In the manufacture of paper, logs are cut into small chips, which are stirred into an alkaline solution that dissolves several of the chemical constituents of wood but not cellulose. the slurry of undissolved chips in solution is further processed to recover most of the original solution constituents and dried wood pulp. in one such process, wood chips with a specific gravity of 0.640 containing 45.0 wt% water are treated to produce 2000.0 tons/day of dry wood pulp containing 85.0 wt% cellulose. the wood chips contain 47.0 wt% cellulose on a dry basis. estimate the feed rate of logs (logs/min), assuming that the logs have an average diameter of 8.00 inches and an average length of 9.00 feet. 21.67 ulogs/min

The estimated feed rate of logs is 14.3 logs/min.

Explanation:

The product of the process is 2000 tons/day of dry wood pulp, of 85 wt% of cellulose. That represents (2000*0.85)=1700 tons/day of cellulose.

That cellulose has to be feed by the wood chips, which had 47 wt% of cellulose in its composition. That means you need (1700/0.47)=3617 tons/day of wood chips to provide all that cellulose.

Th entering flow is wood chips with 45 wt% of water. This solution has an specific gravity of 0.640.

To know the specific gravity of the wood chips we have to write a volume balance. We also know that Mw=0.45*M and Mc=0.55*M.

The specific gravity of the wood chips is 0.494.

The average volume of a log is

The weight of one log is

To provide 3617 ton/day of wood chips, we need

The feed rate of logs is 14.3 logs/min.

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

it's c. elements are placed in rows based on how may valencies electrons have.