Question

Please correct me if there is anything wrong on my equation.

ConstantsI Periodic Table- Part A A raft is made of 10 logs lashed together. Each is 40 cm in diameter and has a length of 6.0 m How many people can the raft hold before they start getting their feet wet, assuming the average person has a mass of 69 kg? Do not neglect the weight of the logs. Assume the specific gravity of wood is 0.60. Express your answer using two significant figures. Submit X Incorrect, Try Again; 3 attempts remaining < Return to Assignment Provide Feedback

[ (10 * pi * ((40 * (1/100))^2) * (6.0) * (1000-(0.6*1000)) ] / 69 = 174.8 which will be 170 because the answer has to be two sigfig.

Answer 1

The volume of the logs is:

10pi * (0.40 / 2)^2 * 6

= 7.54 m^3

The mass of the logs an n people is:

m1 = 7.54 * 10^3 * 0.60 + 69n kg.

The mass of water displaced is:

m2 = 7.54 * 10^3 kg.

Feet will remain dry provided:

m1 < m2

n < 7.54 (1 - 0.60) * 10^3 / 69

n < 43.7

The maximum number of people is 43