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2 years ago

Which of the following best describes entropy

Physics

1 answer:

UkoKoshka [18]2 years ago

3 0

Answer:

Exothermic reactions increase the entropy of the surroundings. Simply put, entropy measures the dispersal of energy. Since ΔH is negative in an exothermic reaction, this must mean that ΔS will take on a positive value, indicating an increase in entropy.

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At a certain instant a particle is moving in the +x direction with momentum +8 kg m/s. During the next 0.13 seconds a constant f

jeka94

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp = change in momentum = final momentum - initial momentum

F = constant force.

Δt = time interval.

Let´s calculate the x-component of the momentum after the 0.13 s:

final momentum - 8 kg m/s = -7 N · 0.13 s

final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s

final momentum = 7.09 kg m/s

Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

final momentum = 5 kg m/s² · 0.13 s

final momentum = 0.65 kg m/s

Then, the mometum vector will be as follows:

p = (7.09 kg m/s, 0.65 kg m/s)

The magnitude of this vector is calculated as follows:

$|p| = \sqrt{(7.09 kg m/s)^{2} + (0.65 kg m/s)^{2}} = 7.12 kg m/s$

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

4 0

3 years ago

how can you tell, as you walk close to a parked car, if it had been running recently? describe your reasoning in terms of energy

Blababa [14]

This question is probably referring to heat energy transferring from the car to its surroundings.

4 0

3 years ago

Radiation in the ultraviolet region of the electromagnetic spectrum is quite energetic. It is this radiation that causes dyes to

N76 [4]

The energy of photon in kJ/mol is 329kJ/mol.

Wavelength of radiation is 370nm. The frequency of given wavelength is

ν = c / λ

ν = 3×10^8 / 370×10^-9

ν = 8.11 × 10^14 s^-1

Now the energy of photon is:

E = hν

E = 6.63×10^-34 J.s/photon × 8.11×10^14s^-1

E = 5.41× 10^-19 J/photon

To find in mole

E = 5.41× 10^-19 × 6.022×10^23

E = 3.29 ×10^ 5 J/mol

So, the energy of mole of photon is equal to 329 kJ/mol.

Learn more about radiation here:

brainly.com/question/18650102

#SPJ4

5 0

1 year ago

Why some air masses are cool, cold, warm, and hot as well as moist or dry?

DENIUS [597]

Answer:

Explanation:

Often times, the property of air masses is a function of where they originate from especially with respect to latitude.

An air mass is certain amount of air with some unique set of temperature and vapor component.

Air masses differs from places to places based on where they originate.
Continental air masses are usually dry and cold because there is little to no water on a land mass.
Air masses that originates from the surface of the ocean are predominantly moist.
In polar regions, the air mass is cold.
Around the tropics, they are usually warm due to insolation here.

6 0

3 years ago

Read 2 more answers

For this problem, we assume that we are on planet-i. the radius of this planet is r =4200 km, the gravitational acceleration at

Minchanka [31]

The expression commonly used for potential gravitational energy is just simplification. It is actually just the first term in Taylor expansion of the real expression.
In general, the potential energy of gravitational field is defined as:
$U=-G \frac{mM}{r}$
Where G is universal gravitational constant, and r is the distance between the objects centers of mass. Negative sign represents the bound state.
Since we are not given the mass of the planet we have to calculate it.
$F_g=G\frac{mM}{r_p^2}\\ mg=G\frac{mM}{r_p^2}\\ g=G\frac{M}{r_p^2}$
This formula can be used for any planet. It gives you the gravitational acceleration on the planet's surface. We can use it to calculate the planet's mass:
$g=G\frac{M}{r_p^2}\\ M=\frac{gr_p^2}{G}=2.41\cdot 10^{24}kg$
Now we can calculate the potential energy of that cannonball when it reaches its maximum height.
$U=-G \frac{mM}{r}\\ U=-G \frac{mM}{r_p+h}$
When we plug in the numbers we get:
$U=-4.99\cdot 10^{10} J$
The potential energy has to be equal to the kinetic energy.
$E_k=4.99\cdot 10^{10} J$

3 0

3 years ago