$\begin{array}{1 1}-13.5 ^{\large\circ}\\-23.7 ^{\large\circ}\\-26.8 ^{\large\circ}\\-11.3 ^{\large\circ}\end{array} $

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Answer : $-13.5 ^{\large\circ}$

With v = 6.00 m/s and r = 500 m, we find that the radial acceleration is

$a_r=-\large\frac{v^2}{r}$

$\Rightarrow -\large\frac{(6.00m/s)^2}{500m}$$=-0.0720m/s^2$

The radial acceleration vector is directed straight downward while the tangential acceleration vector has magnitude $0.300 m/s^2$ and is horizontal.

Because $a=a_r+a_t$,the magnitude of a is

$a=\sqrt{a_r^2+a_t^2}$

$\Rightarrow \sqrt{(-0.0720)^2+(0.300)^2}m/s^2$

$\Rightarrow 0.309m/s^2$

If $4\pi$ is the angle between a and the horizontal,then

$\phi=\tan^{-1} \large\frac{a_r}{a_t}$

$\Rightarrow \tan^{-1}\big(\large\frac{-0.0720m/s^2}{0.300m/s^2}\big)$

$\Rightarrow -13.5^{\large\circ}$

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