Which variable is not required to calculate the gibbs free-energy change?

When an object such as a plastic comb is charged by rubbing it with a cloth, the net charge is typically a few microcoulombs. If

masya89 [10]

The concept required to solve this problem is quantization of charge.

First the number of electrons will be calculated and then the total mass of the charge.

With these data it will be possible to calculate the percentage of load in the mass.

$Q= ne$

Here Q is the charge, n is the number of electrons and e is the charge on the electron

$n = \frac{Q}{e}$

Replacing,

$n = \frac{4*10^{-6}C}{1.6*10^{-19}}$

$n = 2.5 * 10^{13}77$

According to the quantization of charge the charge is defined as product of the number of electron and the charge on the electron

The total mass of the charge is

$m= nm_e$

Here,

m = Mass of the charge

n = Number of electrons

$m_e$ = Mass of the electron

$\text{Percentage change} = \frac{nm_e}{M}*100$

Replacing we have

$\text{Percentage change} = \frac{(2.5*10^13)(9.1*10^{-28})}{33}*100$

$\text{Percentage change} = 6.9*10^{-14} \%$

6 0

4 years ago

A machine part has the shape of a solid uniform sphere of mass 250 g and a diameter of 4.30 cm. It is spinning about a frictionl

zysi [14]

Answer: $\alpha =9.302\ rad/s^2$

Explanation:

Given

mass of sphere $m=250\ gm$

diameter of sphere $d=4.30\ cm$

radius $r=\frac{4.30}{2}\ cm$

$f=0.0200\ N$

friction will provide resisting torque so

$f\times r=I\times \alpha$

where $I=\text{moment of Inertia}$

$f=\text{friction force}$

$\alpha =\text{angular acceleration}$

$I=\frac{2}{5}mr^2$

$0.02\times r=\frac{2}{5}mr^2\times \alpha$

$\alpha =\frac{5}{2r}\times f$

$\alpha =\frac{5}{2}\times \frac{2}{4.3\times 10^{-2}}\times 0.02$

$\alpha =9.302\ rad/s^2$

(b)time taken to decrease its rotational speed by $21\ rad/s$

$t=\dfrac{\Delta \omega }{\alpha }$

$t=\dfrac{21}{9.302}$

$t=2.25\ s$

6 0

3 years ago

How to find the magnitude of a centripetal force?<br>(formula)

frozen [14]

Answer:

F=mv^2÷r

Explanation:

i know every thing

the magnitude f of the centripetal force is equal to the mass m of the body times it veloctiy squared v^2 divided by the radius r of its path

4 0

3 years ago

An lc circuit oscillates at a frequency of 2000 Hz. What will the frequency be if the inductance is quadrupled?

Irina-Kira [14]

<h2>Answer:</h2>

An LC circuits if formed by an inductor and a capacitor. The charge on the capacitor and the current through the inductor both vary sinusoidally with time. Also, energy is transferred between magnetic energy in the inductor and electrical energy in the capacitor. But <em>what happens with the frequency if the inductance is quadrupled? </em>that is, if initially the inductance is $L$ and the frecuency $f=2000Hz$ if now $L_{New}=4L$ What will the frequency be? Well, we know that the frequency, inductance and capacitance are related as:

$f=\frac{1}{2\pi}\sqrt{\frac{1}{LC}}$

and this equals 2000Hz. If now L is quadrupled:

$f_{1}=\frac{1}{2\pi}\sqrt{\frac{1}{4LC}}=\frac{1}{2\pi}\sqrt{\frac{1}{4}}\sqrt{\frac{1}{LC}} \\ \\ \therefore f_{1}=(\frac{1}{2})\frac{1}{2\pi}\sqrt{\frac{1}{LC}} \\ \\ \\ Since \ f=\frac{1}{2\pi}\sqrt{\frac{1}{LC}} \ then: \\ \\ f_{1}=\frac{1}{2}f \therefore f_{1}=\frac{2000}{2} \\ \\ \therefore \boxed{f_{1}=1000Hz}$

<em>Finally, if L is quadrupled the frequency is half the original frequency and equals 1000Hz</em>

3 0

3 years ago

In a series RLC ac circuit, a second resistor is connected in series with the resistor previously in the circuit. As a result of

AnnyKZ [126]

Answer:

* The first thing we observe is that the frequency response does not change

* The current that circulates in the circuit decreases due to the new resistance at the resonance point,

Z = R + R₂

Explanation:

The impedance of a series circuit is

Z₀² = R² + (X_L-X_C) ²

when we place another resistor in series the initial resistance impedance changes to

Z² = (R + R₂) ² + (X_L - X_C) ²

let's analyze this expression

* The first thing we observe is that the frequency response does not change

* The current that circulates in the circuit decreases due to the new resistance at the resonance point,

Z = R + R₂

8 0

3 years ago