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labwork [276]
3 years ago
12

A Van de Graaff generator can be used to create a large amount of A. current. B. static charge. C. electrical resistance. D. mix

ed positive and negative charges on its surface.
Chemistry
2 answers:
Klio2033 [76]3 years ago
8 0

Answer: The correct answer is B. Static Charge :)

Have a great day!

Eduardwww [97]3 years ago
4 0
I think the answer is B. static charge. I hope this helps! :)
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You are requested to reduce the size of 50 ton/hr of a given solid. The size of the feed is such 80% passes a 4-in (76.2 mm) scr
defon

Answer:

1) The power needed to process 50 ton/hr is 135.4 HP.

2) The void fraction of the bed is 0.37.

Explanation:

1) For this type of milling operations, we can estimate the power needed for an operation according to the work index (Ei), the passing size of the circuit feed (F80) and the passing size of the product (P80).

We assume the units of Ei are kWh/t.

The equation that relates this parameters and the power is (size of particles in μm):

W=Ei*(\frac{10}{\sqrt{P80}} -\frac{10}{\sqrt{F80}} )\\\\W=9.45*(\frac{10}{\sqrt{3175\mu m}} -\frac{10}{\sqrt{76200mm}} )\\\\\\W=9.45*(0.1774+0.0362)=2.019 kWh/t

The power needed to process 50 ton/hor is

P=2.0194\frac{kWh}{Ton}*\frac{50Ton}{h}*\frac{1.341HP}{1kW}=   135.4 \, HP

2) The density of the packed bed can be expressed as

\rho=f_v*\rho_v+f_s*\rho_s

being f the fraction and ρ the density of every fraction. We know that the density of the void is 0 (ρv=0) and that fv=1-fs (the sum  of the fractions ois equal to the total space).

Then we can rearrange

\rho=f_v*\rho_v+f_s*\rho_s\\\\\rho=f_v*0+(1-f_v)*\rho_s\\\\\rho/\rho_s=1-f_v\\\\f_v=1-\rho/\rho_s=1-990/1570=1-0.63=0.37

The void fraction of the bed is 0.37.

6 0
3 years ago
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