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Ira Lisetskai [31]
3 years ago
11

Does equation remain valid if charge is located to negative

Physics
1 answer:
Drupady [299]3 years ago
7 0

negative is answer a d l hope it helps you

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A pin fin of uniform, cross-sectional area is fabricated of an aluminum alloy (k = 160 W/m-K). The fin diameter is D = 4 mm, and
frozen [14]

Answer:

Given that

D= 4 mm

K = 160 W/m-K

h=h = 220 W/m²-K

ηf = 0.65

We know that

m=\sqrt{\dfrac{hP}{KA}}

For circular fin

m=\sqrt{\dfrac{4h}{KD}}

m=\sqrt{\dfrac{4\times 220}{160\times 0.004}}

m = 37.08

\eta_f=\dfrac{tanhmL}{mL}

0.65=\dfrac{tanh37.08L}{37.08L}

By solving above equation we get

L= 36.18 mm

The effectiveness for circular fin given as

\varepsilon =\dfrac{2\ tanhmL}{\sqrt{\dfrac{hD}{K}}}

\varepsilon =\dfrac{2\ tanh(37.08\times 0.03618)}{\sqrt{\dfrac{220\times 0.004}{160}}}

ε = 23.52

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3 years ago
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What types of electromagnetic radiation has a shorter wavelength than ultraviolet?
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4 0
3 years ago
Which two parts do the capillaries surround?
Elis [28]

Answer:

do we have to choose 2 answers here?

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Newton’s first and second laws interactive reader
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4 years ago
Light from a laser strikes a diffraction grating that has 5500 lines per centimeter. The central and first-order maxima are sepa
WINSTONCH [101]

.Answer:

491.4 nm

Explanation:

The distance between central and first maxima is,

y=0.455m

And the distance between screen abnd grating is,

L=1.62 m

Now the angle can be find as,

tan\theta=\frac{y}{L} \\\theta=tan^{-1}(\frac{0.455}{1.62})  \\\theta=15.68^{\circ}

Now the grating distance is,

d=\frac{1}{5500} cm\\d=1.82\times 10^{-6}m

Now with m=1 condition will become,

\lambda=dsin\theta

So,

\lambda=1.82\times 10^{-6}m\times sin(15.68^{\circ})\\\lambda=1.82\times 10^{-6}m\times 0.270\\\lambda=491.4\times 10^{-9}m\\\lambda=491.4 nm

Therefore the wavelength of laser light is 491.4 nm.

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3 years ago
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