# 线性模型和方差分析|Linear Models and the Analysis of Variance代写STAT0006代考

This is a course in building and using linear models, though we will also briefly discuss theanalysis of variance which is connected with linear models. Linear models provide a veryflexible and useful way of analysing data, and though this is an introductory course (so we’llonly see one very specific type of model), it should give you a suite of tools that you can usefor future data analysis or even as a basis for more advanced modelling courses.The course is a practical one, and though there is some maths to read and digest this is notthe central theme of the course.

Steps
(a) 1. The ball is moving in a horizontal circle at constant speed. The acceleration is in the centripetal direction.
(b) 1. Draw a free-body diagram for the ball (Figure 5-29). Choose as the $+x$ direction the direction of the ball’s acceleration (toward the center of the circular path).

1. Apply $\sum F_{y}=m a_{y}$ to the ball and solve for the tension $T$.
(c) 1. Apply $\sum F_{x}=m a_{x}$ to the ball.
2. Substitute $m g / \cos \theta$ for $T$ and solve for $v$.

The acceleration is horizontal and directed from the ball toward the center of the circle it is moving in.
$$\Sigma F_{y}=m a_{y} \Rightarrow T \cos \theta-m g=0$$
so $T=\frac{m g}{\cos \theta}$
\begin{aligned} &\Sigma F_{x}=m a_{x} \Rightarrow T \sin \theta=m \frac{v^{2}}{r} \ &\frac{m g}{\cos \theta} \sin \theta=m \frac{v^{2}}{r} \Rightarrow g \tan \theta=\frac{v^{2}}{r} \end{aligned}
$$\operatorname{se} v=\sqrt{\sqrt{r \tan 0}}$$

TAKING IT FURTHER This example involves only kinematic parameters, so the resulting relation is a strictly kinematic relation. The equation $\vec{a} \cdot \vec{v}=\frac{1}{2} d\left(v^{2}\right) / d t$ has unrestricted validity (unlike some kinematic equations we have studied that are valid only if the acceleration remains constant).
$$\vec{a} \cdot \vec{v}=\frac{1}{2} \frac{d}{d t} v^{2}=\frac{d}{d t}\left(\frac{1}{2} v^{2}\right)$$