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
Heat normally travels from areas of higher heat to areas of lower heat. For example, if you were to be in a classroom and all the windows and doors were closed, and then you opened a door, then the reason why the room becomes cooler is not because cold air comes inside the room. However, it is because when you open the door, the heat from the classroom you're in escapes the room. This is why it feels cooler when you open a door, or even a window. Heat is traveling from areas of higher heat to areas of lower heat.
Answer:
1.65 h
121.39 km
Explanation:
Given that
speed of the driver, v = 99.5 km/h
time spent resting, t = 26 min
Average speed of the driver = 73.6 km/h
check attachment for calculation and how I arrived at the answer