whats suface tension

Answers

Answer 1

Answer: "Surface tension is a film of a liquid caused by the attraction of the particles in the surface layer by the bulk of the liquid, which tends to minimize surface area."

Hope this helps!

Answer 2

Answer:

Explanation:

Surface tension is the property of a liquid surface. It is an effect where the surface of the liquid is strong.

example - small insects can walk on water as they do not have enough weight to penetrate it.

This image might help you

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Whats Suface Tension

Related Questions

What two factors determine how much potential energy an object has?

Answers

Answer:

The mass of the object and its height in the gravitational field of the Earth.

Explanation:

If we are talking about gravitational potential energy which is defined as:

[tex]U=m\,*\,g\,*\,h[/tex]

being "m" the object's mass, "g" the acceleration due to gravity, and "h" the height at which the object is located relative to the conventionally picked level for zero of potential energy.

As long as the value of "g" is constant, the only two variables that determine the gravitational potential energy are the mass (m) of the object and its relative height (h).

Answer:

The objects weight and height above Earth's surface

Explanation:

K12 :)

Matter is anything that has mass and takes up
space.
Which of the following is an example of
matter?
A. ear phones
B. music
C. sunlight
D. heat

Answers

Answer: ear phones

Explanation:

You can physically hold ear phones, but you can't hold music, sunlight, or heat.

The answer is Earphones because the other options aren’t a solid thing.

The ways to measure the mass and volume of irregular object​

Answers

Answer:

When we have irregular objects, it may become very hard to calculate the volume of the object, as we actually can not use any simple equation to find it.

The mass is less tricky, just find a scale and wheight it, now we know the mass of the irregular object.

One way to measure the volume of the object is using water... how we do it?

Get some recipient with water, measure the height of the water.

Introduce your object into the water and totally submerge it, now the level of the water will rise. This is because as you introduce the object under the water, you are displacing up a given volume of water that has the same volume as the irregular object.

Now that you know the height of the water before and after you put your object, you can easily calculate the volume of water displaced, and that will be the volume of the object (the tricky part may be totally submerging the object if, for example, is wood and it floats, here you can use a thin wire to push it down but it will affect a little bit the measures.)

An object accelerates to a velocity of 230 m/s over a time of 2.5 s. The acceleration it experienced was 42 m/s2. What was its initial velocity?

Answers

Answer:

230 = x + 105

x= 125

Explanation:

v = v0 + at

i)Distinguish between different methods of charging. ii) You are provided with a positively charged gold leaf electroscope. State and explain what happens when a. a glass rod rubbed with silk is brought near the disc of electroscope. b. an ebonite rod rubbed with fur is brought near the disc of electroscope. c. an uncharged metal rod is brought near the disc of electroscope d. a glass rod rubbed with silk is rolled on the disc of electroscope.

Answers

Answer:

Explanation:

On rubbing a glass rod with silk, the electrons from the glass rod get transferred to the silk. The silk now has an excess of electrons and so is negatively-charged. On the other hand, the glass rod is deficient in electrons and hence is positively-charged.

In the above case, the silk undergoes negative electrification.

Now, when the positively charged glass rod is touched on the disc of a negatively charged gold leaf electroscope, the electrons shifts towards rod, hence amount of charge on gold leaves decreases and the divergence between the gold leaves decreases as unlike charges attract each other.

Hence, the divergence decreases when a glass rod rubbed with silk is brought near the disc of negatively charged electroscope.

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A 1.00-kg glider attached to a spring with a force constant 25.0 N/m oscillates on a frictionless, horizontal air track. At t = 0, the glider is released from rest at x = -2.80 cm (that is, the spring is compressed by 2.80 cm). (a) Find the period of the glider's motion. How does the period depend on the mass and the spring constant? Does it depend on the amplitude of oscillation? s (b) Find the maximum values of its speed and acceleration. speed m/s acceleration m/s2 (c) Find the position, velocity, and acceleration as functions of time. (Where position is in m, velocity is in m/s, acceleration is in m/s2, and t is in s. Use the following as necessary: t.) x(t) = v(t) = a(t) =

Answers

Answer:

a)  T = 1.26 s , b)  v_max = 0.14 m / s ,  a_max = 0.7 m / s²

c) x = 0.028 cos (5 t) ,    v = - 0.14 sin 5t,   a = - 0.7 cos 5t

Explanation:

This is a simple harmonic motion exercise that is described by the equation

    x = A cos (wt +Ф)

with

          w = √ (k / m)

let's apply this expression to our case

a) Angular velocity is related to frequency

          w = 2π f

frequency and period are related

          f = 1 / T

we substitute

         2π / T = √ (k / m)

         T = 2π √(m / k)

let's calculate

         T = 2π √(1/25)

          T = 1.26 s

In the expression for the period, the amplitude does not appear, therefore there is no dependence, as long as Hooke's law is fulfilled, which is correct for small amplitudes.

b) in the initial equation we have the position as a function of time, let's use the definition of speed and acceleration

           v = dx / dt

           v = - A w sin (wt + Ф)

the speed is maximum when the sine is -1

            v_max = A w

            w = √ (k / m)

            w = √ 25/1

            w = 5 rad / s

the amplitude of the movement is equal to the maximum compression of the spring

            A = 2.8 cm = 0.028 m

             

we substitute

            v_max = 0.028 5

            v_max = 0.14 m / s

acceleration

             a = dv / dt

             a = - A w² cos (wt + Ф)

the acceleration is maximum when the cosine is -1

             a_max = A w²

let's calculate

             a_max = 0.028 5²

             a_max = 0.7 m / s²

c) let's start by finding the phase constant

              v = -A w cos (wt + Ф)

at t = 0 they indicate that the system has v = 0

              0 = -A w sin (0 + Ф)

              Ф = sin⁻¹ 0

              Ф = 0

we write the equation

            x = 0.028 cos (5 t)

           v = - A w sin (wt + Ф)

           v = - 0.028 5 sin (5t + 0)

           v = - 0.14 sin 5t

acceleration

           a = - A w² cos (wt + Ф)

           a = - 0.028 5 2 cos (5 t + 0)

           a = - 0.7 cos 5t

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