x_j}\left(q_i\right) \ & =\rho \frac{\partial}{\partial t}\left(\frac{q_i q_i}{2}\right)+\rho q_i f_i+\rho u_j q_i \frac{\partial}{\partial x_j}\left(q_i\right)\end{aligned}

where we have used the continuity equation $\partial \rho/\partial t + \partial(\rho u_j)/\partial x_j = 0$ and the equation of motion $\rho D u_i/Dt = \rho f_i – \partial \sigma_{ij}/\partial x_j$ to simplify the second term. Using the identity $\partial (q_i^2)/\partial x_j=2q_i\partial q_i/\partial x_j$ and the Einstein summation convention, we can rewrite the last term as $\partial(q_i q_j)/\partial x_j=q_i f_i+\partial(\sigma_{ij} q_i)/\partial x_j-\sigma_{ij}\partial q_i/\partial x_j$. Putting everything together, we get:

$\rho \frac{D}{D t}\left(\frac{q_i q_i}{2}\right)=\rho f_i q_i+\frac{\partial\left(\sigma_{i j} q_i\right)}{\partial x_j}-\sigma_{i j} \frac{\partial q_i}{\partial x_j}$

as desired.

Derive the explicit expression for the last term
$$\Phi=\sigma_{i j} \frac{\partial q_i}{\partial x_j}$$
for a two dimensionnal flow in term of $u, v$ and $x, y$ and comment on the sign of $\Phi$.

In two-dimensional flow, the stress tensor $\sigma_{ij}$ has only three non-zero components: $\sigma_{xx}$, $\sigma_{yy}$, and $\sigma_{xy}=\sigma_{yx}$. Therefore, we can write

$\Phi=\sigma_{x x} \frac{\partial q}{\partial x}+\sigma_{y y} \frac{\partial p}{\partial y}+2 \sigma_{x y} \frac{\partial q}{\partial y}$

where $q$ and $p$ are the velocity potential and stream function, respectively. In two-dimensional flow, the velocity components can be written as $u=\frac{\partial \phi}{\partial y}$ and $v=-\frac{\partial \phi}{\partial x}$, where $\phi=q+ip$ is the complex potential.

Using the Cauchy-Riemann equations, we can write $\frac{\partial q}{\partial x}=\frac{\partial p}{\partial y}$ and $\frac{\partial q}{\partial y}=-\frac{\partial p}{\partial x}$. Substituting these relations and using the fact that $\sigma_{xx}=2\mu \frac{\partial u}{\partial x}$, $\sigma_{yy}=2\mu \frac{\partial v}{\partial y}$, and $\sigma_{xy}=\sigma_{yx}= \mu \left(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}\right)$ (where $\mu$ is the dynamic viscosity), we get

$\Phi=2 \mu\left(\frac{\partial u}{\partial x} \frac{\partial p}{\partial y}+\frac{\partial v}{\partial y} \frac{\partial p}{\partial x}\right)+2 \mu\left(\frac{\partial u}{\partial y} \frac{\partial q}{\partial y}-\frac{\partial v}{\partial x} \frac{\partial q}{\partial y}\right)$

Using the definition of the complex potential, we can write

$\frac{\partial u}{\partial x}=\frac{\partial}{\partial x}\left(\frac{\partial \phi}{\partial y}\right)=\frac{\partial^2 \phi}{\partial x \partial y}, \quad \frac{\partial u}{\partial y}=-\frac{\partial}{\partial y}\left(\frac{\partial \phi}{\partial x}\right)=-\frac{\partial^2 \phi}{\partial x \partial y}$

and

$\frac{\partial v}{\partial x}=-\frac{\partial}{\partial x}\left(\frac{\partial \phi}{\partial x}\right)=-\frac{\partial^2 \phi}{\partial x^2}, \quad \frac{\partial v}{\partial y}=\frac{\partial}{\partial y}\left(\frac{\partial \phi}{\partial y}\right)=\frac{\partial^2 \phi}{\partial y^2}$

For long-scale slow motion show that mass conservation requires
$$\frac{\partial h}{\partial t}+\frac{\partial q}{\partial x}=0$$
where $q$ is the discharge rate.
$$q=\int_0^h u(x, y, t) d y$$
Assume the profile obtained for uniform flow with $h$ and $h_o$ depending on $x, t$. Obtain a set of equations for $h(x, t)$ and $h_o(x, t)$.

The control volume is defined by two cross-sectional areas, $A_1$ and $A_2$, and a length, $\Delta x$. The volume of fluid entering the control volume at $x$ per unit time is $q(x,t)$, and the volume leaving the control volume at $x+\Delta x$ per unit time is $q(x+\Delta x,t)$. The change in volume of the fluid within the control volume per unit time is $\frac{\partial h}{\partial t} \Delta x$, where $h(x,t)$ is the height of the fluid at $x$. By the principle of conservation of mass, we have:

$\frac{\partial h}{\partial t} \Delta x+q(x, t)-q(x+\Delta x, t)=0$

Dividing by $\Delta x$ and taking the limit as $\Delta x \rightarrow 0$, we obtain:

$\frac{\partial h}{\partial t}+\frac{\partial q}{\partial x}=0$

where $q(x,t)$ is the discharge rate at $x$ and time $t$, defined as:

$q(x, t)=\int_0^{h(x, t)} u(x, y, t) d y$

where $u(x,y,t)$ is the velocity profile of the fluid.

For the case of uniform flow, the velocity profile is constant in the $y$ direction, and we have:

$u(x, y, t)=\frac{Q}{h(x, t)}$

where $Q$ is the discharge per unit width of the channel. Substituting this into the definition of $q(x,t)$, we have:

$q(x, t)=\int_0^{h(x, t)} \frac{Q}{h(x, t)} d y=Q$

Therefore, for uniform flow, the discharge rate is constant along the channel. Substituting this into the mass conservation equation, we obtain:

$\frac{\partial h}{\partial t}+\frac{\partial Q}{\partial x}=0$

To obtain a set of equations for $h(x,t)$ and $h_o(x,t)$, we need an additional equation that relates the depth of the fluid to the width of the channel. For a rectangular channel of width $b$, the area of the cross-section is $A(x,t) = b h(x,t)$, and the wetted perimeter is $P(x,t) = b + 2h(x,t)$. Using these relations, we can derive the equation of motion for $h(x,t)$ by applying the principle of conservation of momentum to the control volume shown in the figure above.

Assuming the flow is steady and inviscid, the momentum equation reduces to:

$\frac{\partial}{\partial x}\left(\frac{h^2}{2}+g h+\frac{Q^2}{2 g b^2}\right)=0$

where $g$ is the acceleration due to gravity. This equation can be integrated to obtain:

$\frac{h^2}{2}+g h+\frac{Q^2}{2 g b^2}=C(x)$