Answer:
t = 1.098*RC
Explanation:
In order to calculate the time that the capacitor takes to reach 2/3 of its maximum charge, you use the following formula for the charge of the capacitor:
![Q=Q_{max}[1-e^{-\frac{t}{RC}}]](https://tex.z-dn.net/?f=Q%3DQ_%7Bmax%7D%5B1-e%5E%7B-%5Cfrac%7Bt%7D%7BRC%7D%7D%5D)
(1)
Qmax: maximum charge capacity of the capacitor
t: time
R: resistor of the circuit
C: capacitance of the circuit
When the capacitor has 2/3 of its maximum charge, you have that
Q=(2/3)Qmax
You replace the previous expression for Q in the equation (1), and use properties of logarithms to solve for t:
![Q=\frac{2}{3}Q_{max}=Q_{max}[1-e^{-\frac{t}{RC}}]\\\\\frac{2}{3}=1-e^{-\frac{t}{RC}}\\\\e^{-\frac{t}{RC}}=\frac{1}{3}\\\\-\frac{t}{RC}=ln(\frac{1}{3})\\\\t=-RCln(\frac{1}{3})=1.098RC](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B2%7D%7B3%7DQ_%7Bmax%7D%3DQ_%7Bmax%7D%5B1-e%5E%7B-%5Cfrac%7Bt%7D%7BRC%7D%7D%5D%5C%5C%5C%5C%5Cfrac%7B2%7D%7B3%7D%3D1-e%5E%7B-%5Cfrac%7Bt%7D%7BRC%7D%7D%5C%5C%5C%5Ce%5E%7B-%5Cfrac%7Bt%7D%7BRC%7D%7D%3D%5Cfrac%7B1%7D%7B3%7D%5C%5C%5C%5C-%5Cfrac%7Bt%7D%7BRC%7D%3Dln%28%5Cfrac%7B1%7D%7B3%7D%29%5C%5C%5C%5Ct%3D-RCln%28%5Cfrac%7B1%7D%7B3%7D%29%3D1.098RC)
The charge in the capacitor reaches 2/3 of its maximum charge in a time equal to 1.098RC
Answer:
c) True. If the coating cancels the light requested by the reflection, so there is more energy to enter the cell and therefore its efficiency increases
Explanation:
This exercise asks to analyze the effect of the antireflective coating on the efficiency of solar cells.
Let's start by expressing the expression for the interference of two light beams taken at when
* the phase change introduced when passing from air to 180º film
* the wavelength change by the refractive index of the film ln = lo / n
therefore the expression for destructive interference is
2 n t = m λ
where m is an integer
with these concepts we can analyze the different statements
a) False. Phase shift does not change the wavelength of light
b) False. The refractive index of the solar cell is not affected by the refractive index of the film since the two materials do not mix.
c) True. If the coating cancels the light requested by the reflection, so there is more energy to enter the cell and therefore its efficiency increases
d) false. In solar cells the incidence is almost normal, therefore the effect of refraction (separation of colors for different angles) is very small
Answer:
13.8 N
Explanation:
Pressure on the one end of the hydraulic system = Pressure on the other end
Pressure = Force / Area where Force is in Newton, area is in m²
so Force of one end (F1) / area of that end = force of the other end (F2) / area of that end
3112 / ( 707 /10000) in m² = F2 / ( 3.14 / 10000) in m²
cross multiply
44016.97 × 0.000314 = 13.82 N
Answer:
μsmín = 0.1
Explanation:
- There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
- This friction force has a maximum value, that can be written as follows:
where μs is the coefficient of static friction, and Fn is the normal force,
perpendicular to the wall and aiming to the center of rotation.
- This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
- This force has the following general expression:
where ω is the angular velocity of the riders, and r the distance to the
center of rotation (the radius of the circle), and m the mass of the
riders.
Since Fc is actually Fn, we can replace the right side of (2) in (1), as
follows:
- When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:
- (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
- Cancelling the masses on both sides of (4), we get:
- Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:
- Replacing by the givens in (5), we can solve for μsmín, as follows:
Answer:
The correct answer is "53.15 days".
Explanation:
Given that:
Half life of 
,
- Let the initial activity be "
". - and, activity to time t be "R".
To find t when R will be "1%" of , then
⇒ 
As we know,
⇒ 
or,
∴ 
By putting the values, we get
We know that,
Decay constant, 
hence,
⇒ 
