<p>From chap 1 on measurements:</p>
<ul>
<li><p>A vertical container with base area measuring 14 cm by 17 cm is being filled with identical pieces of candy, each with a volume of 50 mm^3 and a mass of 0.0200 g. Assume that the volume of the empty spaces between the candy is negligible. If the height of the candies in the container increases at the rate of 0.250 cm/s, at what rate (kg/min) does the mass of the candies in the container increase? </p></li>
<li><p>I have no idea if this is a related rates problem or how to start it, does anyone know how to do this? The answer is supposed to be 1.43 kg/min.</p></li>
</ul>
<p>This is not a related rates problem or calc at all. Just look at the chapter name!
Ok, to start with, the base area is 14 by 17 cm. I’m assuming that the height of candies will increase only if they occupy the base area first. Kinda like when filling a tank of water. Before the height of the water increases, the bottom part must be filled up first.
That being said, since the height increases only if the base area is filled up first, we can convert the rate shown. Instead if the height, we can convert to the volume of candies in the container, since the base area is uniform throughout the container.
So we multiply the height rate by the base area, .25 x 14 x 17, giving us
59.5 cm^3/s as the volume rate. We then convert the volume into mass, because each candy has a mass of .02g for every 50 mm^3. So we convert 59.5 cm^3/s to mm^3/s first, giving us 59500mm^3/s, then multiply by
.02g/50mm^3, giving us 23.8g/s
Finally, we convert to kg/min, which should be multiplied by 60 then divided by 1000, giving us 1.43kg/min</p>
<p>Thank you so much for the help. This is for the 8th edition and the solution manual sucks. If anyone has a different approach/check, I would like to see as well.</p>
<p>All right, you should know that 10 mm = 1cm, so (10 mm )^3 = (1cm)^3
and 1000 mm^3 = 1 cm^3. We multiply 59.5 cm^3/s by 1000 mm^3/ 1 cm^3, cancelling out the cm^3</p>
<p>For the 02g/50mm^3 part, since .02g of candies have a volume of 50mm^3, we could say for this situation that .02g = 50mm^3.</p>
<p>We multiply by 0.02g/50mm^3 to get rid of the mm^3 unit (Technically we didn’t change anything except the units because 0.02g/50mm^3 is equal to 1) and convert to grams/second</p>
<p>Got it, thanks again for the help!</p>