Section 1.7 showed that in 1997 los angeles county air had carbon monoxide (co) levels of 15.0 ppm. an average human inhales about 0.50 l of air per breath and takes about 20 breaths per minute. how many milligrams of carbon monoxide does the average person inhale in an 7 hour period for this level of carbon monoxide pollution? assume that the carbon monoxide has a density of 1.2 g/l. (hint: 15.0 ppm co means 15.0 lco per 106l air.)

75.6 mg
       
First calculate how many liters of air an average person will breath. We have 0.5 l per breath, times 20 breaths per minute, times 60 minutes per hour, times 7 hours. 0.50 l * 20 * 60 * 7 = 4200 l Since there's 15 ppm CO, multiply the volume breathed by 15x10^-6, getting 4200 * 15x10^-6 = 0.063 l Now, multiply the 0.063 l of CO breathed by its density of 1.2 g/l 0.063 * 1.2 g/l = 0.0756 g Convert from grams to milligrams by multiplying by 1000 0.0756 * 1000 = 75.6 mg

The objects do not reach terminal velocity         there is not enough time from the time the object falls from when it hits the ground to achieve terminal velocity. an object must be placed higher to achieve terminal velocity in order to be in free fall.