Transformer contains two coils: primary and secondary. They allow change of voltage to lower or higher value. In first case we have step-down and in second case we have step-up transformer.
Formula used for transformer is:

Where:N1 = number of turns on primary coilN2 = number of turns on secondary coilV1 = voltage on primary coilV2 = voltage on secondary coil
In a step-down transformer primary coil has more turns than secondary coil. So the ratio 1:38 means that for each turn on secondary coil we have 38 turns on primary coil.
We can solve the equation for V2:

Secondary coil provides voltage of 3.16V.
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:
Explanation:
Height of building
H = 6m
Horizontal speed of first balloon
U1x = 2m/s
Second ballot is thrown straight downward at a speed of
U2y = 2m/s
Time each gallon hits the ground
Balloon 1.
Using equation of free fall
H = Uoy•t + ½gt²
Uox = 0 since the body does not have vertical component of velocity
6 = ½ × 9.8t²
6 = 4.9t²
t² = 6 / 4.9
t² = 1.224
t = √1.224
t = 1.11 seconds
For second balloon
H = Uoy•t + ½gt²
6 = 2t + ½ × 9.8t²
6 = 2t + 4.9t²
4.9t² + 2t —6 = 0
Using formula method to solve the quadratic equation
Check attachment
From the solution we see that,
t = 0.9211 and t = -1.329
We will discard the negative value of time since time can't be negative here
So the second balloon get to the ground after t ≈ 0.92 seconds
Conclusion
The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.
Newtons 3.law: Action = Reaction
If a body exerts a force on a rope of 400 N the rope exerts a force on the body of 400N also. So the tension in the rope is 400N. See pictures below.
I’ve done this before the answer is B