The shape is connected in parallel so;
5.1) Ans;

5.2) Ans;

I hope I helped you^_^
Answer:8.75 s,
136.89 m
Explanation:
Given
Initial velocity
velocity after 5 s is 
Therefore acceleration during these 5 s


therefore time required to stop
v=u+at
here v=final velocity =0 m/s
initial velocity =31.29 m/s


(b)total distance traveled before stoppage


s=136.89 m
V₁(O2) = 6.50<span> L
</span>p₁(O2) = 155 atm
V₂(acetylene) = <span>4.50 L
</span>p₂(acetylene) =?
According to Boyle–Mariotte law (At constant temperature and unchanged amount of gas, the product of pressure and volume is constant) we can compare two gases that have ideal behavior and the law can be usefully expressed as:
V₁/p₁ = V₂/p₂
6.5/155 = 4.5/p₂
0.042 x p₂ = 4.5
p₂ = 107.3 atm
Answer: 2.55meter
Explanation: Using the second equation of motion.
S{hieght} = U*t + {g*t²}/2
Where U is initial velocity =0m/s
g is acceleration due to gravity 10m/s²
t is time 1secs
So we have,
hieght = 0 + {g*t²}/2
hieght = {10*(1)²}/2
Total hieght travelled is 10/2
Which is 5 meter.
But we are asked to find the hieght above the window which as a hieght of 2.45meter.
So,
hieght above window would be
{5 - 2.45}meter
Which is 2.55 meter.
Answer:
i think it might be D or C