10판/3. 벡터

3-4 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 8. 6. 15:55

(a) 20.0° 20.0=20.0π[rad]180=π9[rad]0.349[rad] 20.0^\circ =20.0^\circ \cdot \frac{\pi\ut{rad}}{180^\circ}=\frac{\pi}{9}\ut{rad}\approx0.349\ut{rad}
(b) 50.0° 50.0=50.0π[rad]180=518π[rad]0.873[rad] 50.0^\circ =50.0^\circ \cdot \frac{\pi\ut{rad}}{180^\circ}=\frac{5}{18}\pi\ut{rad}\approx0.873\ut{rad}
(c) 100° 100=100π[rad]180=59π[rad]1.75[rad] 100^\circ =100^\circ \cdot \frac{\pi\ut{rad}}{180^\circ}=\frac{5}{9}\pi\ut{rad}\approx1.75\ut{rad}
(d) 0.330 rad 0.330[rad]=0.330[rad]180π[rad]=2975π18.9 0.330\ut{rad} = 0.330\ut{rad} \cdot \frac{180^\circ}{\pi\ut{rad}}=\frac{297}{5\pi}^\circ\approx18.9^\circ
(e) 2.30 rad 2.30[rad]=2.30[rad]180π[rad]=414π132 2.30\ut{rad} = 2.30\ut{rad} \cdot \frac{180^\circ}{\pi\ut{rad}}=\frac{414}{\pi}^\circ\approx132^\circ
(f) 7.70 rad 7.70[rad]=7.70[rad]180π[rad]=1386π441 7.70\ut{rad} = 7.70\ut{rad} \cdot \frac{180^\circ}{\pi\ut{rad}}=\frac{1386}{\pi}^\circ\approx441^\circ