Assignment-daixieTM为您提供格拉斯哥大学University of Glasgow PHYSICS 1 PHYS1001物理代写代考和辅导服务！
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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.
物理代写|PHYSICS 1 PHYS1001 Cardiff University Assignment
问题 1.
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: $$ \int_C f(x, y, z) d s=\int_{s_1}^{s_2} f[x(s), y(s), z(s)] d s $$
证明 .
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 $C$.
问题 2.
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!
证明 .
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 $[\rho]=\frac{C}{m^3}$.
问题 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!
证明 .
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:
Assignment-daixieTM为您提供卡迪夫大学Cardiff University PX1121 Mechanics and Matter代写代考和辅导服务！
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.
机械学和物质代写|Mechanics and Matter PX1121 Cardiff University Assignment
问题 1.
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 $R$ ?
证明 .
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 $$ P_{\text {film }}-P_o=P_{\text {interior }}-P_{\text {film }}=\frac{2 \mathcal{S}}{R} $$ leading to $$ P_{\text {interior }}-P_o=\frac{4 \mathcal{S}}{R} $$ Hence, the air pressure inside the bubble is larger than atmospheric pressure by $4 \mathcal{S} / R$.
问题 2.
(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).
证明 .
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 $$ \mathcal{S}{a w} \leq \mathcal{S}{a s}-\mathcal{S}{w s} . $$
问题 3.
(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.
证明 .
(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, $$ S \cdot 4 \pi R^2 \approx m g R=\rho V g R=\frac{4 \pi}{3} R^4 g $$
leading to $$ R \approx \sqrt{\frac{3 S}{\rho g}} \approx \sqrt{\frac{3 \cdot 7 \times 10^{-2} \mathrm{~N} / \mathrm{m}}{10^3 \mathrm{~kg} / \mathrm{m}^3 \times 10 \mathrm{~m} / \mathrm{s}^2}} \approx 1.5 \times 10^{-3} \mathrm{~m}=1.5 \mathrm{~mm} . $$ 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, $$ R \approx\left(\frac{2 \times 7 \times 10^{-2} \mathrm{~N} / \mathrm{m}}{10^5 \mathrm{~N} / \mathrm{m}^2}\right) \approx 10^{-6} \mathrm{~m} . $$
这是一份2023年的卡迪夫大学Cardiff University 机械学和物质PX1121代写的成功案例
Assignment-daixieTM为您提供卡迪夫大学Cardiff University PX1221 Electricity, Magnetism and Waves电、磁和波代写代考和辅导服务！
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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.
电、磁和波代写|Electricity, Magnetism and Waves PX1221 Cardiff University Assignment
问题 1.
(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.
证明 .
To find the total charge $Q$ on the rectangular surface, we need to integrate the charge density $\sigma(x, y)$ over the surface area:
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 $\boxed{\frac{k a^2 b^2}{4}}$.
问题 2.
(b) Find the total charge on a circular plate of radius $R$ if the charge distribution is $\sigma(r, \theta)=k r(1-\sin \theta)$.
证明 .
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 $\frac{2}{3}\pi k R^3$.
问题 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.
证明 .
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:
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:
$$\boxed{E(r) = \frac{Q}{2\pi\epsilon_0 rl}}$$
这是一份2023年的卡迪夫大学Cardiff University 电、磁和波PX1221代写的成功案例
Assignment-daixieTM为您提供卡迪夫大学Cardiff University EA1300 World of Dynamic Environments动态环境代写代考和辅导服务！
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
动态环境代写|World of Dynamic Environments EA1300 Cardiff University Assignment
问题 1.
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.
证明 .
Starting from the left-hand side of the equation, we have:
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:
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:
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 $\Phi$.
证明 .
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
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
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 $h_o(x, t)$.
证明 .
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:
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:
$\frac{h^2}{2}+g h+\frac{Q^2}{2 g b^2}=C(x)$
where $C(x)$ is the integration constant. The value of $C(x)$ can be determined from the boundary conditions. For example, if the depth of the fluid is $
这是一份2023年的卡迪夫大学Cardiff University EA1300动态环境代写的成功案例
Assignment-daixieTM为您提供东英吉利大学University of East Anglia PHY-4005AAPPLIED ELECTROMAGNETISM AND WAVES电磁学和电波代写代考和辅导服务！
Instructions:
Electricity and Magnetism: Electricity and magnetism are two closely related phenomena that together form the foundation of the study of electromagnetism. The theory of electricity and magnetism deals with the behavior of charged particles and their interactions with electric and magnetic fields.
Electric charge is a fundamental property of matter, and there are two types of charge, positive and negative. When like charges are brought close together, they repel each other, while opposite charges attract each other. The force between two charged particles is known as the Coulomb force and is proportional to the product of the charges and inversely proportional to the distance between them.
When charges are in motion, they create a magnetic field, and the motion of charges in a magnetic field creates a force on the charges. This is known as the Lorentz force and is proportional to the velocity of the charges and the strength of the magnetic field.
Electric and magnetic fields are related and can create each other. When an electric charge is moving, it creates a magnetic field, and when a magnetic field changes, it creates an electric field. This is known as electromagnetic induction and is the basis of many electrical devices, including generators, transformers, and motors.
电磁学和电波代写ELECTROMAGNETISM AND WAVES|PHY-4005A University of East Anglia Assignment
问题 1.
In this problem, we consider several equivalent situations for a plane wave propagation in the $\hat{z}$ direction. Let the electric field be $\hat{x}$ directed. $$ \bar{E}=\hat{x} E_0 e^{i k z}, \quad \bar{H}=\hat{y} \frac{1}{\eta} E_0 e^{i k z} $$ and the region of interest be $z>0$. (a) Put an electric current sheet with $\bar{J}_s=A \hat{x}$. What is the value of $A$ so that the same field is preserved in the region of interest?
证明 .
(a) The electric current sheet produces a magnetic field $\bar{H}_s = \hat{y} \frac{A}{\epsilon_0} \delta(z)$. To preserve the same field in the region of interest, this magnetic field must cancel the magnetic field produced by the plane wave. Since $\bar{H}_s$ only exists at $z=0$, we need to consider the $z>0$ region where the plane wave is propagating. Using the relationship $\bar{E} = \eta \bar{H}$, we have $\bar{H} = \frac{1}{\eta} \hat{y} E_0 e^{ikz}$ for $z>0$. The magnetic field produced by the electric current sheet at $z=0$ is $\bar{H}_s = \hat{y} \frac{A}{\eta}$. Therefore, we need $A = E_0$ for the same field to be preserved in the region of interest.
问题 2.
(b) Put a magnetic current sheet with $\bar{M}_s=B \hat{y}$. What is the value of $B$ so that the same field is preserved in the region of interest?
证明 .
(b) The magnetic current sheet produces an electric field $\bar{E}_s = \hat{x} \frac{B}{\mu_0} \delta(z)$. To preserve the same field in the region of interest, this electric field must cancel the electric field produced by the plane wave. Since $\bar{H} = \frac{1}{\eta} \bar{E}$, we have $\bar{E} = \eta \hat{x} E_0 e^{ikz}$ for $z>0$. The electric field produced by the magnetic current sheet at $z=0$ is $\bar{E}_s = \hat{x} \frac{B}{\mu_0}$. Therefore, we need $B = \frac{\mu_0}{\eta} E_0$ for the same field to be preserved in the region of interest.
问题 3.
(c) Replace the region $z<0$ with a perfect conductor. Place in front of the conductor an electric sheet with $\bar{J}_s=C \hat{x}$ and a magnetic current sheet with $\bar{M}_s=D \hat{y}$. What is the value of $C$ and $D$ so that the same field is preserved in the region of interest?
证明 .
(c) The perfect conductor reflects the incident wave, so the reflected wave will interfere with the incident wave. To preserve the same field in the region of interest, we need to ensure that the interference results in no net field. The electric current sheet produces a reflected magnetic field $\bar{H}{sr} = -\hat{y} \frac{C}{\epsilon_0} \delta(z)$. The magnetic current sheet produces a reflected electric field $\bar{E}{sr} = \hat{x} \frac{D}{\mu_0} \delta(z)$. The total magnetic field at $z>0$ is the sum of the incident and reflected fields:
$\bar{H}_t=\hat{y} \frac{1}{\eta} E_0 e^{i k z}-\hat{y} \frac{C}{\eta \epsilon_0} e^{-i k z}$
The total electric field at $z>0$ is the sum of the incident and reflected fields:
$\bar{E}_t=\hat{x} \eta E_0 e^{i k z}+\hat{x} \frac{D}{\mu_0} e^{-i k z}$
For the same field to be preserved, we need $\bar{H}_t = 0$ and $\bar{E}_t = 0$ for $z>0$. Solving for $C$ and $D$, we get:
$C=E_0, \quad D=-\frac{\mu_0}{\eta} E_0$
Therefore, to preserve the same field in the region of interest, we need to place an electric current sheet with $A=E_0$
这是一份2023年的东英吉利大学University of East Anglia PHY-4005A电磁学和电波代写的成功案例
Assignment-daixieTM为您提供东英吉利大学University of East Anglia PHY-4005AHEAT, ATOMS AND SOLIDS热原子和固体代写代考和辅导服务！
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Thermodynamics is the study of energy and its transformation. It deals with macroscopic systems and how they respond to changes in energy. Condensed matter physics is the study of materials in which the atoms are close together, and they interact strongly with each other. This includes solids, liquids, and some gases.
Electronic Structure: The Sommerfeld and Band Theories
In the free-electron Sommerfeld theory, electrons are treated as free particles moving in a periodic potential created by the ions. This model explains some properties of metals, such as their electrical conductivity and heat capacity. However, it fails to account for the diversity of properties observed in different materials.
The band theory of solids describes the electronic structure of materials in terms of energy bands. In a periodic crystal lattice, the wave functions of the electrons overlap, forming bands of allowed energies. The electrons can occupy these bands, up to a maximum called the Fermi energy. This theory explains the diversity of electronic properties of materials, including their conductivity, magnetism, and optical properties.
Phonons and Heat Capacity
Phonons are quantized vibrations of the atoms in a crystal lattice. They contribute to the heat capacity of a solid by carrying energy and increasing the number of ways in which energy can be distributed among the atoms.
Structure, Bonding, and Properties of Solids
The properties of solids depend on their atomic structure and bonding. The types of bonding include ionic, covalent, and metallic bonding. The electronic structure and bonding determine the electrical conductivity, magnetism, and other properties of the material.
热、原子和固体代写HEAT, ATOMS AND SOLIDS|PHY-4005A University of East Anglia Assignment
问题 1.
(a) Prove the finite temperature version of the fluctuation dissipation theorem $$ \chi^{\prime \prime}(q, \omega)=\frac{1}{2}\left(e^{-\beta \omega}-1\right) S(q, \omega), $$ and $$ S(q, \omega)=-2\left(n_B(\omega)+1\right) \chi^{\prime \prime}(q, \omega) $$ where $S(q, \omega)=\int d \vec{x} d t e^{-i \vec{q}-\vec{x}} e^{i \omega x}\langle\rho(\vec{x}, t) \rho(0,)\rangle_T$ and $n_B(\omega)=\left(e^{\beta \omega}-1\right)^{-1}$ is the Bose occupation factor.
证明 .
The fluctuation-dissipation theorem relates the correlation function of a fluctuating variable to its response to an external perturbation. Specifically, it relates the imaginary part of the susceptibility, $\chi^{\prime \prime}(q,\omega)$, to the power spectrum of the fluctuations, $S(q,\omega)$.
Consider a system described by a Hamiltonian $H$ at temperature $T$. Let $\rho(\vec{x},t)$ be an operator that measures the local density fluctuations of the system. Then, the susceptibility $\chi(q,\omega)$ is defined as the linear response of the density fluctuations to a spatially varying potential $V(\vec{x})$ of wavevector $\vec{q}$ and frequency $\omega$:
where $\text{c.c.}$ denotes the complex conjugate of the first term.
问题 2.
(b) Show that $\chi^{\prime \prime}(q, \omega)=-\chi^{\prime \prime}(-q,-\omega)$ and $S(-q,-\omega)=e^{-\beta \omega} S(q, \omega)$. In terms of the scattering probability, show that this is consistent with detailed balance.
证明 .
To show that $\chi^{\prime \prime}(q, \omega)=-\chi^{\prime \prime}(-q,-\omega)$, we start with the definition of the dynamic structure factor:
$\chi^{\prime \prime}(q, \omega)=-\frac{1}{\pi} \operatorname{Im} \int_{-\infty}^{\infty} d t e^{i \omega t} \int d^3 r e^{-i q \cdot r}\langle\hat{\rho}(\mathbf{r}, t) \hat{\rho}(0,0)\rangle$
where $\hat{\rho}(\mathbf{r}, t)$ is the density operator at position $\mathbf{r}$ and time $t$, and $\langle \rangle$ denotes the statistical average over an ensemble of systems in thermal equilibrium. Using the fact that the correlation function $\left\langle\hat{\rho}(\mathbf{r}, t) \hat{\rho}(0,0)\right\rangle$ is real, we can rewrite the integral as
$\chi^{\prime \prime}(q, \omega)=\frac{1}{\pi} \int_{-\infty}^{\infty} d t \sin (\omega t) \int d^3 r e^{-i q \cdot r}\langle\hat{\rho}(\mathbf{r}, t) \hat{\rho}(0,0)\rangle$
Changing variables $t\rightarrow -t$ and $r\rightarrow -r$ gives
$\begin{aligned} \chi^{\prime \prime}(-q,-\omega) & =\frac{1}{\pi} \int_{-\infty}^{\infty} d t \sin (-\omega t) \int d^3 r e^{i(-q) \cdot r}\langle\hat{\rho}(-\mathbf{r},-t) \hat{\rho}(0,0)\rangle \ & =\frac{1}{\pi} \int_{-\infty}^{\infty} d t \sin (\omega t) \int d^3 r e^{-i q \cdot r}\langle\hat{\rho}(\mathbf{r}, t) \hat{\rho}(0,0)\rangle \ & =-\chi^{\prime \prime}(q, \omega)\end{aligned}$
Thus, we have shown that $\chi^{\prime \prime}(q, \omega)=-\chi^{\prime \prime}(-q,-\omega)$.
Next, we show that $S(-q,-\omega)=e^{-\beta \omega} S(q, \omega)$. Starting with the definition of the dynamic structure factor,
(a) Using linear response theory, derive the following expression for the magnetic susceptibility $\chi_{|}=\partial M_z / \partial H_z$. $$ \chi_{|}=\lim _{q \rightarrow 0} \int \frac{d \omega}{2 \pi}\left\langle S_z(q, \omega) S_z(-q,-\omega)\right\rangle \frac{\left(1-e^{-\frac{\hbar \omega}{k T}}\right)}{\omega} $$
证明 .
Linear response theory is a framework for understanding the response of a physical system to small perturbations. In this case, we want to calculate the magnetic susceptibility $\chi_{|}$, which is the response of the magnetization $M_z$ to a small magnetic field $H_z$. We can use the Kubo formula to relate the susceptibility to the spin-spin correlation function:
where $\hat{S}z(\mathbf{r},t)$ is the $z$-component of the spin density operator at position $\mathbf{r}$ and time $t$, and $\left\langle\cdots\right\rangle{H_z=0}$ denotes the thermal average in the absence of an external magnetic field. The Fourier transform of the spin density operator is defined by
where $\hat{\chi}_{zz}(\mathbf{q},\omega)$ is the Fourier transform of the spin-spin correlation function $\left\langle\hat{S}z(\mathbf{r},t)\hat{S}z(\mathbf{0},0)\right\rangle{H_z=0}$. Substituting this into the expression for $\chi{|}$ and taking the limit $\omega\rightarrow 0$, we obtain
Assignment-daixieTM为您提供莱斯特大学University of Leicester PA1140 Waves and Quanta波与量子代写代考和辅导服务！
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Waves and quanta are two fundamental concepts in physics that are closely related but also distinct.
A wave is a disturbance that travels through space and time, carrying energy without the transfer of matter. Waves can take many forms, including electromagnetic waves (such as light and radio waves), sound waves, and water waves. Waves are characterized by their wavelength, frequency, and amplitude, which determine their properties such as speed and energy.
On the other hand, a quantum is a discrete packet of energy that behaves both as a particle and as a wave. In the quantum world, particles can exhibit wave-like behavior, and waves can behave like particles. This duality is known as wave-particle duality. The behavior of quanta is described by quantum mechanics, a branch of physics that deals with the behavior of matter and energy at the atomic and subatomic level.
The relationship between waves and quanta is that waves can be thought of as a collection of quanta. For example, electromagnetic waves consist of a stream of photons, which are quanta of electromagnetic radiation. Similarly, matter waves, which describe the behavior of particles such as electrons, can be thought of as a collection of quanta.
Overall, waves and quanta are both fundamental concepts in physics that are used to explain and understand the behavior of matter and energy in our universe.
波与量子代写Waves and Quanta|PA1140 University of Leicester Assignment
问题 1.
A long, uniform taut string (mass per unit length $\rho$, tension $T$ ) along $-\infty<x<\infty$ is supported on an elastic foundation of stiffness $\alpha$, and a point mass $M$ is attached at $x=0$. Suppose that a time-harmonic vertical force $$ F \cos \Omega t $$ is applied to the mass at $x=0$. Determine the steady-state displacement response of the string for $-\infty<x<\infty$.
证明 .
The equation of motion for the string is given by the wave equation:
where $u(x,t)$ is the displacement of the string from its equilibrium position at point $x$ and time $t$. The first term on the right-hand side represents the tension in the string, and the second term represents the elastic foundation. We can assume a solution of the form:
$u(x, t)=U(x) e^{i \omega t}$
where $\omega = \Omega + i\beta$ is the complex frequency and $U(x)$ is the complex amplitude of the displacement.
Substituting this solution into the wave equation and simplifying, we get:
$\frac{d^2 U}{d x^2}+k^2 U=0$
where $k^2 = \frac{\omega^2 \rho}{T} – i\frac{\alpha}{T}$. This is a second-order homogeneous differential equation with constant coefficients, which has solutions of the form:
$U(x)=A e^{i k x}+B e^{-i k x}$
where $A$ and $B$ are complex constants to be determined by the boundary conditions.
At $x=0$, the displacement of the string must be equal to the displacement of the mass, which is given by:
where $k = \sqrt{\frac{\omega^2 \rho}{T} – i\frac{\alpha}{T}}$. This is the steady-state displacement response of the string for $-\infty < x < \infty$.
问题 2.
The propagation of free uni-directional surface waves of small amplitude on moderately shallow water is governed by the equation $$ \frac{\partial \eta}{\partial t}+c_0 \frac{\partial \eta}{\partial x}+\beta \frac{\partial^3 \eta}{\partial x^3}=0 $$ where $\eta(x, t)$ is the free-surface elevation and $c_0$ and $\beta$ are constants. (a) Suppose that an external localized pressure disturbance traveling with constant speed $V$ acts on the free surface. Determine the wavenumber(s) of the excited steadystate radiating wave(s), depending on the forcing speed $V$. Sketch the position of these waves relative to the forcing. (Take $c_0>0$ and consider $\beta>0$ and $\beta<0$ as well as $V>0$ and $V<0$.)
证明 .
To determine the wavenumber(s) of the excited steady-state radiating wave(s), we start by assuming a solution of the form:
$\eta(x, t)=A e^{i(k x-\omega t)}$
where $A$ is the amplitude of the wave, $k$ is the wavenumber, and $\omega$ is the frequency. Substituting this solution into the given equation yields:
$-i \omega A+i c_0 k A-i \beta k^3 A=0$
Dividing through by $-iA$ and solving for $k$, we get:
Now suppose that an external localized pressure disturbance traveling with constant speed $V$ acts on the free surface. This means that the phase velocity of the wave excited by the disturbance will be $V$, i.e. $\omega/k=V$. Substituting this into the above expression for $k^2$ yields:
where $c=\sqrt{c_0^2+\beta k^2}$ is the phase velocity of the wave.
Note that $k_1$ and $k_2$ have opposite signs, which means that the two waves have opposite directions of propagation. Furthermore, the two waves have different wavelengths and frequencies. The wave with wavenumber $k_1$ has a longer wavelength and a lower frequency than the wave with wavenumber $k_2$.
To sketch the position of these waves relative to the forcing, we note that the wave with wavenumber $k_1$ travels in the same direction as the external disturbance, while the wave with wavenumber $k_2$ travels in the opposite direction. The amplitude of the waves decreases with distance from the source, so the wave with wavenumber $k_1$ will be closer to the source than the wave with wavenumber $k_2$.
Case 2: $\beta<0$
In this case, there is only one possible value of $k$ given
问题 3.
(b) Suppose at $t=0$ a localized initial wave disturbance is introduced in the vicinity of $x=0$. Sketch qualitatively the time history of the response for $t>0$ at a fixed station $x=L>0$, far from the region of the initial disturbance. Sketch qualitatively a snapshot of the disturbance for $-\infty<x<\infty$ at time $t=T$, long after the initial excitation. Justify your answers. (Again, $c_0>0$ and consider $\beta>0$ and $\beta<0$.)
证明 .
To sketch qualitatively the time history of the response for $t>0$ at a fixed station $x=L>0$, we first need to find the solution to the given equation. We can use the method of characteristics to solve this equation.
Let $dx/dt=c_0$ and $d^2x/dt^2=0$, which gives $x=c_0t+x_0$ and $x_0=L$ for $t=0$. Then, we have
We assume a solution of the form $\eta(x,t)=f(x-ct)$, where $c$ is the wave speed. Substituting this into the above equation and integrating three times with respect to $x$, we get
$\eta(x, t)=A e^{-k(x-c t)} \cos (k(x-c t)+\phi)$
where $k=\sqrt[3]{\beta/c_0}$, $A$ and $\phi$ are constants determined by the initial conditions.
Now, we can sketch the time history of the response for $t>0$ at a fixed station $x=L>0$. At $x=L$, the free surface elevation is given by
$\eta(L, t)=A e^{-k(L-c t)} \cos (k(L-c t)+\phi)$
As $t$ increases, the amplitude of the cosine function decays exponentially with a decay rate of $k(c t-L)$. Therefore, the free surface elevation decreases in amplitude and oscillates with a decreasing frequency as time progresses. The decay rate is faster for larger values of $k$ (i.e., for larger values of $\beta$ or smaller values of $c_0$). If $\beta<0$, then $k$ is imaginary, which means the wave amplitude decays exponentially without oscillation.
To sketch qualitatively a snapshot of the disturbance for $-\infty<x<\infty$ at time $t=T$, long after the initial excitation, we can use the same solution as above, but with $t=T$. Thus, we have
The cosine function represents a wave with wavelength $\lambda=2\pi/k$ and wave number $k$. The wave is propagating to the right if $c>c_0$, or to the left if $c<c_0$. The amplitude of the wave decreases exponentially as $x$ increases or decreases from the initial position of the disturbance. The decay rate is faster for larger values of $k$ (i.e., for larger values of $\beta$ or smaller values of $c_0$). If $\beta<0$, then $k$ is imaginary, which means the wave amplitude decays exponentially without oscillation. The phase velocity of the wave is given by $c/k$, which decreases as $k$ increases. If $\beta>0$, then the group velocity of the wave is given by $c-(2/3)k^2c_0$, which is smaller than the phase velocity. If $\beta<0$, then the group velocity is undefined because $k$ is imaginary.
这是一份2023年的莱斯特大学University of Leicester PA1140波与量子代写的成功案例
Assignment-daixieTM为您提供莱斯特大学University of Leicester PA1120 Light and Matter光与质代写代考和辅导服务！
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Light is a form of electromagnetic radiation that travels through space as a wave. It has both particle-like and wave-like properties, and is typically described in terms of its wavelength, frequency, and energy. The wavelength and frequency of light are related by the speed of light, which is constant in a vacuum and is denoted by the symbol c. The energy of light is directly proportional to its frequency, and is given by the equation E = hf, where h is Planck’s constant.
Matter, on the other hand, is composed of atoms and molecules, which are made up of protons, neutrons, and electrons. These particles have mass and occupy space, and interact with each other through various forces, such as electromagnetic forces and the strong and weak nuclear forces. The behavior of matter is governed by the laws of quantum mechanics, which describe how particles move, interact, and are affected by external forces.
In the context of optics, matter can interact with light in a variety of ways, including absorption, reflection, and refraction. When light is absorbed by matter, it can excite electrons to higher energy levels, which can lead to various physical and chemical changes. Reflection occurs when light bounces off a surface, while refraction occurs when light passes through a material and changes direction due to a change in its speed.
光与质代写Light and Matter|PA1120 University of Leicester Assignment
问题 1.
A state function for a $\operatorname{Van}$ der Waals gas is given by an equation between thermodynamic variables that depend on model parameters $A, B$, and a physical constant $R$ : $$ \left(P+\frac{A N^2}{V^2}\right)(V-N B)=N R T $$ where $A N^2 / V^2$ is referred to as the internal pressure due to the attraction between molecules and $N B$ is an extra volume, sometimes associated with the the volume per molecule.
Write out a differential expression for $d N$ in terms of differentials of the thermodynamic variables.
证明 .
A state function for a Van der Waals gas is given by an equation between thermodynamic variables that depend on model parameters $A, B$, and a physical constant R: $$ \left(P+\frac{A N^2}{V^2}\right)(V-N B)=N R T $$ where $A N^2 / V^2$ is referred to as the internal pressure due to the attraction between molecules and $N B$ is an extra volume, sometimes associated with the volume per molecule.
Write out a differential expression for $d N$ in terms of differentials of the thermodynamic variables.
The solution is pretty straightforward. One way is to differentiate the entire expression and group the terms corresponding to $d N, d P, d T$, and $d V$. Another way to do is by implicit differentiation. The real gas equation can be rewritten such that, $$ \begin{aligned} N & =N(T, V, P) \ d N & =\left(\frac{\partial N}{\partial P}\right){T, V} d P+\left(\frac{\partial N}{\partial V}\right){T, P} d V+\left(\frac{\partial N}{\partial T}\right){P, V} d T \end{aligned} $$ In an equivalent way, you could have written the function $P=P(V, T, N)$ and extract $d N$ from the following. $$ d P=\left(\frac{\partial P}{\partial N}\right){T, V} d N+\left(\frac{\partial P}{\partial V}\right){T, N} d V+\left(\frac{\partial P}{\partial T}\right){N, V} d T $$ For instance, the first term $\left(\frac{\partial P}{\partial N}\right){T, V}$ can be evaluated as $$ \left(\frac{\partial P}{\partial T}\right){P, V}=\frac{N R}{V-N B} $$ Using any one of the methods you would get $$ d N=\frac{(V-B N) d P+\left(P-\left(A N^2 / V^2\right)+\left(2 A B N^3 / V^3\right)\right) d V-R N d T}{B P+R T+\left(3 A B N^2 / V^2\right)-(2 A N / V)} $$
问题 2.
Each day a certain amount of water evaporates from the oceans, lakes, and earth surface and forms water vapor and clouds in the atmosphere. Each day a certain amount of rain falls back to the earth. Make the reasonable assumption that, on the average, the energy consumed by evaporating and lifting the water is equal to the energy produced by condensation and rain falling back to earth.
To evaporate one mole of water, approximately 41090 joules of heat are required and an equivalent amount is expelled when a mole of water condenses.
Estimate the amount of work required each day to produce the rain that raineth. That is, the work to evaporate and lift the water.
You may need to find data to help make this estimate such as the average height of a rain cloud, state what those data are and from whence they came. Discuss what is supplying this daily energy. Compare this daily energy consumption with the energy produced each year in metropolitan Boston.
证明 .
Each day a certain amount of water evaporates from the oceans, lakes, and earth surface and forms water vapor and clouds in the atmosphere. Each day a certain amount of rain falls back to the earth. Make the reasonable assumption that, on the average, the energy consumed by evaporating and liftin the water is equal to the energy produced by condensation and rain falling back to earth. To evaporate one mole of water, approximately 41090 joules of heat
to evaporate and lift the water. You may need to find dat a to help make this estimate such as the average height of a rain cloud, state what those data are and from whence they came. Discuss what is supplying this daily energy. Compare this daily energy consumption with the energy produced each year in metropolitan Boston. From the problem statement, The energy to lift water $=$ energy produced by condensation and rain The first step would be to find the volume of water evaporated in the surface of earth. Since all the water is assumed to precipitate and et urned as rain, an average rainfall data and earth surface area would be needed to estimate this. A simplifying assumption for the calculation of volume of water precipitated could be that the curvature of surface is very small compared to the average height of precipitation. (The following data are from problem solutions by the group Ryan Clark, Daniel MacPhee, Taylor Schildgen, Alex Bradley – reported taken from climate prediction center, 1979-2000 dat a). global precipitation average per day $=3 \mathrm{~mm}$ radius of earth $=6371 \mathrm{kms}$ surface area of earth $=5.1^* 10^{14} \mathrm{~m}^2$ volume of precipitation $=$ surface area $$ average precipitation $=3.0^ 10^{-3} * 5.1 * 10^{14} \mathrm{~m}^3=$ $15.3^* 10^{12} \mathrm{~m}^3$ mass of precipitation $=$ density $$ volume $=15.3^ 10^{15} \mathrm{~kg}$ (density of water $\left.=1 \mathrm{gm} / \mathrm{cc}\right)$ number of moles of water $=$ total mass $/$ molar mass $=15.3 / 18^* 10^{15}=8.5^* 10^{14}$
Next step is to calculate the energy spent on lifting this estimated volume of water to the atmosphere. The energy needed to evaporate water is given in the problem statement as 41090 joules per gram. you would also need some information on the average height these drops are lifted or the average cloud height to calculate its potential energy change. The total energy per day day can be estimated from these calculations. This amount of energy is consumed everyday by the water that rainet $h$. A reasonable distance at which clouds are formed could be around $5000 \mathrm{kms}$. The energy needed to lift the mass of water to this height $=\mathrm{m}^* \mathrm{~g}^* \mathrm{~h}=15.3^* 10^{15} * 9.8^$ 5000 Joules $=7.497^ 10^{20}$ joules
The energy needed to evaporate the entire volume of water $=$ number of moles $*$ energy for evaporation per mole $=8.5 * 10^{14} * 41090$ Joules $=3.49265 * 10^{19}$ Joules. The total energy to evaporate and lift $=(7.497+.85)^* 10^{20}$ Joules $=8.347^* 10^{20}$ Joules. Discussion on what is supplying the energy to lift this amount of water can be on fraction of sun’s energy reaching earth, on the fraction of water surface available, average global temperature, ozone layers and anything else that might affect the supply of energy to evaporate and lift water vaopur.
问题 3.
There are four state variables for an arbitrary pure gas: total moles of gas molecules, pressure, volume, and temperature. How many independent state variables exist for a mole of an ideal gas? Using different combinations of independent variables, write as many unique (but numerically equivalent) expressions for the molar internal energy of an ideal gas as possible.
证明 .
One mole of an ideal gas has 2 state variables. The four degrees of freedom (n, $P, V$, and $T$ ) for an arbitrary gas are decreased by the constraints, $n=1$ and $P V=n R T$.
We can write $\bar{U}$ in terms of each combination of two independent variables: $\bar{U}(T, \bar{V}), \bar{U}(T, P), \bar{U}(P, \bar{V})$. The differential form for the molar internal energy of an ideal gas is given by $$ d \bar{U}=\overline{C_v} d T $$ Integration gives: $$ \bar{U}(T)=C_v\left(T-T_0\right)=U(T, \bar{V})=U(T, P) $$ Substitution using the ideal gas law gives: $$ \overline{U(P, \bar{V})}=\frac{\overline{C_v}}{R}\left(\bar{V} P-\bar{V}_0 P_0\right) $$
这是一份2023年的莱斯特大学University of Leicester PA1120光与质代写的成功案例
Assignment-daixieTM为您提供莱斯特大学University of Leicester PA1710 Mathematical Physics 1.1数学物理代写代考和辅导服务！
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Physics is a mathematical subject that relies heavily on mathematical techniques and methods to describe and understand the physical world. As a result, students studying physics typically receive training in a wide range of mathematical topics, including calculus, linear algebra, differential equations, probability theory, and more.
Integral calculus, in particular, is a core mathematical tool in physics that is used to model and analyze a wide range of physical phenomena, such as motion, forces, and energy. Series and convergence are also important concepts in physics, as they are used to represent functions and data in various forms, such as Fourier series and Taylor series.
Vector analysis is another important mathematical topic in physics that is used to describe physical quantities such as velocity, acceleration, and force. Probability distributions are also crucial in physics, as they allow us to model and understand the statistical behavior of physical systems, such as the distribution of particles in a gas or the distribution of energies in a solid.
Overall, a strong foundation in mathematics is essential for success in physics, and many of the mathematical techniques that are taught in the first year of a physics degree program provide the necessary tools for understanding and analyzing physical phenomena.
物理数学代写Mathematical Physics 1.1|PA1710 University of Leicester Assignment
问题 1.
Use the definition of the derivative as a limit of difference quotients to compute the derivative of $y=x+\frac{1}{a}$ for all points $x>0$. Show all work.
证明 .
Denote by $f(x)$ the function $x+\frac{1}{x}$. By definition, the derivative of $f(x)$ at $x=a$ is, $$ f^{\prime}(a)=\lim {h \rightarrow 0} \frac{f(a+h)-f(a)}{h} . $$ The increment $f(a+h)-f(a)$ equals, $$ \left((a+h)+\frac{1}{a+h}\right)-\left(a+\frac{1}{a}\right)=h+\left(\frac{1}{a+h}-\frac{1}{a}\right) . $$ To compute the second term, clear denominators, $$ \frac{1}{a+h}-\frac{1}{a}=\frac{1}{a+h} \frac{a}{a}-\frac{1}{a} \frac{a+h}{a+h}=\frac{a-(a+h)}{a(a+h)}=\frac{-h}{a(a+h)} . $$ Thus the increment $f(a+h)-f(a)$ equals, $$ h-\frac{h}{a(a+h)} . $$ Factoring $h$ from each term, the difference quotient equals, $$ \frac{f(a+h)-f(a)}{h}=1-\frac{1}{a(a+h)} . $$ Thus the derivative of $f(x)$ at $x=a$ equals, $$ f^{\prime}(a)=\lim {h \rightarrow 0}\left(1-\frac{1}{a(a+h)}\right)=1-\frac{1}{a(a+0)}=1-\frac{1}{a^2} . $$ Therefore the derivative function of $f(x)$ equals, $$ f^{\prime}(x)=1-\frac{1}{x^2} . $$
问题 2.
For the function $f(x)=e^{-x^2 / 2}$, compute the first, second and third derivatives of $f(x)$.
证明 .
Set $u$ equals $-x^2 / 2$ and set $v$ equals $e^u$. So $v$ equals $f(x)$. By the chain rule, $$ \frac{d v}{d x}=\frac{d v}{d u} \frac{d u}{d x} . $$ Since $v$ equals $e^u, d v / d u$ equals $\left(e^u\right)^{\prime}=e^u$. Since $u$ equals $-x^2 / 2, d u / d x$ equals $-(2 x) / 2=-x$. Thus, back-substituting, $$ f^{\prime}(x)=\frac{d v}{d x}=\left(e^u\right)(-x)=e^{-x^2 / 2}(-x)=-x e^{-x^2 / 2} . $$ For the second derivative, let $u$ and $v$ be as defined above, and set $w$ equals $-x v$. So $w$ equals $f^{\prime}(x)$. By the product rule, $$ \frac{d w}{d x}=(-x)^{\prime} v+(-x) v^{\prime}=-v-x \frac{d v}{d x} . $$ By the last paragraph, $$ \frac{d v}{d x}=-x e^{-x^2 / 2} $$ Substituting in, $$ f^{\prime \prime}(x)=\frac{d w}{d x}=-e^{-x^2 / 2}-x\left(-x e^{-x^2 / 2}\right)=-e^{-x^2 / 2}+x^2 e^{-x^2 / 2}=\left(x^2-1\right) e^{-x^2 / 2} . $$ For the third derivative, take $u$ and $v$ as above, and set $z$ equals $\left(x^2-1\right) v$. So $z$ equals $f^{\prime \prime}(x)$. By the product rule, $$ \frac{d z}{d x}=\left(x^2-1\right)^{\prime} v+\left(x^2-1\right) v^{\prime}=2 x v+\left(x^2-1\right) \frac{d v}{d x} . $$ By the first paragraph, $$ \frac{d v}{d x}=-x e^{-x^2 / 2} $$ Substituting in, $$ f^{\prime \prime \prime}(x)=\frac{d z}{d x}=2 x e^{-x^2 / 2}+\left(x^2-1\right)\left(-x e^{-x^2 / 2}\right)=2 x e^{-x^2 / 2}+\left(-x^3+x\right) e^{-x^2 / 2}=\left(-x^3+3 x\right) e^{-x^2 / 2} . $$
问题 3.
A function $y=f(x)$ satisfies the implicit equation, $$ 2 x^3-9 x y+2 y^3=0 . $$ The graph contains the point $(1,2)$. Find the equation of the tangent line to the graph of $y=f(x)$ at $(1,2)$.
证明 .
Differentiating both sides of the equation gives, $$ \frac{d}{d x}\left(2 x^3-9 x y+2 y^3\right)=\frac{d}{d x}(0)=0 $$ Because the derivative is linear, $$ \frac{d}{d x}\left(2 x^3-9 x y+2 y^3\right)=2 \frac{d\left(x^3\right)}{d x}-9 \frac{d(x y)}{d x}+2 \frac{d\left(y^3\right)}{d x} . $$ Of course $d\left(x^3\right) / d x$ equals $3 x^2$. By the product rule, $$ \frac{d(x y)}{d x}=\frac{d(x)}{d x} y+x \frac{d y}{d x}=y+x \frac{d y}{d x} $$ For the last term, the chain rule gives, $$ \frac{d\left(y^3\right)}{d x}=\frac{d\left(y^3\right)}{d y} \frac{d y}{d x}=3 y^2 \frac{d y}{d x} $$ Substituting in gives, $$ \frac{d}{d x}\left(2 x^3-9 x y+2 y^3\right)=2\left(3 x^2\right)-9\left(y+x \frac{d y}{d x}\right)+2\left(3 y^2\right) \frac{d y}{d x}=\left(6 x^2-9 y\right)+\left(6 y^2-9 x\right) \frac{d y}{d x} $$ By the first paragraph, $d / d x\left(2 x^3-9 x y+2 y^3\right)$ equals 0 . Substituting in gives the equation, $$ \left(6 x^2-9 y\right)+\left(6 y^2-9 x\right) \frac{d y}{d x}=0 $$ Subtracting the first term from each side gives, $$ \left(6 y^2-9 x\right) \frac{d y}{d x}=\left(9 y-6 x^2\right) . $$ Dividing both sides by $\left(6 y^2-9 x\right)$ gives, $$ \frac{d y}{d x}=\frac{9 y-6 x^2}{6 y^2-9 x}=\frac{3 y-2 x^2}{2 y^2-3 x} $$
Finally, plugging in $x$ equals 1 and $y$ equals 2 gives, $$ \frac{d y}{d x}=\frac{3(2)-2(1)^2}{2(2)^2-3(1)}=\frac{6-2}{8-3}=\frac{4}{5} . $$ Therefore, the equation of the tangent line is, $$ y=\frac{4}{5}(x-1)+2, $$ which simplifies to, $$ y=(4 / 5) x+6 / 5 $$
这是一份2023年的莱斯特大学University of Leicester PA1710数学物理代写的成功案例
Assignment-daixieTM为您提供莱斯特大学University of Leicester PA1130 Electricity and Magnetism电磁学代写代考和辅导服务！
Instructions:
Maxwell’s theory of electromagnetism is a set of equations that describe the behavior of electric and magnetic fields and their interactions with charged particles. The theory provides a unified framework for understanding various phenomena related to electricity and magnetism, such as how charges produce electric fields, how moving charges create magnetic fields, and how changing magnetic fields produce electric fields.
Maxwell’s equations also predict the existence of electromagnetic waves, which are waves of oscillating electric and magnetic fields that can propagate through space at the speed of light. These waves are the basis for many modern technologies, including radio communication, television, and cellular phones.
The importance of Maxwell’s theory of electromagnetism cannot be overstated, as it forms the basis for our understanding of the fundamental workings of many devices and technologies that we rely on in our daily lives.
电磁学代写Electricity and Magnetism|PA1130 University of Leicester Assignment
问题 1.
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: $$ \int_C f(x, y, z) d s=\int_{s_1}^{s_2} f[x(s), y(s), z(s)] d s $$
证明 .
As an example, let us consider the following integral in two dimensions: $$ I=\int_C(x+y) d s $$ where $C$ is a straight line from the origin to $(1,1)$, as shown in the figure. Let $s$ be the arc length measured from the origin. We then have $$ \begin{aligned} & x=s \cos \theta=\frac{s}{\sqrt{2}} \ & y=s \sin \theta=\frac{s}{\sqrt{2}} \end{aligned} $$
The endpoint $(1,1)$ corresponds to $s=\sqrt{2}$. Thus, the line integral becomes $$ I=\int_0^{\sqrt{2}}\left(\frac{s}{\sqrt{2}}+\frac{s}{\sqrt{2}}\right) d s=\sqrt{2} \int_0^{\sqrt{2}} s d s=\left.\sqrt{2} \cdot \frac{s^2}{2}\right|_0 ^{\sqrt{2}}=\sqrt{2} $$
问题 2.
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.
证明 .
To use Gauss’s Law, we need to choose a Gaussian surface that encloses the charge distribution. In this case, we can choose a cylindrical Gaussian surface with radius $r$ and height $l$ (same as the height of the cylinders). The axis of the Gaussian surface is aligned with the axis of the cylinders.
By symmetry, the electric field is radial and has the same magnitude at every point on the Gaussian surface. Thus, we can write Gauss’s Law as:
where $S$ is the Gaussian surface, $d\mathbf{A}$ is a differential element of area on the surface, $\mathbf{E}$ is the electric field, $Q_{enc}$ is the total charge enclosed by the surface, and $\epsilon_0$ is the permittivity of free space.
The left-hand side of this equation can be simplified to:
where $E(r)$ is the magnitude of the electric field at radius $r$, and the integral on the right-hand side is the total area of the cylindrical surface.
The charge enclosed by the Gaussian surface is equal to the charge on the inner cylinder, $Q$, since there is no charge outside the inner cylinder:
$$Q_{enc} = Q$$
Substituting these expressions into Gauss’s Law, we get:
$$E(r) \cdot 2\pi rl = \frac{Q}{\epsilon_0}$$
Solving for the electric field magnitude $E(r)$, we get:
$$E(r) = \frac{Q}{2\pi\epsilon_0 rl}$$
The direction of the electric field is radial, pointing outward from the inner cylinder and inward towards the outer cylinder. Therefore, the electric field points in the $+r$ direction for $a<r<b$, and in the $-r$ direction for $r>b$ and $r<a$.
问题 3.
The voltage difference between the cylinders, $\Delta V$, is defined to be the work done per test charge against the electric field in moving a test charge $q_t$ from the inner cylinder to the outer cylinder $$ \Delta V \equiv V(a)-V(b)=-\int_b^a \overrightarrow{\mathbf{E}} \cdot d \overrightarrow{\mathbf{s}} \text {. } $$ Find an expression for the voltage difference between the cylinders in terms of the charge $Q$, the radii $a$ and $b$, the height $l$, and any other constants that you may find necessary.
证明 .
Let’s assume that the cylinders are coaxial and have a uniform charge distribution with charge density $\rho$.
From Gauss’s Law, the electric field inside a cylinder with uniform charge density $\rho$ is given by:
$E(r)=\frac{\rho}{2 \epsilon_0} r$, for $r \leq a$
where $\epsilon_0$ is the electric constant (permittivity of free space).
Similarly, the electric field outside the cylinder is given by:
$E(r)=\frac{\rho a^2}{2 \epsilon_0 r}$, for $a \leq r \leq b$
Therefore, the voltage difference between the cylinders can be expressed as: \begin{align*} \Delta V &= -\int_b^a \overrightarrow{\mathbf{E}} \cdot d \overrightarrow{\mathbf{s}} \ &= -\int_b^a E(r) dr \ &= -\int_b^a \frac{\rho a^2}{2\epsilon_0 r} dr \ &= \frac{\rho a^2}{2\epsilon_0} \ln{\frac{b}{a}} \end{align*}
where we have used the fact that the integral of $1/r$ is $\ln{r}$.
Now, we can express the charge density $\rho$ in terms of the total charge $Q$ and the volume of the cylinder:
$\rho=\frac{Q}{\pi l\left(b^2-a^2\right)}$
Substituting this expression for $\rho$ in the equation for $\Delta V$, we get: \begin{align*} \Delta V &= \frac{Q}{2\pi\epsilon_0 l} \frac{a^2}{b^2 – a^2} \ln{\frac{b}{a}} \ &= \frac{Q}{2\pi\epsilon_0 l} \ln{\frac{b}{a}} \left(\frac{a}{b+a}\right) \end{align*}
Therefore, the voltage difference between the cylinders is given by $\Delta V = \frac{Q}{2\pi\epsilon_0 l} \ln{\frac{b}{a}} \left(\frac{a}{b+a}\right)$, in terms of the total charge $Q$, the radii $a$ and $b$, the height $l$, and the electric constant $\epsilon_0$.
这是一份2023年的莱斯特大学University of Leicester PA1130电磁学代写的成功案例
