a) Atleta com velocidade constante: $\vec{F}_{\text {resultante }}=0$\n$||$ ao plano inclinado: $m g \operatorname{sen} \theta-\left|\vec{F}_{\text {atr }}\right|-\left|\vec{F}_{\text {res\_ar }}\right|=0$\n$\perp$ ao plano inclinado: $|\vec{N}|-m g \cos \theta=0$\nAlém disso: $\left|\vec{F}_{\text {atr }}\right|=\mu_{c}|\vec{N}|$\nDessa forma: $\left|\vec{F}_{\text {res\_ar }}\right|=m g \operatorname{sen} \theta-\mu_{c} m g \cos \theta=m g\left(\operatorname{sen} \theta-\mu_{c} \cos \theta\right)$\n$$ \begin{gathered} \left|\vec{F}_{\text {res\_ar }}\right|=(90 \mathrm{~kg}) \times\left(10 \mathrm{~m} / \mathrm{s}^{2}\right) \times[0,6-0,05 \times 0,8]=(900 \mathrm{~N}) \times(0,6-0,04)=(900 \mathrm{~N}) \times(0,56) \\ \left|\vec{F}_{\text {res\_ar }}\right|=504 \mathrm{~N} . \end{gathered} $$\nb)\n$\beta=(10 \mathrm{~dB}) \times \log \left(\frac{I\left(r_{0}\right)}{I\left(r_{1}\right)}\right)=40 \mathrm{~dB} \Rightarrow \log \left(\frac{I\left(r_{0}\right)}{I\left(r_{1}\right)}\right)=4 \Rightarrow \frac{I\left(r_{0}\right)}{I\left(r_{1}\right)}=10^{4}$ ou $I\left(r_{1}\right)=\frac{I\left(r_{0}\right)}{10^{4}}$\nSubstituindo $I\left(r_{1}\right)=\frac{c t e}{r_{1}^{2}}$ e $I\left(r_{0}\right)=\frac{c t e}{r_{0}^{2}}$:\n$\frac{c t e}{r_{1}^{2}}=\frac{c t e}{10^{4} r_{0}^{2}} \Rightarrow r_{1}=\sqrt{10^{4}} r_{0}=100 r_{0}=100 \times 0,6 \mathrm{~m}=60 \mathrm{~m}$.