To solve this problem we will apply the expression of charge per unit of time in a capacitor with a given resistance. Mathematically said expression is given as
![q(t) = e^{-\frac{t}{(R*C)}}](https://tex.z-dn.net/?f=q%28t%29%20%3D%20e%5E%7B-%5Cfrac%7Bt%7D%7B%28R%2AC%29%7D%7D)
Here,
q = Charge
t = Time
R = Resistance
C = Capacitance
When the charge reach its half value it has passed 10ms, then the equation is,
![\frac{1}{2}*q_{final} = e^{-\frac{0.01}{(R*C)}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%2Aq_%7Bfinal%7D%20%3D%20e%5E%7B-%5Cfrac%7B0.01%7D%7B%28R%2AC%29%7D%7D)
![Ln(\frac{1}{2}) = -\frac{0.01}{RC}](https://tex.z-dn.net/?f=Ln%28%5Cfrac%7B1%7D%7B2%7D%29%20%3D%20-%5Cfrac%7B0.01%7D%7BRC%7D)
![- RC = \frac{0.01}{Ln(1/2)}](https://tex.z-dn.net/?f=-%20RC%20%3D%20%5Cfrac%7B0.01%7D%7BLn%281%2F2%29%7D)
![RC = 0.014s](https://tex.z-dn.net/?f=RC%20%3D%200.014s)
We know that RC is equal to the time constant, then
![T = RC = 0.014s = 14ms](https://tex.z-dn.net/?f=T%20%3D%20RC%20%20%3D%200.014s%20%3D%2014ms)
Therefore the time constant for the process is about 14ms
Answer:
hope this answer helps this is what I understand
The problem is solved and the questions are answered below.
Explanation:
a. To calculate the speed of the 0.66 kg ball just before the collision
V₀ + K₀ = V₁ + K₁
= mgh₀ = 1/2 mv₁²
where, h= r - r cosθ
V = ![\sqrt{2gh}](https://tex.z-dn.net/?f=%5Csqrt%7B2gh%7D)
V = 2.42 m/s
b. Calculate the speed of the 0.22 kg ball immediately after the collision
y = y₀ + Vy₀t - 1/2 gt²
0 = 1.2 - 1/2 gt²
t = 0.495 s
x = x₀ + Vx₀t
1.4 = 0 + vx₀ (0.495)
Vx₀ = 2.83 m/s
C. To Calculate the speed of the 0.66 kg ball immediately after the collision
m₁ v₁ = m₁ v₃ + m₂ v₄
(0.66)(2.42) = (0.66) v₃ + (0.22)(2.83)
V₃ = 1.48 m/s
D. To Indicate the direction of motion of the 0.66 kg ball immediately after the collision is to the right.
E. To Calculate the height to which the 0.66 kg ball rises after the collision
V₀ + k₀ = V₁ + k₁
1/2 mv₀² = mgh₁
h₁ = v₀²/2 g
= 0.112 m
F. Based on your data, No the collision is not elastic.
Δk = 1/2 m₁v₃² =1/2 m₂v₄² - 1/2 m₁v₁²
= 1/2 (0.66)(1.48)² + 1/2 (0.22)(2.83)² - 1/2 (0.66)(2.42)²
= - 0.329 J
Hence, kinetic energy is not conserved.
E = hf, and h is the Planck's constant. When larger frequency is needed, more energy will also be needed. Since the blue light has the higher frequency, it would be the<span> level X to Y's transition which is the one that has the highest energy difference.
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<span>I am hoping that
this answer has satisfied your query and it will be able to help you in your
endeavor, and if you would like, feel free to ask another question.</span>
A scientific notation is a large number for example this number 12.000000000000
You can calculate it as 12x10 to the power of 14