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3 years ago

Explain briefly where the energy come from when a liquid Rises against Gravity in a capillary tube

Physics

1 answer:

SSSSS [86.1K]3 years ago

3 0

Answer:

Surface tension

Explanation:

When liquid rises against gravity in a capillary tube, the energy comes from surface tension.

This is because surface tension is the energy that's needed to increase the liquid surface area.

As a result of hydrogen bonding present in Water, it usually has high surface tension which makes it to possess a tough skin that can make it not to break despite high forces applied to it.

The liquid will be in contact with the capillary tube and as such experiences surface tension which in turn makes the capillary tube to experience an upward force that makes the liquid begin to rise up.

The more the liquid keeps rising, the more it gets to the point where the surface tension becomes balanced from the weight of the liquid.

You might be interested in

Which data set has the largest range?

MrRissso [65]

Answer:

Option C

Explanation:

We have to check range of all options first

For A:

Largest Value: 5

Smallest Value: 1

So range = Largest value - smallest value

5-1 = 4

For B:

Largest Value: 6

Smallest Value: 4

Range = 6-4 = 2

For C:

Largest Value: 9

Smallest Value: 1

Range = 9-1 = 8

For D:

Largest Value = 9

Smallest Value = 3

Range = 9-3=6

So, the data set in option C has the largest range

4 0

2 years ago

How does the lever arm change if you decrease the angle of the force?

Vladimir79 [104]

Answer:

The lever arm could decrease or increase depending of the initial angle.

Explanation:

The lever arm d is calculated by:

d = rsin(θ)

where r is the radius and θ the angle between the force and the radius.

So, the increse or decrees of d depends of the sin of the angle θ, if the initial angle is greather than 90° and the angle decrease to an angle closer to 90°, the lever arm will increase but if the initial angle is 90° or lower and the angle decrease, the lever arm will decrease.

8 0

2 years ago

A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0. The wheel has

irga5000 [103]

Answer:

The total linear acceleration is approximately 0.246 meters per square second.

Explanation:

The total linear acceleration ( $a$ ) consist in two components, <em>radial</em> ( $a_{r}$ ) and <em>tangential</em> ( $a_{t}$ ), in meters per square second:

$a_{r} = \omega^{2}\cdot r$ (1)

$a_{t} = \alpha \cdot r$ (2)

Since both components are orthogonal to each other, the total linear acceleration is determined by Pythagorean Theorem:

$a = \sqrt{a_{r}^{2}+a_{t}^{2}}$ (3)

Where:

$r$ - Radius of the wheel, in meters.

$\omega$ - Angular speed, in radians per second.

$\alpha$ - Angular acceleration, in radians per square second.

Given that wheel accelerates uniformly, we use the following kinematic equation:

$\omega = \omega_{o}+ \alpha\cdot t$ (4)

Where:

$\omega_{o}$ - Initial angular speed, in radians per second.

$t$ - Time, in seconds.

If we know that $r = 0.1\,m$ , $\alpha = 2\,\frac{rad}{s^{2}}$ , $\omega_{o} = 0\,\frac{rad}{s}$ and $t = 0.60\,s$ , then the total linear acceleration is:

$\omega = \omega_{o}+ \alpha\cdot t$

$\omega = 1.2\,\frac{rad}{s}$

$a_{r} = \omega^{2}\cdot r$

$a_{r} = 0.144\,\frac{m}{s^{2}}$

$a_{t} = \alpha \cdot r$

$a_{t} = 0.2\,\frac{m}{s^{2}}$

$a = \sqrt{a_{r}^{2}+a_{t}^{2}}$

$a \approx 0.246\,\frac{m}{s^{2}}$

The total linear acceleration is approximately 0.246 meters per square second.

3 0

2 years ago

A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein

RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of $8m^{3}/s$ . As you may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi r^{2}h$

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

$\frac {r}{h}=\frac{4}{5}$