Complete question:
A college dormitory room measures 14 ft wide by 13 ft long by 6 ft high. Weight density of air is 0.07 lbs/ft3. What is the weight of air in it under normal conditions?
Answer:
the weight of the air is 76.44 lbs
Explanation:
Given;
dimension of the dormitory, = 14 ft by 13 ft by 6 ft
density of the air, = 0.07 lbs/ft³
The volume of the air in the dormitory room = 14 ft x 13 ft x 6 ft
= 1092 ft³
The weight of the air = density x volume
= 0.07 lbs/ft³ x 1092 ft³
= 76.44 lbs
Therefore, the weight of the air is 76.44 lbs
Answer:
a) 
b) The second runner will win
c) d = 10.54m
Explanation:
For part (a):

For part (b) we will calculate the amount of time that takes both runners to cross the finish line:


Since it takes less time to the second runner to cross the finish line, we can say the she won the race.
For part (c), we know how much time it takes the second runner to win, so we just need the position of the first runner in that moment:
X1 = V1*t2 = 239.46m Since the finish line was 250m away:
d = 250m - 239.46m = 10.54m
Please find attached photograph for your answer. Please do comment whether it is useful or not.
The answer that is got 8.7 . I got that because if you divide 200 by 23 you get <span>8.69565217391 and if you round that you get 8.7</span>
Well 200 doubled or (x2)=400 if that’s what it means