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3 years ago

Which of the following would be the best thermal conductor?

Physics

2 answers:

Elis [28]3 years ago

4 0

The best thermal conductor is b. metal

r-ruslan [8.4K]3 years ago

3 0

Answer: Option (b) is the correct answer.

Explanation:

A material through which energy can travel in the form of heat without any movement of material itself is known as a thermal conductor.

It is known that metals are good conductors of heat and electricity. Hence, they act as good thermal conductors.

Thus, we can conclude that out of the given options metal would be the best thermal conductor.

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If a sample of a radioactive isotope has a half-life of 1 year, how much of the original sample will be left at the end of the s

Tasya [4]

Answer:

1/4 of the original

Explanation:

That would be TWO half lives:

1/2 * 1/2 = 1/4 <======= 1/4 would be left

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3 years ago

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Sonbull [250]

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2 years ago

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The standard deviation of Eric’s data is 0.8°C. Martha conducted the same experiment. Her average temperature was 35.1 with a st

REY [17]

Answer:

less precise than

Explanation:

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3 years ago

Consider an optical cavity of length 40 cm. Assume the refractive index is 1, and use the formula for Icavity vs wavelength to p

Bad White [126]

Answer:

Diode Lasers

Consider a InGaAsP-InP laser diode which has an optical cavity of length 250

microns. The peak radiation is at 1550 nm and the refractive index of InGaAsP is

4. The optical gain bandwidth (as measured between half intensity points) will

normally depend on the pumping current (diode current) but for this problem

assume that it is 2 nm.

(a) What is the mode integer m of the peak radiation?

(b) What is the separation between the modes of the cavity? Please express your

answer as Δλ.

(d) What is the reflection coefficient and reflectance at the ends of the optical

cavity (faces of the InGaAsP crystal)?

(e) The beam divergence full angles are 20° in y-direction and 5° in x-direction

respectively. Estimate the x and y dimensions of the laser cavity. (Assume the

beam is a Gaussian beam with the waist located at the output. And the beam

waist size is approximately the x-y dimensions of the cavity.)

Solution:

(a) The wavelength λ of a cavity mode and length L are related by

λ = , where m is the mode number, and n is the refractive index.

So the mode integer of the peak radiation is

1290

1055.1

10250422

= ×

××× == −

−

m .

(b) The mode spacing is given by nL

c f 2

=Δ . As

c f = , λ

Δ−=Δ 2

c f .

Therefore, we have nm

nL f

20.1

)10250(42

)1055.1(

2 || 6

2 2 26

= ×××

× ==Δ=Δ −

− λλ λ .

bandwidth could fit in two possible modes.

For mode integer of 1290, nm

nL 39.1550

1290

10250422 6

= ××× ==

−

Take m = 1291, nm

nL 18.1549

1291

10250422 6

= ××× ==

−

Or take m = 1289, nm

nL 59.1551

1289

10250422 6

= ××× ==

−

λ .

Explanation:

8 0

3 years ago