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3 years ago

40 POINTS!!

Physics

2 answers:

Mazyrski [523]3 years ago

5 0

Answer:

1-put the black can in the temperature thing.

2-put the silver can well put water in it and leave it out side for a bit.

3-judge of which one will have a larger increase.

Explanation:

Mademuasel [1]3 years ago

3 0

Answer:

1 put the black can in the temperature thing.

2.put the silver can well put water in it and leave it out side for a while.

3.Then you can be the judge of witch one will have a larger increase.

hope this helps :)

happy to help any time:)

Explanation:

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g In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period

NeTakaya

The question is incomplete. The complete question is :

In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period of the oscillations is 1.2 microseconds (1.2*10^-6). What is Q?

Solution :

The underdamped RLC circuit

$v_{t} = ve^{-\frac{R}{2L}t} \cos \omega t$

$\omega = \sqrt{\frac{1}{LC}-\frac{R^2}{4L^2}}= \frac{2 \pi}{T}$

We know in one time period, v = 2v, at t = T, $v_t = 3.8 v$

so, $9.8 = 5 e^{-\frac{R}{2L}T} \cos \frac{2 \pi}{T}T$

$e^{-\frac{R}{2L}T} = \frac{3.8}{5} \times 1$

$\frac{R}{2L}T= \ln \frac{5}{3.8}$

$\frac{R}{L}= \frac{2}{1.2 \times 10^{-6}} \ln \frac{5}{3.8}$

$\frac{R}{L} = 457.3 \times 10^3$

Now, Q value $= \frac{1}{R}\sqrt{\frac{L}{C}}$

$=\sqrt{\frac{L}{R^2C}\times \frac{L}{L}}$

$=\sqrt{(\frac{L}{R})^2 \times \frac{1}{LC}}$

$\frac{1}{LC}=27.43 \times 10^{12}$

∴ $Q=\sqrt{\left(\frac{1}{457.3 \times 10^3}\right)^2 \times 27.43 \times 10^{12}}$

$Q=\sqrt{131.166}$

= 11.45

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larisa [96]

Hi,

I've found a link that should assist you or answer your question.
http://click.dji.com/ANbvbbP7bwUWtSACp6U_?pm=link&as=0004

Have a nice day!

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Explanation:

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