3 years ago

What is one use for infrared waves? A radar B Medical imaging C Thermal imaging cameras

Physics

1 answer:

Vilka [71]3 years ago

7 0

Answer:

Im answering for free points sry

Explanation:

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A capacitor is charged to a potential of 12.0 V and is then connected to a voltmeter having an internal resistance of 3.90 MΩ. A

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Explanation:

(a) Formula to calculate the capacitance is as follows.

$V_{c} = V_{o} (e^(\frac{-t}{RC}))$

Now, putting the given values into the above formula as follows.

$V_{c} = V_{o} (e^(\frac{-t}{RC}))$

$3.5 V = 12 V (e^(\frac{-3.50 sec}{3.90 \times C}))$

$e^(\frac{-3.50 sec}{3.90 \times C})$ = 0.291

or, $\frac{-3.50 sec}{3.90 \times C}$ = ln (0.291)

$\frac{-3.50 sec}{3.90 \times C}$ = -1.23

C = 0.729 F

Hence, the value of capacitance is 0.729 F.

(b) Formula to calculate the constant of circuit is as follows.

T = $R \times C$

= $3.1 \times 0.729$

= 2.259 sec

Therefore, the time constant of the circuit is 2.259 sec.

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