If a current of one- or two-tenths of an ampere were to flow into one of your hands and out the other, you would probably be ele
2 answers:
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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:
so maximum velocity for walk on the surface of europa is 0.950999 m/s
Explanation:
Given data
legs of length r = 0.68 m
diameter = 3100 km
mass = 4.8×10^22 kg
to find out
maximum velocity for walk on the surface of europa
solution
first we calculate radius that is
radius = d/2 = 3100 /2 = 1550 km
radius = 1550 × 10³ m
so we calculate no maximum velocity that is
max velocity = √(gr) ...............1
here r is length of leg
we know g = GM/r² from universal gravitational law
so G we know 6.67 × N-m²/kg²
g = 6.67 × ( 4.8×10^22 ) / ( 1550 × 10³ )
g = 1.33 m/s²
now
we put all value in equation 1
max velocity = √(1.33 × 0.68)
max velocity = 0.950999 m/s
so maximum velocity for walk on the surface of europa is 0.950999 m/s