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3 years ago

Explain and derive the equation for capillary action in the phenomenon of surface tension

Physics

1 answer:

Mars2501 [29]3 years ago

8 0

Answer:

Explanation:Capillary action is the ability of a liquid to flow in narrow spaces without the assistance of, ... This article is about the physical phenomenon. ... If the diameter of the tube is sufficiently small, then the combination of surface tension (which is caused by cohesion ... They derived the Young–Laplace equation of capillary action.

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4) (5 points) Given are the magnitudes and orientations (with respect to x-axis) of 3

Kazeer [188]

Expand each vector into their component forms:

$\vec A=(4.5\,\mathrm N)(\cos\theta_A\,\vec\imath+\sin\theta_A\,\vec\jmath)=(2.58\,\vec\imath+3.69\,\vec\jmath)\,\mathrm N$

Similarly,

$\vec B=(-1.23\,\vec\imath+0.860\,\vec\jmath)\,\mathrm N$

$\vec C=(-3.44\,\vec\imath-4.91\,\vec\jmath)\,\mathrm N$

Then assuming the resultant vector $\vec R$ is the sum of these three vectors, we have

$\vec R=\vec A+\vec B+\vec C$

$\vec R=(-2.09\,\vec\imath-0.368\,\vec\jmath)\,\mathrm N$

and so $\vec R$ has magnitude

$\|\vec R\|=\sqrt{(-2.09)^2+(-0.368)^2}\,\mathrm N\approx2.12\,\mathrm N$

and direction $\theta_R$ such that

$\tan\theta_R=\dfrac{-0.368}{-2.09}\implies\theta_R=-170^\circ=190^\circ$

5 0

3 years ago

A horizontal spring with spring constant 85 N/m extends outward from a wall just above floor level. A 5.5 kg box sliding across

Aloiza [94]

Answer: The box was moving with a velocity of 0.256m/s when it hit the spring

Explanation: Please see the attachments below

6 0

3 years ago

Psuedopod in a sentence

mixas84 [53]

Does it have to be that exact word. cause it is just another term for psuedopodium

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3 years ago

Two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision

Sidana [21]

a) The velocity after the collision.is 11.456 m/s.

b) The kinetic energy lost due to the collision is 44.564 J.

<h3>What is conservation of momentum principle?</h3>

When two bodies of different masses move together each other and have head on collision, they travel to same or different direction after collision.

The external force is not acting here, so the initial momentum is equal to the final momentum. For inelastic collision, final velocity is the common velocity for both the bodies.

m₁u₁ +m₂u₂ =(m₁ +m₂) v

Given are the two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision. Their initial velocities along the one-dimension path are vi1 = 32.4 m/s [right] and vi2 = 8.6 m/s [left].

(a) Substitute the values, then the final velocity will be

0.6 x32.4 +4.4 x 8.6 = (0.6+4.4)v

v = 11.456 m/s

Thus, the velocity after collision is 11.456 m/s.

(b) Kinetic energy lost due to collision will be the difference between the kinetic energy before and after collision.

= [1/2m₁u₁² +1/2m₂u₂² ] - [1/2(m₁ +m₂) v²]

Substitute the value, we have

= [1/2 x 0.6 x32.4² + 1/2 x4.4 x 8.6²] - [1/2 x(0.6+4.4)11.456²]

= 44.564 J

Thus, the kinetic energy lost due to the collision is 44.564 J.

Learn more about conservation of momentum principle

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4 0

2 years ago

Read 2 more answers

A rifle bullet with mass 8.00g strikes and embeds itself in a block with mass 0.992kg that rests on a frictionless, horizontal s

mafiozo [28]

Answer:

block velocity v = 0.09186 = 9.18 10⁻² m/s and speed bollet v₀ = 11.5 m / s

Explanation:

We will solve this problem using the concepts of the moment, let's try a system formed by the two bodies, the bullet and the block; In this system all scaffolds during the crash are internal, consequently, the moment is preserved.

Let's write the moment in two moments before the crash and after the crash, let's call the mass of the bullet (m) and the mass of the Block (M)

Before the crash

p₀ = m v₀ + 0

After the crash

$p_{f}$ = (m + M) v

p₀ = $p_{f}$

m v₀ = (m + M) v (1)

Now let's lock after the two bodies are joined, in this case the mechanical energy is conserved, write it in two moments after the crash and when you have the maximum compression of the spring

Initial

Em₀ = K = ½ m v2

Final

E $m_{f}$ = Ke = ½ k x2

Emo = E $m_{f}$

½ m v² = ½ k x²

v² = k/m x²

Let's look for the spring constant (k), with Hook's law

F = -k x

k = -F / x

k = - 0.75 / -0.25

k = 3 N / m

Let's calculate the speed

v = √(k/m) x

v = √ (3/8.00) 0.15

v = 0.09186 = 9.18 10⁻² m/s

This is the spped of the block plus bullet rsystem right after the crash

We substitute calculate in equation (1)

m v₀ = (m + M) v

v₀ = v (m + M) / m

v₀ = 0.09186 (0.008 + 0.992) /0.008

v₀ = 11.5 m / s

6 0

3 years ago

Explain and derive the equation for capillary action in the phenomenon of surface tension​

Explain and derive the equation for capillary action in the phenomenon of surface tension