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
Answer:
Steven has to row at a speed to reach the same horizontal spot at the other side of the river is, V = 6 m/s
Explanation:
Given data,
The river flowing south at the rate, v = 3 m/s
To reach the other side directly across the river, he aims the raft, Ф = 30°
The speed of his raft across the river is given by the formula,
V = v / Sin Ф
= 3 / Sin 30°
= 6 m/s
Steven has to row at a speed to reach the same horizontal spot at the other side of the river is, V = 6 m/s
D. physical property
the bonds between molecules of mercury are breaking so it's physical and it's not changing the chemical composition of the substance
Dang dude you are a soldier! Good job