Why is energy transferred from the substance to the surroundings when a substance freezes

What is true about the relationship between the kinetic energy of molecules in an object and the objects temperature?

vekshin1

As the temperature increases the kinetic energy of the molecules increases, if u add more heat you get more kinetic energy.

6 0

3 years ago

A disk-shaped merry-go-round of radius 2.13 m and mass 175 kg rotates freely with an angular speed of 0.651 rev/s. A 55.4 kg per

GREYUIT [131]

Answer:

Explanation:

To find out the angular velocity of merry-go-round after person jumps on it , we shall apply law of conservation of ANGULAR momentum

I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω

I₁ is moment of inertia of disk , I₂ moment of inertia of running person , I is the moment of inertia of disk -man system , ω₁ and ω₂ are angular velocity of disc and man .

I₁ = 1/2 mr²

= .5 x 175 x 2.13²

= 396.97 kgm²

I₂ = m r²

= 55.4 x 2.13²

= 251.34 mgm²

ω₁ = .651 rev /s

= .651 x 2π rad /s

ω₂ = tangential velocity of man / radius of disc

= 3.51 / 2.13

= 1.65 rad/s

I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω

396.97 x .651 x 2π + 251.34 x 1.65 = ( 396.97 + 251.34 ) ω

ω = 3.14 rad /s

kinetic energy = 1/2 I ω²

= 3196 J

8 0

3 years ago

which two gases in earths atmosphere are believed by scientists to be greenhouse gases that are major contributors to global war

Olenka [21]

Nitrogen and carbon dioxide??

4 0

1 year ago

The outer shell electrons in metals are not tightly bound to the nuclei of their atoms they are free to roam throughout the mate

Yuliya22 [10]

The pumps which supplies energy to move the water from the ground to a high elevation. The charges that flow throughout the wires.

6 0

3 years ago

At the very end of Wagner's series of operas The Ring of Nibelung, Brunnhilde takes the golden ring form the finger of the dead

Blababa [14]

Answer:

a) 404 m² b) apparent height = 7.5 m

Explanation:

This question is about refraction and total internal refraction.

Here I will take refractive index of air and water

$n_{air}=1\\ n_{water}=1.33=4/3$

Now let's look at the diagram I have attached here

At some angle A, the light from the ring (yellow point) under water will be totally internally refracted (B = 90°), which means that rays of light (yellow arrow) that make large enough angle A will not be able to escape from the water. Since we assumed that the ring is a point, there will be a critical cone of angle A with the ring at its apex which traces a circle of radius R on the surface of water, which, beyond this radius, no light could escape.

According to snell's law

$\frac{sin(B)}{sin(A)} = \frac{n_{water}}{n_{air}} = 4/3$

At critical angle B = 90°

$\frac{3)}{4}sin(B) = [tex]\frac{3}{4} sin(90^\circ ) = 0.75 = sin(A)$

Therefore

$A = 48.6^\circ$

With this, we can find the radius of the circle (refer to my diagram)

$h* tan (A) = R\\R =11.3 m$

And with that we can find the area

$A = \pi R^2=404\ m^2$

Additional Problem

For apparent depth from above, we can think that, since we are accustomed to seeing light at the speed of c in air, our brain interpret light from <em>any</em> source to be traveling at c. This causes light that originated under water, which has the speed of

$v_{water} = \frac{c}{n_{water}} = 0.75c$

to appear as if it has traveled with the same duration as light with speed c

In order for this to happen our brain perceive shortened length which is the apparent depth.

To put it in mathematical term

$t_{travel}=\frac{h_{apparent}}{v_{water}} =\frac{h}{c}$

So we get apparent depth

$h_{apparent}=0.75h = 7.5\ m$

4 0

3 years ago