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ankoles [38]
1 year ago
9

A process with a negative change in enthalpy and a negative change in entropy will generally be:________

Physics
1 answer:
valentina_108 [34]1 year ago
3 0

A process with a negative change in enthalpy and a negative change in entropy will generally be: <u>spontaneous</u>.

<h3>Gibbs free energy:</h3>

Since the Gibbs free energy is a parameter that tells us whether a chemical reaction is spontaneous (Gibbs free energy less than 0) or nonspontaneous (Gibbs free energy greater than 0) in this situation, we can describe it mathematically as:

                                      ΔG = ΔH - TΔS

Therefore, any process with a negative change in enthalpy and a positive change in entropy will be spontaneous. If the enthalpy and the entropy are both negative, the subtraction becomes always negative, for which the Gibbs free energy is also negative.

One of the most crucial thermodynamic functions for the characterization of a system is the Gibbs free energy. It influences results like the voltage of an electrochemical cell and the equilibrium constant for a reversible reaction, among others.

Learn more about spontaneous here:

brainly.com/question/16975806

#SPJ4

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A 37 cm long solenoid, 1.8 cm in diameter, is to produce a 0.50 T magnetic field at its center. If the maximum current is 4.4 A,
taurus [48]

Answer:

33,458.71 turns

Explanation:

Given: L = 37 cm = 0.37 m, B= 0.50 T, I = 4.4 A, n= number of turn per meter

μ₀ = Permeability of free space = 4 π × 10 ⁻⁷

Solution:

We have B = μ₀ × n × I

⇒ n = B/ (μ₀ × I)

n = 0.50 T / ( 4 π × 10 ⁻⁷ × 4.4 A)

n = 90,428.94 turn/m

No. of turn through 0.37 m long solenoid = 90,428.94 turn/m × 0.37

= 33,458.71 turns

8 0
3 years ago
Read 2 more answers
Energy that is passed on from one trophic level to the next is called what?
yan [13]
Energy decreases as it moves uptrophic levels because energy is lost as metabolic heat when the organisms from one trophic level are consumed by organisms from the next level.Trophic level transfer efficiency (TLTE) measures the amount of energy that is transferred between trophic levels.
8 0
3 years ago
Pls help me with this fast. I will mark brainiest
denis23 [38]

Answer:

a) 70, 95

b) 95-70= 25cc

c) density= mass/volume

102/25

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5 0
3 years ago
About 65 million years ago an asteroid struck Earth in the area of the Yucatán Peninsula and wiped out the dinosaurs and many ot
9966 [12]

Answer:

2.44156\times 10^{13}\ m^3

29010.53917 m

Explanation:

\rho = Density of asteroid = 2 g/cm³

V = Volume

d = Diameter = 10 km

r = Radius = \dfrac{d}{2}=\dfrac{10}{2}=5\ km

v = Velocity = 11 km/s

H_v = Heat vaporization of water = 2.26\times 10^6\ J/kg

\Delta T = Change in temperature = 100-20

Mass is given by

m=\rho V\\\Rightarrow m=\rho\dfrac{4}{3}\pi r^3\\\Rightarrow m=2000\dfrac{4}{3}\times \pi\times 5000^3\\\Rightarrow m=1.0472\times 10^{15}\ kg

The kinetic energy is

K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1.0472\times 10^{15}\times 11000^2\\\Rightarrow K=6.33556\times 10^{22}\ J

Heat is given by

Q=mc\Delta T+mH_v\\\Rightarrow 6.33556\times 10^{22}=m\times (4186\times (100-20)+2.26\times 10^6)\\\Rightarrow m=\dfrac{ 6.33556\times 10^{22}}{4186\times (100-20)+2.26\times 10^6}\\\Rightarrow m=2.44156\times 10^{16}\ kg

Mass of water is 2.44156\times 10^{16}\ kg

Volume is \dfrac{2.44156\times 10^{16}}{10^3}=2.44156\times 10^{13}\ m^3

Amount of water is 2.44156\times 10^{13}\ m^3

If it were a cube

h=V^{\dfrac{1}{3}}\\\Rightarrow h=(2.44156\times 10^{13})^{\dfrac{1}{3}}\\\Rightarrow h=29010.53917\ m

The height of the water would be 29010.53917 m

4 0
3 years ago
to 10 Hz. Superimposed on this signal is 60-Hz noise with an amplitude of 0.1 V. It is desired to attenuate the 60-Hz signal to
givi [52]

Answer:

G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1

If we square both sides we got:

G^2 (1+\frac{f}{f_c})^{2n}= 1

We divide both sides by G^2 and we got:

(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}

Now we can apply log on both sides and we got:

2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})

And solving for n we got:

n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}

And replacing we got:

n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}

n = \frac{4.60517}{3.8918}=1.18

And since n needs to be an integer the correct answer would be n=2 for the filter order.

Explanation:

For this case we can use the formula for the Butterworth filter gain given by:

[tec] G = \frac{1}{\sqrt{1 +(\frac{f}{f_c})^{2n}}}[/tex]

Where:

G represent the transfer function and we want that G =0.1 since the desired signal is less than 10% of it's value

f_c = 10 Hz represent the corner frequency

f= 60 Hz represent the original frequency

n represent the filter order and that's the variable that we need to find

G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1

If we square both sides we got:

G^2 (1+\frac{f}{f_c})^{2n}= 1

We divide both sides by G^2 and we got:

(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}

Now we can apply log on both sides and we got:

2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})

And solving for n we got:

n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}

And replacing we got:

n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}

n = \frac{4.60517}{3.8918}=1.18

And since n needs to be an integer the correct answer would be n=2 for the filter order.

7 0
3 years ago
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