First question: 800J
Second question: 20.4m
Explanation:
First we will convert the given mass from lb to kg as follows.
157 lb = 
= 71.215 kg
Now, mass of caffeine required for a person of that mass at the LD50 is as follows.

= 12818.7 mg
Convert the % of (w/w) into % (w/v) as follows.
0.65% (w/w) = 
= 
= 
Therefore, calculate the volume which contains the amount of caffeine as follows.
12818.7 mg = 12.8187 g = 
= 1972 ml
Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.
Answer:
vf = 3.27[m/s]
Explanation:
In order to solve this problem we must analyze each body individually and find the respective equations. The free body diagram of each body (box and bucket) should be made, in the attached image we can see the free body diagrams and the respective equations.
With the first free body diagram, we determine that the tension T should be equal to the product of the mass of the box by the acceleration of this.
With the second free body diagram we determine another equation that relates the tension to the acceleration of the bucket and the mass of the bucket.
Then we equalize the two stress equations and we can clear the acceleration.
a = 3.58 [m/s^2]
As we know that the bucket descends 1.5 [m], this same distance is traveled by the box, as they are connected by the same rope.
![x = \frac{1}{2} *a*t^{2}\\1.5 = \frac{1}{2}*(3.58) *t^{2} \\t = 0.91 [s]](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2Aa%2At%5E%7B2%7D%5C%5C1.5%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2A%283.58%29%20%2At%5E%7B2%7D%20%5C%5Ct%20%3D%200.91%20%5Bs%5D)
And the speed can be calculated as follows:
![v_{f}=v_{o}+a*t\\v_{f}=0+(3.58*0.915)\\v_{f}= 3.27[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3Dv_%7Bo%7D%2Ba%2At%5C%5Cv_%7Bf%7D%3D0%2B%283.58%2A0.915%29%5C%5Cv_%7Bf%7D%3D%203.27%5Bm%2Fs%5D)
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