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2 years ago

Why does water feel cool when it evaporates

2 answers:

5 0

Basically, when water evaporates, it always absorbs energy. What happens is that a molecule steals surface energy for all molecules until it is hot enough to evaporate. This is how the evaporation process happens. For this reason, the surface cools because the molecules steal the heat from the medium. This can easily be noticed when we sweat and after the water molecules evaporate from our body, it cools.

Illusion [34]2 years ago

4 0

First, it makes your skin feel cooler when it's wet. And when it evaporates it removes some heat. But sweat will only evaporate in an environment where there isn't muchwater in the air. In a place with high humidity, there're already lots of water molecules in the air.

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How is an image formed by a plane mirror

masha68 [24]

Is always virtual (meaning that the light rays do not actually come from the image), upright, and of the same shape and size as the object it is reflecting. A virtual image is a copy of an object formed at the location from which the light rays appear to come.

8 0

2 years ago

The electromagnetic wave shown above is made of vibrating

amid [387]

The electromagnetic wave shown above is made up of vibrating Electric fields.

I assume this is why electromagnetic waves can travel through a vacuum, due to lack of matter.

8 0

3 years ago

For a chair to be balanced on one leg, where must the center of gravity be located? A. in the center of the seat B. toward the f

mixer [17]

Answer:

The correct answer is c

Explanation:

When a body is in rotational equilibrium, the relationship must be met

∑ τ = 0

Σ F d = 0

therefore the forces that are exerted on the body that are represented in the center of gravity are different from zero, therefore, to maintain balance the distance must be zero, therefore the center of gravity must be above the leg where the body is balancing

The correct answer is c

6 0

3 years ago

Ou are driving on the highway at a speed of 40 m/s (which is over the speed limit) when you notice a cop in front of you. To avo

harina [27]

Answer:

What is the acceleration of your vehicle?

+ 2.0 m/s^2✔

- 2.0 m/s^2

+ 8.0 m/s^2

- 6.0 m/s^2

7 0

2 years ago

A 1.40-kg block is on a frictionless, 30 ∘ inclined plane. The block is attached to a spring (k = 40.0 N/m ) that is fixed to a

Oliga [24]

Answer:

the mass drop by 6.5cm before coming to rest.

Explanation:

Given that:

the mass of the block M= 1.40 kg

angle of inclination θ = 30°

spring constant K = 40.0 N/m

mass of the suspended block m = 60.0 g = 0.06 kg

initial downward speed = 1.40 m/s

The objective is to determine how far does it drop before coming to rest?

Let assume it drops at y from a certain point in the vertical direction;

Then :

Workdone by gravity on the mass of the block is:

$w_1g = 1.4*9.81*y*sin30$ = - 6.867y

Workdone by gravity on the mass of the suspended block is:

$w_2g = 0.06*9.81$ = 0.5886

The workdone by the spring = -1/2ky²

= -0.5 × 40y²

= -20 y²

The net workdone is = -20 y² - 6.867y + 0.5886y

According to the work energy theorem

Net work done = Δ K.E

-20 y² - 6.867y + 0.5886 = 1/2 × 0.06 × 1.4²

-20 y² - 6.867y + 0.5886 = 0.5 × 0.1176

-20 y² - 6.867y + 0.5886 = 0.0588

-20 y² - 6.867y + 0.5886 - 0.0588

-20 y² - 6.867y + 0.5298 = 0

multiplying through by (-)

20 y² + 6.867y - 0.5298 = 0

Using the quadratic formula:

$\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

where;

a = 20 ; b = 6.867 c= - 0.5298

$\dfrac{-(6.867) \pm \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)}$

$= \dfrac{-(6.867) + \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)} \ \ \ OR \ \ \ \dfrac{-(6.867) - \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)}$ = 0.0649 OR −0.408

We go by the positive integer

y = 0.0649 m

y = 6.5 cm

Therefore; the mass drop by 6.5cm before coming to rest.

7 0

3 years ago