Answer:
32.1 g
Step-by-step explanation:
In each 3 grams of C, there are 2 grams of A and 1 gram of B. So, for some amount C, the amount remaining of A is 40 -(2C/3), and the amount remaining of B is (50 -C/3). Since the reaction rate is proportional to the product of these amounts, we have ...
C' = k(40 -2C/3)(50 -C/3) = (2k/9)(60 -C)(150 -C) . . . for some constant k
This is separable differential equation with a solution of the form ...
ln((150 -C)/(60 -C)) = at + b
We know that C(0) = 0, so b=ln(150/60) = ln(2.5). And we know that C(10) = 20, so ln(130/40) = 10a +ln(2.5) ⇒ a = ln(1.3)/10
Then our equation for C is ...
ln((150 -C)/(60 -C)) = t·ln(1.3)/10 +ln(2.5)
__
For t=20, this is ...
ln((150 -C)/(60 -C)) = 2ln(1.3) +ln(2.5) = ln(2.5·1.3²) = ln(4.225)
Taking antilogs, we have ...
(150 -C)/(60 -C) = 4.225
1 +90/(60 -C) = 4.225
C = 60 -90/3.225 ≈ 32.093 . . . . . grams of product in 20 minutes
In 20 minutes, about 32.1 grams of C are formed.
, where
is the final velocity in a direction after the acceleration is applied,
is the initial velocity in that direction before the acceleration is applied, a is the acceleration applied in such direction, and t is the amount of time during where that acceleration was applied.
5320 m/s, the acceleration (the applied by the engine) in x direction is
1.79 m/s2 and, the time during the acceleration was applied (the time during the engines were fired) of the is 739 s. Then:
0 m/s, the acceleration in y direction is
7.18 m/s2, and the time is the same that in the x direction, 739 s, then for the final velocity in the y direction: