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
= 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} ]](https://tex.z-dn.net/?f=Q%20%3D%20CV%5B1-e%5E%7B-t%2FRC%7D%20%5D)
The capacitor is fully charge when t approaches infinity, therefore:
![Q = \lim_{t \to \infty} a_n CV[1-e^{-t/RC} ]](https://tex.z-dn.net/?f=Q%20%3D%20%20%5Clim_%7Bt%20%5Cto%20%5Cinfty%7D%20a_n%20CV%5B1-e%5E%7B-t%2FRC%7D%20%5D)
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}](https://tex.z-dn.net/?f=Q%20%3D%20%282.66%2A10%5E%7B-6%7D%29%20F%20%2A%20%2812.0%29V%20%2A%5B1%20-%200%5D%20%3D%203.192%20%2A%2010%5E%7B-5%7D)
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](https://tex.z-dn.net/?f=%2814.3%20%2A%2010%20%5E%206%29%20C%20%3D%20%282.66%2A10%5E%7B-6%7D%29F%20%2A%2812.0%29V%2A%5B1-e%5E%7B-t%2F%282.85s%29%7D%5D%5C%5C0.552%20%3D%20e%5E%7B-t%2F%282.85s%29%7D%5C%5Ct%20%3D%20-1%20%2A%202.85%20%2A%20ln%280.552%29%20%5C%5Ct%20%3D%201.69120678%20s)
t = 1.691 s
Hope this helps! I'm not sure what the units you want, so convert to the desired units.
When hot tea is mixed to chilled coke, tea loses heat and coke gains heat. Thus, tea cools down but coke gets heated. Because it is liquid and liquid does not totally cool down to the ambient temperature, it and the iced drink will eventually reach the same temperature.
Kinetic energy of the rock just before it hits the ground=KE=33000 J
Explanation:
Weight= 2200N
mg=2200
m(9.8)=2200
m=224.5 kg
initial velocity=0
final velocity =V
using kinematic equation V²=Vi²+2gh
V²=0+2 (9.8)(15)
V=17.1 m/s
now kinetic energy= 1/2 mV²
KE= 1/2 (224.5)(17.1)²
KE=33000 J
Thus the kinetic energy of the rock just before it hits the ground=33000 J
To develop this problem it is necessary to use the expression that allows us to convert the degrees Celsius to Fahrenheit. The expression that allows to realize it is given mathematically by:

In this way for 33.2 °C:



In this way for 38.2°C



Expressed in a range term, we can say that the measure in degrees Fahrenheit is:
![[91.76\° F , 100.76\° F]](https://tex.z-dn.net/?f=%5B91.76%5C%C2%B0%20F%20%2C%20100.76%5C%C2%B0%20F%5D)