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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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3 years ago
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A 1.1 kg ball is attached to a ceiling by a 2.16 m long string. The height of the room is 5.97 m . The acceleration of gravity i
nydimaria [60]

1. -23.2 J

The gravitational potential energy of the ball is given by

U=mgh

where

m = 1.1 kg is the mass of the ball

g = 9.8 m/s^2 is the acceleration of gravity

h is the height of the ball, relative to the reference point chosen

In this part of the problem, the reference point is the ceiling. So, the ball is located 2.16 m below the ceiling: therefore, the heigth is

h = -2.16 m

And the gravitational potential energy is

U=(1.1 kg)(9.8 m/s^2)(-2.16 m)=-23.2 J

2. 41.1 J

Again, the gravitational potential energy of the ball is given by

U=mgh

In this part of the problem, the reference point is the floor.

The height of the ball relative to the floor is equal to the height of the floor minus the length of the string:

h = 5.97 m - 2.16 m = 3.81 m

And so the gravitational potential energy of the ball relative to the floor is

U=(1.1 kg)(9.8 m/s^2)(3.81 m)=41.1 J

3. 0 J

As before, the gravitational potential energy of the ball is given by

U=mgh

Here the reference point is a point at the same elevation of the ball.

This means that the heigth of the ball relative to that point is zero:

h = 0 m

And so the gravitational potential energy is

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4 years ago
A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed
Tju [1.3M]

Answer:

v = 2.85 m/s

Explanation:

By the law of similarity of the triangle we can say

\frac{height\: of\: man}{distance\: of \: man\: from \: spotlight} = \frac{height \:of \:image}{distance\: of \:wall\: from\: spotlight}

here we know that

height of man = 2 m

let the distance of man from spotlight = x

distance of wall from spotlight = 12 m

height of image is let say "y"

so we will have

\frac{2}{x} = \frac{y}{12}

y = \frac{24}{x}

now we have

\frac{dy}{dt} = \frac{24}{x^2}\frac{dx}{dt}

here we know

\frac{dx}{dt} = speed = 1.9 m/s

x = 4 m

now we have

\frac{dy}{dt} = \frac{24}{4^2}(1.9)

v = 2.85 m/s

8 0
4 years ago
An object is projected at 25m/s from the top of a building of height 50m. At the same instant,another object is projected from t
docker41 [41]

A) The objects have the same vertical position after 2 seconds

B) The objects have same vertical position at y = 30.4 m (but they do not collide since they have different x-position)

Explanation:

The motion of the first object along the vertical direction is a uniformly accelerated motion, so we can write its position at time t using the following equation:

y_1(t)=h+u_1 t + \frac{1}{2}gt^2

where:

h = 50 m is the initial height

u_1=0 is the initial vertical velocity (the object is projected horizontally, so the vertical velocity is zero at the beginning)

g=-9.8 m/s^2 is the acceleration of gravity

So, its vertical position can be rewritten as

y_1(t)=50-4.9t^2

The position of object 2 instead can be written as

y_2(t)=(u_2 sin \theta)t + \frac{1}{2}gt^2

where

u_2 sin \theta is the initial vertical velocity, where

u_2 = 50 m/s is the initial velocity

\theta=30^{\circ} is the angle of projection

Substituting, we get:

y_2(t)=(50)(sin 30^{\circ})t+\frac{1}{2}(-9.8)t^2=25t-4.9t^2

The two objects collide when their vertical position is the same, so:

y_1(t)=y_2(t)\\50-4.9t^2 = 25t-4.9t^2

And solving for t, we find:

50=25t\\t= 2 s

Note that this means that the two object at t = 2 s have the  same vertical position: however, this is not true for the horizontal position.

B)

In order to find the point where they collide, we have to substitute the time of the collision that we found in part A into one of the expressions of the vertical position.

Substituting into the expression of object 2, we find:

y_2(t) = 25t-4.9t^2=25(2.0)-4.9(2.0)^2=30.4 m

We can verify that at the same time, the vertical position of object 1 is the same:

y_1(t)=50-4.9t^2=50-4.9(2.0)^2=30.4 m

This means that the two objects have the same vertical position at 30.4 m.

However, in reality, the two objects do not collide. In fact, object 1 is moving in the horizontal direction with constant velocity

v_{1x}=25 m/s

So its horizontal position at t = 2.0 s is

x_1(2.0)=v_{1x}t=(25)(2.0)=50 m

While object 2 is moving in the horizontal plane with velocity

v_{2x}=u_2 cos \theta=(50)(cos 30^{\circ})=43.3 m/s

So its horizontal position  at t = 2.0 s is

x_2(2.0)=v_{2x}t=(43.3)(2.0)=86.6 m

So in reality, the two objects do not collide, if they start from the same x-position.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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3 years ago
A car is travelling at a constant speed of 50 km/h. How long will it take to travel 20 km? *
Sholpan [36]

20km / 50km/h = 0.4h = 24mins

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3 years ago
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