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Joined on May 27, 2020

  • 專題說明 $\qquad$碰撞是高中物理的一部份,也是生活常見到的物理現象,請觀看以下以片,我們接下來來探討影片中的現象,便對其關係做延伸討論 影片網址:https://www.youtube.com/watch?v=HEfHFsfGXjs {%youtube HEfHFsfGXjs %} 分析原因 $\qquad$讓我們開始分析為何會有這樣的結果。首先,速度以右為正,並且我們標記最右側(質量較大)的方塊$a$質量為$m_1$,中間的方塊$b$質量為$m_2$,$a、b$接下來的速度標記為$v_{a_k}、v_{b_k}$以及初始速度為$v_{a_i}$,碰撞過程中變化的速度,並且假設全成為完全彈性碰撞。 $\qquad$根據動量守恆,我們在接下來的碰撞過程中可以得到下列結果。 \begin{gather*} m_1v_{a_k}+m_2v_{b_k}=m_1v_{a_i}
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  • # 材料力學
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  • # 流體力學
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  • Problem 1-9 ans: $90km/h=25m/s,d=25\times 0.5=12.5m$ ans: $(a)v=\frac{2\times 73.2}{\frac{73.2}{1.22}\ + \frac{73.2}{2.85}}=\frac{146.4}{85.684}=1.709m/s$ $(b)v=\frac{1.22\times 60+3.05\times 60}{120}=2.135m/s$
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  • Problem 1-6 Earth is approximately a sphere of radius $6.37\times 10^6m$. what are (a) its circumference in kilometers, (b) its surface area in square kilometers,and (c) its volume in cubic kilometers? formula : $circumference=2r\pi、sureface\ area=4r^2\pi、volume=\frac{4}{3}r^3\pi$ Ans:(a)$4.002\times 10^4km$、(b)$5.099\times 10^8km^2$、(c)$1.083\times 10^{12}$ A $gry$ is an old English measure for length, defined as 1/10 of a $line$, where $line$ is another English measure for length, defined as 1/12 inch. A common measure for length in the publishing business is a $point$, defined as 1/72 inch. What is an area of 0.75 $gry^2$ in $points\ squared$ ($points^2$) $1\ gry=1/10\ line=1/120\ inch=3/5\ point\rightarrow 0.75\ gry^2=0.75*(3/5)^2points^2=0.27points^2$
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  • 名詞介紹 CPU(central processing unit) 電腦的主要裝置之一,功能主要是解釋電腦指令以及處理電腦軟體中的資料。電腦的可程式化性主要是指對中央處理器的編程,廣義上指一系列可以執行複雜的電腦程式的邏輯機器。 CPU具備下列功能: 抓取指令:CPU必須具備到記憶體抓取指令的功能。 解譯指令:CPU必須看得懂指令以便採取動作。 抓取資料:當CPU執行指令時, ++必須所有的運算元(operands)皆到齊++, 因此,它必須有從I/O設備或記憶體抓取資料的功能。 處理資料:當資料齊備後, CPU必須能處理這些資料, 可能是算術運算(加、減、乘、除), 也可能是邏輯運算(AND、OR等)。
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  • 名詞介紹 布林代數(Boolean algebra) 能呈現兩種數值(0與1)其中一種的代數形式。 訊號邊緣(signal edge) 或稱訊號邊沿,是數位訊號在兩種邏輯準位(0或1)之間狀態的轉變。由於數位訊號準位由方波來表示,因此這種狀態的變化被稱為「邊緣」。 競逐狀態 電位輸出訊號不穩定的狀態
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  • #include<stdio.h> #include<stdlib.h> # define N 20 void initial(double [N][N] ,double [N][N], double* ); void clear(double[N][N],double [N][N],double [N]); void clear2(double[N][N],double [N][N], double[N]); double matrixsolver(double[N][N], double [N][N],double[N]); int main() { double u = 0; double A[N][N] = { {} };
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  • ###### tags: `introduction of computer` # 正反器
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  • #include <iostream> using namespace std; int num_array[100]; int kick_array[100]; int t = 100; //表示剩下幾個數 void Num_array(); void rotation();
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  • #include <iostream> #include <math.h> #define N 20 using namespace std; double A[N][N] = {}; double e[N][N] = {}; //column of Q 等於Q的轉置 double R[N][N] = {}; double f[N] = {}; //f=Rx=Q^t乘上F double F[N] = {};
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  • #include<stdio.h> #include<stdlib.h> #include<math.h> #define N 4 int main() { int A[N] = { 50,10,5,1 }; int j = 0; for (int a = 0; a < 2; a++) { for (int b = 0; b < 7; b++) { for (int c = 0; c < 14; c++) {
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  • hw-10-(2)-(b) Question: $\qquad$Let $A$ be a Hermitian matrix on $M_{n\times n}(\mathbb{F})$, $\lambda_{i}$ be eigenvalues of $A$ and $E_{\lambda_{i}}$ be the corresponding eigenspace of $\lambda_{i}$. $\qquad$Let $P_{\lambda_i}$ : $\mathbb{R}^{3}\rightarrow\mathbb{R}^{3}$ be the orthogonal projection of $\mathbb{R}^{3}$ onto $E_{\lambda_i}$. $\qquad$Show that $A=\sum_{\lambda \in\sigma(A)}\lambda P_{\lambda}$ . Here $\sigma(A)$ is the set of all eigenvalue of $A$ . Solution: $\qquad$Because $A$ is a Hermitian matrix , $A$ can be diagonalized like $Q\Lambda Q^{}$ . $Q$ is unitary matrix and $\Lambda$ is diagonal matrix . $\qquad$let $Q=\begin{pmatrix}q_1,q_2,q_3,\cdots,q_n\end{pmatrix}$ then $Q^=Q^{-1}=\begin{pmatrix}q_1^\q_2^\q_3^\ \vdots\q_n^\end{pmatrix}$ and $A=\begin{pmatrix}q_1,q_2,q_3,\cdots,q_n\end{pmatrix}\begin{pmatrix}\lambda_1\&\ddots\&&\lambda_n\end{pmatrix}\begin{pmatrix}q_1^\q_2^\q_3^\ \vdots\q_n^\end{pmatrix}$ $\rightarrow A=\begin{pmatrix}\lambda_1 q_1,\lambda_2 q_2,\lambda 3q_3,\cdots,\lambda_nq_n\end{pmatrix}\begin{pmatrix}q_1^\q_2^\q_3^\ \vdots\q_n^\end{pmatrix}=\sum{i=1}^n\lambda_iq_i q_i^$
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  • #include<stdio.h> #include<stdlib.h> #include<time.h> /*number of times of test*/ #define N 10000 /*number2 of times of test*/ #define n 1000 /*width of interval*/ #define I 5 int main() {
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  • #include<stdio.h> #include<stdlib.h> #include<math.h> #define N 8 void initial(double [N][N]); void matrixmulvector(double [N][N], double [N]); double computernorm2(double [N]); double linearlyapproach(double [N][N],double [N][N],double [N],double [N]); int main() { double A[N][N] = { {} };
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  • #include<stdio.h> #include<stdlib.h> #define N 100 #define n 2 int valuefind(int ,int*, int*, int*); int main(){ /* use the array to display the polynomial*/ int a[N] = {0}; int b[n] = {0}; int c[N] = {0};
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  • #include<stdio.h> #include<stdlib.h> #include<math.h> #define N 5 #define A 50 void ref(int ,int [],int []); void rot(int ,int [],int []); void refandrot(int,int[],int [],int[]); void rxry(int,int,int[],int[],int[]); int main() {
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  • #include<stdio.h> #include<stdlib.h> #define I 4 #define J 4 double matrixmul(double [I][J],double [I][J]); int main() { double A[I][J] = { {0.1,0.2,0.4,0.3},{0.5,0.2,0.1,0.2},{0.3,0.4,0.2,0.1},{0.15,0.25,0.2,0.4} }; double B[I][J] = { {0} }; /*give value of entries of B[I][J]*/
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