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Answer:
0.15625 grams
Explanation:
Half life: It is related to the decay of radioactive material. The duration in which half of the material will be degraded/decayed. That means after half life 50% of the radioactive material will be left. Here the half life is 28 years.
Initial quantity of the sample: 2.5 grams.
After 28 years, the leftover quantity = 1.25 grams
After 56 years, the leftover quantity = 0.625 grams
After 84 Years, the leftover quantity = 0.3125 grams
After 112 years, the leftover quantity = 0.15625 grams
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)]](https://tex.z-dn.net/?f=Q%20%3D%20C%20V%20%5B1%20-%20e%5E%28-t%2FRC%29%5D)
= 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:
L_new =L+x^2 = L_new = 0.54_m.
Explanation:
Given data:
Force in the first case,
F_1 = 5N
Force in the second case,
F_2 = 20 N
Natural length of spring,
L= 0.5
Extension in the first case,
x_1 = 0.01m
Let the force constant of the spring be k.
Thus,
F_1=kx_1
5 = k × 0.01
⇒ k = 500 N/m.
The extension in the spring in the second case can be given as,
F_2=kx_2
20 = 500x_2
⇒ x_2 = 0.04 m.
Thus, the effective length of the spring would be,
L_new =L+x^2
L_new = 0.5+0.04
L_new = 0.54_m.
