If the absolute pressure of a gas is 550.280 kPa, its gage pressure is
<span>
a. 101.325 kPa.
b. 448.955 kPa.
c. 651.605 kPa.
d. 277.280 kPa.</span>
The answer is B.
Answer:
Second drop: 1.04 m
First drop: 1.66 m
Explanation:
Assuming the droplets are not affected by aerodynamic drag.
They are in free fall, affected only by gravity.
I set a frame of reference with the origin at the nozzle and the positive X axis pointing down.
We can use the equation for position under constant acceleration.
X(t) = x0 + v0 * t + 1/2 * a *t^2
x0 = 0
a = 9.81 m/s^2
v0 = 0
Then:
X(t) = 4.9 * t^2
The drop will hit the floor when X(t) = 1.9
1.9 = 4.9 * t^2
t^2 = 1.9 / 4.9

That is the moment when the 4th drop begins falling.
Assuming they fall at constant interval,
Δt = 0.62 / 3 = 0.2 s (approximately)
The second drop will be at:
X2(0.62) = 4.9 * (0.62 - 1*0.2)^2 = 0.86 m
And the third at:
X3(0.62) = 4.9 * (0.62 - 2*0.2)^2 = 0.24 m
The positions are:
1.9 - 0.86 = 1.04 m
1.9 - 0.24 = 1.66 m
above the floor
Explanation:
We will calculate the gravitational potential energy as follows.

= 
= 1164000 J
or, = 1164 kJ (as 1 kJ = 1000 J)
Now, we will calculate the change in potential energy as follows.

=
= 
= -873000 J
or, = -873 kJ
Thus, we can conclude that change in gravitational potential energy is -873 kJ.
Answer is B. ABAB. Hope it helped you, and have a great day.
-Charlie