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.
Answer:
B. moving faster than car B, but not necessarily accelerating
Explanation:
Velocity is the speed of something. So car A's velocity is greater than car B but does not mean car A is accelerating.
Two Sets of solid double yellow lines with spaced two or more feet apart indicates that Do not drive on or over the barrier.
<h3>What are the line colors in lane control?</h3>
Solid yellow lines - it is marked at the center of a road and used for two-way traffic.
Broken yellow lines - It indicate that you may pass if the broken line is next to your driving lane.
Two solid yellow lines - It indicates no passing or Do not drive on or over the barrier.
Two sets of solid double yellow lines are considered a barrier , make a left turn, or a U-turn to cross it.
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Starting velocity (u)=0
acceleration (a) = 9.8
time (t) = 1.5
final velocity (v) =?
v=at+u
v=9.8(1.5)+0
v=14.7m/s2
Answer:
<em>a) 3344 N</em>
<em>b) 3344 N</em>
Explanation:
This is the complete question
1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 5000 N. Rolling friction can be neglected. A. What is the magnitude of the force of the car on the truck? Express your answer to two significant figures and include the appropriate units. B. What is the magnitude of the force of the truck on the car?
Mass of the car = 1100 kg
Mass of the truck = 2200 kg
Force exerted on the ground by the car = 5000 N
The total mass in the system = 1100 + 2200 = 3300 Kg
Total force in the system = 5000 N
Recall that the force in the system = mass x acceleration
therefore,
5000 = 3300 x a
Total acceleration in the system = 5000/3300 = 1.52 m/s^2
The force on the truck individually fro the car, will be the product of this acceleration and its mass
Force on the truck = 2200 x 1.52 = <em>3344 N</em>
b) Force on the car From the truck will be equal to this force but will act in the opposite direction.
Force on the car from the truck is <em>3344 N </em>