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2 years ago

How might the distance between the needle and the water droplets have affected the movement of the droplets?

Physics

2 answers:

ElenaW [278]2 years ago

8 0

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shutvik [7]2 years ago

5 0

Answer:

The behavior of droplets trapped in geometric structures is essential to droplet manipulation applications such as for droplet transport. Here we show that directional droplet movement can be realized by a V-shaped groove with the movement direction controlled by adjusting the surface wettability of the groove inner wall and the cross sectional angle of the groove. Experiments and analyses show that a droplet in a superhydrophobic groove translates from the immersed state to the suspended state as the cross sectional angle of the groove decreases and the suspended droplet departs from the groove bottom as the droplet volume increases. We also demonstrate that this simple grooved structure can be used to separate a water-oil mixture and generate droplets with the desired sizes. The structural effect actuated droplet movements provide a controllable droplet transport method which can be used in a wide range of droplet manipulation applications.

Explanation:

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What is force x time is called

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Answer:

In words, it could be said that the force times the time equals the mass times the change in velocity. In physics, the quantity Force • time is known as impulse. And since the quantity m•v is the momentum, the quantity m•Δv must be the change in momentum. The equation really says that the Impulse = Change in momentum.

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2 years ago

The sum of the potential energy and the kinetic energy of an object is its thermal energy. _______________ *

Law Incorporation [45]

Answer:

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Explanation:

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2 years ago

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2 years ago

Read 2 more answers

A 4.0-m-diameter playground merry-go-round, with a moment of inertia of

HACTEHA [7]

Answer:

$7.1$ ms⁻¹

Explanation:

$d$ = diameter of merry-go-round = 4 m

$r$ = radius of merry-go-round = $\frac{d}{2}$ = $\frac{4}{2}$ = 2 m

$I$ = moment of inertia = 500 kgm²

$w_{i}$ = angular velocity of merry-go-round before ryan jumps = 2.0 rad/s

$w_{f}$ = angular velocity of merry-go-round after ryan jumps = 0 rad/s

$v$ = velocity of ryan before jumping onto the merry-go-round

$m$ = mass of ryan = 70 kg

Using conservation of angular momentum

$Iw_{i} - m v r = (I + mr^{2})w_{f}$

$(500)(2.0) - (70) v (2) = (I + mr^{2})(0)$

$1000 = 140 v$

$v = 7.1$ ms⁻¹

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2 years ago

A hot air balloon is traveling vertically upward at a constant speed of 4.5 m/s. When it is 28 m above the ground, a package is

ella [17]

Answer:

1.97 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow 21=4.5t+\frac{1}{2}\times -9.8\times t^2\\\Rightarrow 21=-4.5t-4.9t^2\\\Rightarrow 4.9t^2+4.5t-28=0\\\Rightarrow 49t^2+45t-280=0$

Solving the above equation we get

$t=\frac{-45+\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}, \frac{-45-\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}\\\Rightarrow t=1.97, -2.89$

So, the time the package was in the air is 1.97 seconds

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2 years ago