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Nookie1986 [14]
3 years ago
15

Concept Simulation 20.4 provides background for this problem and gives you the opportunity to verify your answer graphically. Ho

w many time constants (a decimal number) must elapse before a capacitor in a series RC circuit is charged to 65.0% of its equilibrium charge?
Physics
1 answer:
77julia77 [94]3 years ago
6 0

Answer:

The time constant is 1.049.

Explanation:

Given that,

Charge q{t}= 0.65 q_{0}

We need to calculate the time constant

Using expression for charging in a RC circuit

q(t)=q_{0}[1-e^{-(\dfrac{t}{RC})}]

Where, \dfrac{t}{RC} = time constant

Put the value into the formula

0.65q_{0}=q_{0}[1-e^{-(\dfrac{t}{RC})}]

1-e^{-(\dfrac{t}{RC})}=0.65

e^{-(\dfrac{t}{RC})}=0.35

-\dfrac{t}{RC}=ln (0.35)

-\dfrac{t}{RC}=-1.049

\dfrac{t}{RC}=1.049

Hence, The time constant is 1.049.

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The age of the universe is around 100,000,000,000,000,000s. A top quark has a lifetime of roughly 0.000000000000000000000001s. W
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Answer:

10⁴¹ s quark top lives have been in the history of the universe.

Explanation:

You need to determine how many quark top lives there have been in the history of the universe, that is, what is the age of the universe divided by the lifetime of a top quark. Expressed in a formula, this is:

t\frac{Age of the universe}{Lifetime of a top quark}

Yo know that the "Age of the universe" is 100,000,000,000,000,000  which can also be expressed as 10¹⁷ s .

You also know that the "Lifetime of a top quark" is 0.000000000000000000000001 which can also be expressed as 10⁻²⁴ s.

Then t=\frac{10^{17} }{10^{-24} }

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