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
The best and most correct answer among the choices provided by your question is the first choice or letter A.
Heat transfers associated with phase changes known as latent or "hidden" heats because h<span>eat absorbed or released in a phase change is measured in kJ while temperature is measured in °C.</span>
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Answer:
B
Explanation:
nothing to do with black holes creating star or related
Answer:
Relativistic velocity is of the order of 1/10th of the velocity of light
Explanation:
We define relativistic speed (or velocity) as a speed that is a significant fraction of the speed of light: c = 3*10^8 m/s
Such that for these speeds, the special relativity theory starts to apply (the relativity effects starts to apply).
Usually, we define relativistic speeds as those that are of the order (or larger) of c/10, which is one-tenth of the speed of light.
Then the correct option is C:
Relativistic velocity is of the order of 1/10th of the velocity of light
