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SpyIntel [72]
3 years ago
13

A circuit contains a resistor in series with a capacitor, the series combination being connected across the terminals of a batte

ry. The time constant for charging the capacitor is 4.6 s when the resistor has a resistance of 1.4 x 104 Ω. What would the time constant be if the resistance had a value of 5.6 x 104 Ω?
Physics
1 answer:
Natali5045456 [20]3 years ago
8 0

Answer:

18.4 s

Explanation:

The time constant of an RC circuit is given by

\tau = RC

where

R is the resistance

C is the capacitance

For the first circuit we have

\tau = 4.6 s

R=1.4\cdot 10^4 \Omega

So we can find the capacitance

C=\frac{\tau}{R}=\frac{4.6 s}{1.4\cdot 10^4 \Omega}=3.29\cdot 10^{-4} F

Now in the second circuit, the new resistance is

R=5.6\cdot 10^4 \Omega

So the new time constant will be

\tau = RC=(5.6\cdot 10^4 \Omega )(3.29\cdot 10^{-4} F)=18.4 s

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3 years ago
It is estimated that 1kg of body fat will provide 3.8 * 10^7 J of energy. A 67kg mountain climber decides to climb a mountain 35
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Actually, it's more correct to say that gravity does NEGATIVE work
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                     Work = (mass) x (gravity) x (height) .

For the guy in this problem:

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If he eats no candy bars on the way, and completely depends on
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4 0
3 years ago
QUESTION 4.
n200080 [17]

Answer:

Follows are the solution to this question:

Explanation:

Please find the correct question in the attachment file.

Let:

\overrightarrow{R}= R_i\hat{i}+R_j\hat{j}+R_k\hat{k}\\\\\overrightarrow{S}= S_i\hat{i}+S_j\hat{j}+S_k\hat{k}\\\\

Calculating the value of  \overrightarrow{R} \times \overrightarrow{S}:

\to \left | \begin{array}{ccc}\hat{i}&\hat{j}&\hat{K}\\R_i&R_j&R_k\\S_i&S_j&S_k\end{array}\right | = \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j]

Calculating the value of \overrightarrow{R}  \cdot (\overrightarrow{R} \times \overrightarrow{S}):

\to (R_i\hat{i}+R_j\hat{j}+R_k\hat{k}) \cdot ( \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j])

by solving this value it is equal to 0.

8 0
3 years ago
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