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The surface tension of a liquid is to be measured using a liquid film suspended on a U-shaped wire frame with an 12-cm-long movable side. If the force needed to move the wire is 0.096 N, determine the surface tension of this liquid in air.

Answers

Answer 1

Answer:

Answer:

σ = 0.8 N/m

Explanation:

Given that

L = 12 cm

We know that 1 m = 100 cm

L = 0.12 m

The force ,F= 0.096 N

Lets take surface tension = σ

We know that surface tension is given as

$\sigma =(F)/(L)\n\sigma =(F)/(L)\nNow\ by\ putting\ the\ values\n\sigma =(0.096)/(0.12)\ N/m\n\sigma=0.8\ N/m$

Therefore the surface tension σ will be 0.8 N/m .

σ = 0.8 N/m

Answer 2

Answer:

Final answer:

The surface tension of the liquid in air is 0.8 N/m.

Explanation:

To determine the surface tension of the liquid, we need to use the formula F = yL, where F is the force needed to move the wire, y is the surface tension, and L is the length of the wire. In this case, F = 0.096 N and L = 12 cm. We can rearrange the formula to solve for y: y = F / L. Plugging in the values, we get y = 0.096 N / 0.12 m = 0.8 N/m. So, the surface tension of the liquid in air is 0.8 N/m.

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Use diagrams and sketches in your explanation and avoid using physics terms like rotational, centripetal, normal forces, tension, friction, acceleration, velocity, vectors, buoyancy, energy, kinetics, potential, heat capacity, etc.

Answers

Answer:

1.) Everything that moves, will eventually come to a stop. Rest is the “natural” state of all objects

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2.) A continuous force is needed for continuous motion

This misconception is a direct consequence of the first one. While this is true, if you are, for example, pushing a grocery cart in a supermarket, again this is only because there is friction involved. The force you apply to keep an object moving is only to counteract the frictional force. If you were to throw a rock on outer space, it would travel with a constant velocity forever, unless it hits something, of course. This is because space is mostly empty (it has trace elements of gas and dust throughout), and there would not be any frictional force acting on that rock.

3.) An object is hard to push because it is heavy

This is one of the most common misconceptions because it’s something we see and feel everyday. While a heavy object is really hard to push, it is not because of its weight, but because of its inertia or mass. Inertia is an objects resistance to change in motion. It is important to note that inertia is resistance to “change motion” rather than just motion itself. When, I was a kid, I imagined that it would be easy to carry and push massive objects when in outer space, but not surprisingly, my younger self was wrong.,

With that said… Since these objects still have mass despite being weightless, this mass represents the object’s inertia.

4.) Planets revolve around the sun because they are pushed by gravity

We have to remember that gravity — the weakest of the four fundamental forces — is an attractive force. The reason why planets revolve around the Sun can be chalked up to the fact that the planets were already spinning within the protoplanetary disk encircling a young Sun. Gravity merely keeps the planets in orbit around the Sun, but it isn’t necessarily the one thing pushing the planets along their orbital plane.

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Again, the problem comes from not being able to identify another force that is involved, which is air resistance. All objects moving through air, and hence, all falling objects, experience air resistance. This force is proportional to the area of the object in the direction of motion. Usually, this force is negligible, but for light objects — with weight comparable to the air resistance, like a feather — it will have a big effect. This is ultimately confirmed by the famous hammer and feather drop experiment on the moon.

6.) There is no gravity in outer space

There is gravity in outer space, it is just weaker than what we experience here on Earth. Astronauts that are orbiting the Earth don’t experience gravity because they are free-falling (yes, you read that right). All satellites, including the moon and the planets, are in a constant state of freefall.

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7.) Planets move in circular orbits around the Sun

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A gate with a circular cross section is held closed by a lever 1 m long attached to a buoyant cylinder. The cylinder is 25 cm in diameter and weighs 200 N. The gate is attached to a horizontal shaft so it can pivot about its center. The liquid is water. The chain and lever attached to the gate have negligible weight. Find the length of the chain such that the gate is just on the verge of opening when the water depth above the gate hinge is 10 m.

Answers

The length of the chain such that the gate is just on the verge of opening is mathematically given as

l=8.58m

What is the length of the chain?

Generally, the equation for the is mathematically given as

$F'=\gamma hA$

Therefore

$9810*10* (\pi)/(4)1^2$

Fh= 77048 N

Where

$ycp-y=((\pi r^4)/(4))/(10*(\pi D^4)/(4))$

ycp-y=0.00625

In conclusion, resultant force

x = F'' - W

x = 9810* 10*( \pi/4 )*0.25^2 *(10-l)-200

x = 4615.5-481.5 l

Therefore

77048* 0.00625 - 1 *(4615.5-481.5 l) = 0

l=8.58m

Answers

Answer:

$SR=949.2N$

Explanation:

From the question we are told that:

Mass $M=84kg$

Speed $V=1.2m/s$

Acceleration Time $t_a=0.73$

Constant speed Time $t_s=5.0s$

Deceleration time $t_d=1.4s$

Generally the equation for Acceleration is mathematically given by

$a=(v)/(t)$

Therefore acceleration for the first 0.80 sec is

$a=(1.2)/(0.80)$

$a=1.5m/s^2$

Therefore

Spring Reading=Normal force -Reaction

$SR=m(g+a)$

$SR=84(9.8+1.5)$

$SR=949.2N$

If the absolute pressure of gas is 550.280 kPa, its gauge pressure is

Answers

pressure absolute = pressure gage + pressure atmosphere

Answer:

650.280

Explanation: 100kpa + 550.280kpa

A 50.0-kg box is being pulled along a horizontal surface by means of a rope that exerts a force of 250 n at an angle of 32.0° above the horizontal. the coefficient of kinetic friction between the box and the surface is 0.350. what is the acceleration of the box?

Answers

The acceleration of the box is 0.81 m/sec².

What is acceleration?

The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration

According to Newton's second law, the resultant of the forces acting on the box is equal to the product between its mass and its acceleration:

$\sum F= ma$ (1)

we are only concerned about the horizontal direction, so there are only two forces acting on the box in this direction:

- the horizontal component of the force exerted by the rope, which is equal to

$F_x = F cos\theta = 250*cos 32 = 212 N$

the frictional force, acting in the opposite direction, which is equal to

$F_f = \mu *mg = 171.7 N$

By applying Newton's law (1), we can calculate the acceleration of the

box,

$F_x - F_f = ma\na = 0.81 m/sec^2$

The acceleration of the box is 0.81 m/sec².

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The actual depth of a shallow pool 1.00 m deep is not the same as the apparent depth seen when you look straight down at the pool from above. How deep (in cm) will it appear to be

Answers

Answer:

d' = 75.1 cm

Explanation:

It is given that,

The actual depth of a shallow pool is, d = 1 m

We need to find the apparent depth of the water in the pool. Let it is equal to d'.

We know that the refractive index is also defined as the ratio of real depth to the apparent depth. Let the refractive index of water is 1.33. So,

$n=(d)/(d')\n\nd'=(d)/(n)\n\nd'=(1\ m)/(1.33)\n\nd'=0.751\ m$

d' = 75.1 cm

So, the apparent depth is 75.1 cm.

The apparent depth of a 1.00-meter-deep pool, when viewed from above, is around 75.2 centimeters. This difference is due to light refraction in water, causing optical distortion.

When observing a shallow pool of 1.00 meter depth from above, the apparent depth is altered by the phenomenon of light refraction in water. Light bends as it passes from air into water, affecting the way objects are perceived underwater.

The apparent depth is less than the actual depth due to this bending of light. To calculate the apparent depth, one can use the Snell's Law formula, which relates the angles of incidence and refraction to the refractive indices of the two media.

However, a simplified formula for the apparent depth (d') in terms of the actual depth (d) is given by d' = d/n, where 'n' is the refractive index of water (approximately 1.33). Therefore, in this case, the pool's apparent depth, when viewed from above, will be approximately 75.2 centimeters, making it shallower than it appears at first glance due to the optical effects caused by light traveling through water.

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Complete question below:

"What is the apparent depth, in centimeters, when looking straight down at a shallow pool that is 1.00 meter deep? Note that the apparent depth is different from the actual depth due to the refraction of light in water."

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