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.

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线性模型和方差分析|Linear Models and the Analysis of Variance代写STAT0006代考
问题 1.


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 $v$.


证明 .

Answers
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 $T=\frac{m g}{\cos \theta}$
$$
\begin{aligned}
&\Sigma F_{x}=m a_{x} \Rightarrow T \sin \theta=m \frac{v^{2}}{r} \
&\frac{m g}{\cos \theta} \sin \theta=m \frac{v^{2}}{r} \Rightarrow g \tan \theta=\frac{v^{2}}{r}
\end{aligned}
$$
$$
\operatorname{se} v=\sqrt{\sqrt{r \tan 0}}
$$

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STAT0006 COURSE NOTES :

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).
From Example 6-9 we have the kinematic relation
$$
\vec{a} \cdot \vec{v}=\frac{1}{2} \frac{d}{d t} v^{2}=\frac{d}{d t}\left(\frac{1}{2} v^{2}\right)
$$
In Section 6-4, this equation is used to derive the work-kinetic-energy theorem for particles moving along curved paths under the influence of forces that are not necessarily constant.



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