MY-ASSIGNMENTEXPERT™可以为您提供cornell.edu PHYS7635 Solid Physics固体物理课程的代写代考和辅导服务！
PHYS7635课程简介
Description
Survey of the physics of solids starting with crystal structures and the band theory of electrons and phonons. Selected topics from semiconductors, magnetism, superconductivity, spin liquids, disordered materials, topology, and mesoscopic physics. The focus is to enable graduate research at the current frontiers of condensed matter physics. Students unable to enroll in open class components during their pre-enroll window can join the waitlist at www.physics.cornell.edu/waitlist. Students will be placed in available spots prior to the start of classes when possible.
Credits
3
Prerequisites
Course information provided by the Courses of Study 2019-2020.
Survey of the physics of solids: crystal structures, X-ray diffraction, phonons, and electrons. Selected topics from semiconductors, magnetism, superconductivity, disordered materials, topological materials, and mesoscopic physics. The focus is to enable graduate research at the current frontiers of condensed matter physics. In addition to the course lectures, students are expected to attend either the LASSP/AEP seminar at 12:20 pm on Tuesdays or the weekly research seminar for their home department. An optional study hall/homework section will be held 2-3 pm on Thursdays.
PHYS7635 Solid Physics HELP（EXAM HELP， ONLINE TUTOR）
A steel wire of length $4.7 \mathrm{~m}$ and cross-sectional area $3.0 \times 10^{-5} \mathrm{~m}^2$ stretches by the same amount as a copper wire of length $3.5 \mathrm{~m}$ and cross-sectional area of $4.0 \times 10^{-5} \mathrm{~m}^2$ under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
Answer:
For steel
$$
l_1=4.7 \mathrm{~m}, \quad A_1=3.0 \times 10^{-5} \mathrm{~m}^2
$$
If $F$ newton is the stretching force and $\Delta l$ metre the extension in each case, then
$$
\begin{aligned}
& Y_1=\frac{F l_1}{A_1 \Delta l} \
\Rightarrow & Y_1=\frac{F \times 4.7}{3.0 \times 10^{-5} \times \Delta l} \
\text { For copper } & l_2=3.5 \mathrm{~m}, A_2=4.0 \times 10^{-5} \mathrm{~m}^2 \
\text { Now, } & Y_2=\frac{F \times 3.5}{4.0 \times 10^{-5} \times \Delta l}
\end{aligned}
$$
Dividing (i) by (ii), we get
$$
\frac{Y_1}{Y_2}=\frac{4.7}{3.0 \times 10^{-5}} \times \frac{4.0 \times 10^{-5}}{3.5}=\frac{4.7 \times 4.0}{3.0 \times 3.5}=1.79 .
$$
Question 9. 2. Figure shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material?
Answer:
(a) 材料的杨氏模量 $(Y)$ 由
$$
\begin{aligned}
& Y=\text { 应力/应变 } \
& =150 \times 10^6 / 0.002 \
& 150 \times 10^6 / 2 \times 10-3 \
& =75 \times 10^9 \mathrm{Nm}^{-2} \
& =75 \times 10^{10} \mathrm{Nm}^{-2}
\end{aligned}
$$
(a) 材料的屈服强度定义为它可以承受的最大应力。从图中可以看出，给定材料的近似屈 服强度
$$
\begin{aligned}
& =300 \times 10^6 \mathrm{Nm}^{-2} \
& =3 \times 10^8 \mathrm{Nm}^{-2} 。
\end{aligned}
$$
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