When the material that exits is lesser in amount than that of the entering material in a black box experiment, the parts of the system need to be changed.
<h3>What happens in a black box experiment?</h3>
In a black box experiment, the experimenters need to make assumptions regarding the drawing of conclusions. One such conclusion is the amount of material that exits.
If such amount is lesser than the one that enters the system, such experiment concludes that it is the time to change the parts of the system.
Hence, option D holds true regarding the black box experiment.
Learn more about black box experiment here:
brainly.com/question/13403296
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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}](https://tex.z-dn.net/?f=%3D2%2A%5Cfrac%7B%5Cpi%20D%5E%7B2%7D%20%7D%7B4%7D%20%2B%20%5Cpi%20DL%5C%5C%3D2%2A%5Cfrac%7B%5Cpi%20%2A%28%5Cfrac%7B0.4%7D%7B100%7D%20%29%5E%7B2%7D%20%7D%7B4%7D%20%2B%20%5Cpi%20%2A%5Cfrac%7B0.4%7D%7B100%7D%20%2A%5Cfrac%7B1.5%7D%7B100%7D)
= 2.136 *10^-4 ![m^{2}](https://tex.z-dn.net/?f=m%5E%7B2%7D)
Heat flux =
= 2808.99 ![W/m^{2}](https://tex.z-dn.net/?f=W%2Fm%5E%7B2%7D)
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](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5Cfrac%7B2%2A%5Cpi%20D%5E%7B2%7D%20%7D%7B4%7D%20%7D%7B%5Cfrac%7B2%2A%5Cpi%20D%5E%7B2%7D%7D%7B4%7D%20%2B%20%5Cpi%20DL%20%7D%20%5C%5C%5C%5C%3D%5C%5C%5Cfrac%7B%5Cfrac%7B2%2A%5Cpi%20%2A%28%5Cfrac%7B0.4%7D%7B100%7D%20%29%5E%7B2%7D%20%7D%7B4%7D%20%7D%7B%5Cfrac%7B2%2A%5Cpi%20%2A%28%5Cfrac%7B0.4%7D%7B100%7D%20%20%29%5E%7B2%7D%20%7D%7B4%7D%20%2B%5Cpi%20%2A%5Cfrac%7B0.4%7D%7B100%7D%20%2A%5Cfrac%7B1.5%7D%7B100%7D%20%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B2.51%2A10%5E%7B-5%7D%20%7D%7B2.136%2A10%5E%7B-4%7D%20%7D%20%5C%5C%5C%5C%5C%3D%200.1175)
=11.75%
That’s too hard for me lol oof
Explanation:
The two types of furnaces used in steel production are:
<u>Basic oxygen furnace </u>
In basic oxygen furnace, iron is combined with the varying amounts of the steel scrap and also small amounts of the flux in the Blast Furnace. Lance is introduced in vessel and blows about 99% of the pure oxygen causing rise in temperature to about 1700°C. This temperature melts scrap and the impurities are oxidized and results in the liquid steel.
<u>Electric arc furnace</u>
Electric arc furnace reuses existing steel. Furnace is charged with the steel scrap. It operates on basis of electrical charge between the two electrodes providing heat for process. Power is supplied through electrodes placed in furnace, which produce arc of the electricity through scrap steel which raises temperature to about 1600˚C. This temperature melts scrap and the impurities can be removed through use of the fluxes and results in the liquid steel.