The figure shows two 1.0 kg blocks connected by a rope. a second rope hangs beneath the lower block. both ropes have a mass of 250
g. the entire assembly is accelerated upward at 3.0 m/s2 by force fâ— .

We need first to use the formula  F=m(a+g), m iis the total mass, a is the acceleration, g is gravity pulling the blocks. So the procedure will be 
m=2kg(both blocks)+500g(both ropes) → m=2.5kg 
a=3.00m/s^2 
g=9.8m/s^2 
F=m(a+g) → F=2.5kg (3.00m/s^2 + 9.8m/s^2) → F=2.5kg (12.8m/s^2) → F=32 N 
To calculate the tension at the top of rope 1 you need to use the formula T=m(a+g) so it will be T=m(a+g) → T=1.5kg(12.8m/s^2) → T=19.2N 
We can now calculate the tension at the bottom of rope 1 using the formula: T=m(a+g) → T=1.25kg(12.8m/s^2) → T=16N 
Now to find the tension at the top of rope 2 we do it like this: 
T=m(a+g) → T=.25kg(12.8m/s^2) → T=3.2

Which of the following are risk factors for emphysema ? correct answer : a and and c          good luck and have a nice benevolentpuppy day! : )