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:
The value is
Explanation:
From the we are told that
The initial speed of the object is
The greatest height it reached is 
Generally from kinematic equation we have that

At maximum height v = 0 m/s
So

=> 
Here H is the height from the initial height to the maximum height
So the initial height is mathematically represented as

=> 
=> 
Generally the time taken for the object to reach maximum height is mathematically evaluated using kinematic equation as follows

At maximum height v = 0 m/s

=> 
Generally the time taken for the object to move from the maximum height to the ground is mathematically using kinematic equation as follows

Here the initial velocity is 0 m/s given that its the velocity at maximum height
Also g is positive because we are moving in the direction of gravity
So

=> 
Generally the total time taken is mathematically represented as

=> 
=>
Answer:
Anticlockwise directions
Please mark me Brainliest to help me
work is distance * force so 15*100=1500
and to find time you know power = diastance * force / time
so 25=15*100/t
25=1500/t
25/1500=t
.016=time
Answer:
because only two electrons can fit in the first orbit around the nucleus, and each period on the table is organized by number of orbits