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Anettt [7]
3 years ago
11

Your classmate says that if all the molecules in a particular liquid had the same speed, and some were able to evaporate, the re

maining liquid would not undergo cooling. Do you agree or disagree and what is your explanation to support your answer.
Physics
2 answers:
ddd [48]3 years ago
5 0

Answer: in an ideal system, your classmate is right

Explanation:

Here I suppose that we are in a closed environment.

If all the particles had the same speed, and some of them are able to evaporate, then you have coexistence between liquid state and gas state, In this coexistence state, there is a line of pressure as a function of temperature where the coexistence is possible.

As some of the liquid evaporates, the volume of the gas is bigger than the one of the liquid, so the pressure in the system will increment a little bit, and the temperature will answer to it until the equilibrium is reached.

In equilibrium, the temperature will not change if we don't do anything to the system (assuming adiabatic walls).

here your classmate may be right

A thing you may know is that the temperature is an intensive quantity, which means that it does not depend on the number of particles in the mixture, so the fact that some of the particles "leave" the liquid does not imply that the liquid changes their temperature. And there is also useful to notice that the change of phase is at a constant temperature.

babymother [125]3 years ago
3 0
The molecules which evaporate presumably take heat away from the liquid. So, I'd disagree with the classmate. Whether the amount of cooling would differ from the usual case wherein the molecules have different speeds is another question. 
I guess the argument goes something along the lines of that the faster moving and therefore most kinetically energetic molecues evaporate and take away most heat. But if there's no faster moving molecules, 'cos they all have the same speed well, then presumably they'd all take away the same amount of heat. So, maybe the cooling would be less. No cooling though ??? Hmmmm dunno .... i think not ....
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Lostsunrise [7]
  • The mechanic did 5406 Joules of work pushing the car.

That's the energy he put into the car.  When he stops pushing, all the energy he put into the car is now the car's kinetic energy.

  • Kinetic energy = (1/2) (mass) (speed²)

And there we have it

  • The car's mass is 3,600 kg.
  • Its speed is 'v' m/s .
  • (1/2) (mass) (v²) =  5,406 Joules

(1/2) (3600 kg) (v²) = 5406 joules

1800 kg (v²) = 5406 joules

v² = (5406 joules) / (1800 kg)

v² = (5406/1800) (joules/kg)

= = = = = This section is just to work out the units of the answer:

  • v² = (5406/1800) (Newton-meter/kg)
  • v² = (5406/1800) (kg-m²/s²  /  kg)
  • v² = (5406/1800)  (m²/s²)

= = = = =

v = √(5406/1800)  m/s

<em>v = 1.733 m/s</em>

4 0
3 years ago
You and a friend each hold a lump of wet clay. Each lump has a mass of 25 grams. You each toss your lump of clay into the air, w
solniwko [45]

Answer:

a) p(total) = <0.05, 0.1, 0.1 > kg m/s

b) p = <0.05, 0.1, 0.1 > kg.m/s

c) v_f = < 1, 2, 2 > m/s

Explanation:

a.)

Mass of each lump = 25 g = 0.025 kg

Velocity of lump 1 = < -2, 0, -7 > m/s

Momentum of lump 1 = Mass×Velocity

                                   = 0.025×< -2, 0, -7 >

                                   = < -0.05, 0, 0.175> kg m/s

Velocity of lump 2 = < 4, 4, -3 > m/s

Momentum of lump 2 = Mass×Velocity

                                    = 0.025×< 4, 4, -3 >

                                    = < 0.1, 0.1, -0.075> kg m/s

Total momentum before impact  =  < -0.05,  0,  0.175 > + < 0.1, 0.1, -0.075>

                                                      = < 0.05, 0.1, 0.1 > kg m/s

⇒p(total) = <0.05, 0.1, 0.1 > kg m/s

b)

As we know that,

By the law of conservation of linear momentum,

The total momentum will be the same before and after the collision.

⇒Momentum of the stuck together  after the collision = Total momentum of the lumps just before impact.

⇒ p = <0.05, 0.1, 0.1 > kg m/s

c)

Let the final velocity =  v_f

Total mass = 0.025 + 0.025 = 0.05 kg

As

Momentum = mass ×velocity

⇒ <0.05, 0.1, 0.1 > = 0.05 ×v_f

⇒ v_f = <0.05, 0.1, 0.1 > / 0.05

          = < 1, 2, 2 > m/s

⇒v_f = < 1, 2, 2 > m/s

7 0
3 years ago
in a falling elevator if you jump at the right time would you be safe? explain, and give a reasonable response.
N76 [4]

Answer:

No

Explanation:

You can easily injury yourself due to the force of gravity.

6 0
3 years ago
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melisa1 [442]
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4 0
3 years ago
When the pivot point of a balance is not at the center of mass of the balance, how is the net torque on the balance calculated
Mashcka [7]

Answer:

It is calculated as Force × perpendicular distance.

Explanation:

Torque is a rotational force and twisting force that can cause an object to rotate in it's axis. This cause angular rotation.

The torque due to gravity on a body about its centre of mass is zero because the centre of mass is the that point of the body at which the force acts by the gravity that is mg.

But if the pivot point of a balance is not at the centre of mass of the balance, it will be FORCE × PERPENDICULAR DISTANCE because at that point, there is no centre in which the force act on a body by gravity. The distance and force will be use to calculate.

6 0
3 years ago
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