## College Physics (4th Edition)

(a) $21,300~Pa$ (b) $3.09~lb/in^2$ (c) $0.21~atm$ (d) $160~torr$
(a) We can convert 160 mm Hg to units of Pa: $160~mm~Hg \times \frac{133.3~Pa}{1~mm~Hg} = 21,300~Pa$ (b) We can convert 160 mm Hg to units of $lb/in^2$: $160~mm~Hg \times \frac{1~lb/in^2}{51.7~mm~Hg} = 3.09~lb/in^2$ (c) We can convert 160 mm Hg to units of atm: $160~mm~Hg \times \frac{1~atm}{760~mm~Hg} = 0.21~atm$ (d) We can convert 160 mm Hg to units of torr: $160~mm~Hg \times \frac{1.0~torr}{1~mm~Hg} = 160~torr$