This question involves the concepts of dynamic pressure, volume flow rate, and flow speed.
It will take "5.1 hours" to fill the pool.
First, we will use the formula for the dynamic pressure to find out the flow speed of water:

where,
v = flow speed = ?
P = Dynamic Pressure = 55 psi
= 379212 Pa
= density of water = 1000 kg/m³
Therefore,

v = 27.54 m/s
Now, we will use the formula for volume flow rate of water coming from the hose to find out the time taken by the pool to be filled:

where,
t = time to fill the pool = ?
A = Area of the mouth of hose =
= 1.98 x 10⁻⁴ m²
V = Volume of the pool = (Area of pool)(depth of pool) = A(1.524 m)
V =
= 100.1 m³
Therefore,

<u>t = 18353.5 s = 305.9 min = 5.1 hours</u>
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<h2>
Entire trip takes 1.22 seconds.</h2>
Explanation:
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Time, t = 0.866 s
Substituting
s = ut + 0.5 at²
s = 0 x 0.866 + 0.5 x 9.81 x 0.866²
s = 3.68 m
Halfway is 3.68 m
Total height = 2 x 3.68 = 7.36 m
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Time, t = ?
Displacement, s = 7.36 m
Substituting
s = ut + 0.5 at²
7.36 = 0 x t + 0.5 x 9.81 x t²
t = 1.22 s
Entire trip takes 1.22 seconds.
Change in temperature = final temperature - Initial temperature
Δt = t₂ - t₁
Δt = 17 - (-6)
Δt = 17 + 6 = 23 f
In short, Your Answer would be Option D
Hope this helps!
Answer:
a ) 11.1 *10^3 m/s = 39.96 Km/h
b) T_{o2} =1.58*10^5 K
Explanation:
a)
= 11.1 km/s =11.1 *10^3 m/s = 39.96 Km/h
b)
M_O2 = 32.00 g/mol =32.0*10^{-3} kg/mol
gas constant R = 8.31 j/mol.K

So, 
multiply each side by M_{o2}, so we have

solving for temperature T_{o2}

In the question given,

T_{o2} =1.58*10^5 K