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2 years ago

Which shows increasing entropy?

Physics

2 answers:

victus00 [196]2 years ago

8 0

Answer:

Falling leaves!

Explanation:

I got it right on EDGE2021. ^^ Hope it helps!~ Remember, I'm proud of you. <3

fomenos2 years ago

6 0

Answer:

mowing a lawn

Explanation:

Entropy is the degree of disorderliness of a system. As a body moves from a more ordered state to a less ordered one, the entropy of the system increases.

Mowing a lawn is a typical example of increasing entropy.
When a lawn is being mowed, the grasses becomes disordered without any fixed orientation.
Folding a clothe is trying to bring orderliness to the clothe patterning.
Washing dishes will make one arrange them in an ordered way.
Falling leaves brings leaves together.

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Please can anybody tell me what are these lab equipments

olga nikolaevna [1]

Answer:

5 is the tripoid stand

Thanks have a bangtastic day

4 0

2 years ago

Read 2 more answers

Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical

lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

$T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}}$ (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

$\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}$

$\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2$ (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio $\frac{T^{2}}{a^{3}}$ is constant for all the planets orbiting the same sun, so we can said that:

$\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}$

$\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}$

$\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}$

$\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83$

6 0

3 years ago

Read 2 more answers

Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.

rjkz [21]

Answer:

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

Explanation:

Given

Before collision:

Momentum of block A = $p_{A1}$ = $-100\ kg\ ms^{-1}$

Momentum of block B = $p_{B1}$ = $-150\ kg\ ms^{-1}$

After collision:

Momentum of block A = $p_{A2}$ = $-200\ kg\ ms^{-1}$

Applying law of conservation of momentum to find momentum of block B after collision $p_{B2}$ .

$p_{A1}+p_{B1}=p_{A2}+p_{B2}$

Plugging in the given values and simplifying.

$-100-150=-200+p_{B2}$

$-250=-200+p_{B2}$

Adding 200 to both sides.

$200-250=-200+p_{B2}+200$

$-50=p_{B2}$

∴ $p_{B2}=-50\ kg\ ms^{-1}$

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

6 0

2 years ago

Which of the following depends on an object's displacement not distance from the starting point?

spayn [35]

The answer is A) velocity, because velocity is speed and displacement.
Hope this helps

6 0

3 years ago

We would be more likely to get an accurate measurement of the distance from Earth to a nearby star if we took the angular measur

NeTakaya

Answer:

6 month interval

Explanation:

The distance to a nearby star in theory is more simple than

one might think! First we must learn about the parallax effect. This is the mechanism our eyes use to perceive things at a distance! When we look at the star from the earth we see it at different angles throughout the earth's movement around the sun similar to how we see when we cover on eye at a time. Modern telescopes and technology can help calculate the angle of the star to the earth with just two measurements (attached photo!) Since we know the distance of the earth from the sun we can use a simple trigonometric function to calculate the distance to the star. The two measurements needed to calculate the angle of the star to the earth caused by parallax (in short angle θ) are shown in the second attached photo.

So using a simple trigonometric function $Sin\theta=\frac{r}{d}$ we can solve for d which is the distance of the earth to the star:

$d=\frac{r}{Sin\theta}$

In the first attached photo a picture where r is the distance to the star and the base of the triangle is the diameter of the earth.

8 0

3 years ago