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2 years ago

You want to see both Michael and Meet Wally Sparks. If you purchase tickets for both

Engineering

1 answer:

Aleonysh [2.5K]2 years ago

6 0

$2.

Both tickets cost $1.50
$1.50 x 2 = $3
$5 - $3 = $2

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Difference between star and delta connection

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2 years ago

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Sonja [21]

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2 years ago

For the Mohr's circle of a plane-strain element, which of the following changes as a result of shear strain change?

OverLord2011 [107]

Answer:

Explanation:

only the radius of the circle as a result of shear strain change.

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2 years ago

What are the different types of documents used to communicate engineering designs?

Ipatiy [6.2K]

Answer:

COMMON ENGINEERING DOCUMENTS

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3 years ago

A cylindrical resistor element on a circuit board dissipates 0.6 W of power. The resistor is 1.5 cm long, and has a diameter of

Burka [1]

Answer:

a. 51.84Kj

b. 2808.99 W/m^2

c. 11.75%

Explanation:

Amount of heat this resistor dissipates during a 24-hour period

= amount of power dissipated * time

= 0.6 * 24 = 14.4 Watt hour

(Note 3.6Watt hour = 1Kj )

=14.4*3.6 = 51.84Kj

Heat flux = amount of power dissipated/ surface area

surface area = area of the two circular end + area of the curve surface

$=2*\frac{\pi D^{2} }{4} + \pi DL\\=2*\frac{\pi *(\frac{0.4}{100} )^{2} }{4} + \pi *\frac{0.4}{100} *\frac{1.5}{100}$

= 2.136 *10^-4 $m^{2}$

Heat flux = $\frac{0.6}{2.136 * 10^{-4} }$ = 2808.99 $W/m^{2}$

fraction of heat dissipated from the top and bottom surface

$=\frac{\frac{2*\pi D^{2} }{4} }{\frac{2*\pi D^{2}}{4} + \pi DL } \\\\=\\\frac{\frac{2*\pi *(\frac{0.4}{100} )^{2} }{4} }{\frac{2*\pi *(\frac{0.4}{100} )^{2} }{4} +\pi *\frac{0.4}{100} *\frac{1.5}{100} } \\\\=\frac{2.51*10^{-5} }{2.136*10^{-4} } \\\\\= 0.1175$

=11.75%

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3 years ago

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