Why are chemical properties harder to observe than physical properties?

Question 9 In an RC series circuit, ε = 12.0 V, R = 1.07 MΩ, and C = 2.66 µF. (a) Calculate the time constant. (b) Find the maxi

meriva

Answer:

a.) τ = 2.85 s b.) Q = 3.19 * 10^-5 C c.) t = 1.691 s

Explanation:

So we are told that it is a RC circuit. We are told $Q = C V [1 - e^(-t/RC)]$ = 12.0 V, R = 1.07 MΩ and C = 2.66 µF.

a.) The time constant for RC circuit, τ = RC. Substituting our known values we get:

τ = RC where R = (1.07 * 10 ^ 6)Ω and C = (2.66 * 10 ^ -6) F

τ = (1.07 * 10 ^ 6)Ω * (2.66 * 10 ^ -6) F = 2.8462 s ≈ 2.85 s

τ = 2.85 s

b.) The relationship between capacitance, potential, charge is given:

$Q = CV[1-e^{-t/RC} ]$

The capacitor is fully charge when t approaches infinity, therefore:

$Q = \lim_{t \to \infty} a_n CV[1-e^{-t/RC} ]$

When t approaches infinity, the term e becomes very small (e^-∞ = 0), therefore we can simplify the equation and plug in our values

$Q = (2.66*10^{-6}) F * (12.0)V *[1 - 0] = 3.192 * 10^{-5}$

Q = 3.19 * 10^-5 C

c.) Using the same equation as before, we can substitute Q in and solve for Q:

$(14.3 * 10 ^ 6) C = (2.66*10^{-6})F *(12.0)V*[1-e^{-t/(2.85s)}]\\0.552 = e^{-t/(2.85s)}\\t = -1 * 2.85 * ln(0.552) \\t = 1.69120678 s$

t = 1.691 s

Hope this helps! I'm not sure what the units you want, so convert to the desired units.

6 0

3 years ago

Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr

soldi70 [24.7K]

$Given:\\T=29.46y\approx 9.29\cdot 10^8s\\M_S\approx2.0\cdot 10^{30}kg\\G=6.67\cdot 10^{-11} \frac{m^3}{kg\cdot s^2} \\\\Find:\\R=?\\\\Solution:\\\\F_g= G\frac{mM_s}{R^2} \\\\F_c= \frac{mv^2}{R} \\\\F_g=F_c\\\\G\frac{mM_s}{R^2}=\frac{mv^2}{R} \Rightarrow G\frac{M_s}{R^2}=\frac{v^2}{R}\\\\v=\omega r\\\\G\frac{M_s}{R^2}= \frac{\omega^2R^2}{R}\Rightarrow G\frac{M_s}{R^2}=\omega^2R \\\\\omega= \frac{2 \pi }{T} \\\\G\frac{M_s}{R^3}= \frac{4 \pi ^2}{T^2}$

$GM_ST^2=4 \pi ^2R^3\Rightarrow R= \sqrt[3]{ \frac{GM_ST^2}{4 \pi ^2} }\\\\\\R= \sqrt[3]{ \frac{6.67\cdot 10^{-11} \frac{m^3}{kg\cdot s^2}\cdot2.0\cdot 10^{30}kg( 9.29\cdot 10^8s)^2}{4\cdot 3.14^2} } \approx 1.42\cdot 10^{12}m$

3 0

4 years ago

An 1100-kg car traveling at 27.0 m/s starts to slow down and comes to a complete stop in 578 m. What is the magnitude of the ave

Trava [24]

Answer:

693.685 N

Explanation:

t = Time taken

u = Initial velocity = 27 m/s

v = Final velocity

s = Displacement = 578 m

a = Acceleration

m = Mass

Equation of motion

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-27^2}{2\times 578}\\\Rightarrow a=-0.63062\ m/s^2$

Force

F = ma

$F=1100\times -0.63062\\\Rightarrow F=-693.6851\ N$

Magnitude of braking force required to stop the car is 693.685 N

7 0

4 years ago

Which has a charge with the greatest magnitude?

pishuonlain [190]

Answer:

an object with charge -3

3 0

3 years ago

Which three characteristics make a glowing piece of charcoal a blackbody radiator? A. The frequencies of EMR it emits depend on

Eva8 [605]

Answer:

A. The frequencies of EMR it emits depend on its temperature

B. It emits only one frequency of EMR

C. It absorbs most of the EMR it receives

Explanation:

A blackbody is an object that absorbs most of the electromagnetic spectrum of energy that falls on it.

According to law to reradiates most of the available energy back on top the outer space at an efficiency of 100% and thus radiation may be in the visible range of temperature than are in 1000K.

4 0

3 years ago