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masha68 [24]
3 years ago
5

How do you calculate the average density of two different fluids?

Physics
1 answer:
ANTONII [103]3 years ago
4 0
-- Take a sample of the first fluid.
-- Measure its mass.
-- Measure its volume.
-- Divide its mass by its volume.
   This gives you the density of the first fluid.

-- Take a sample of the second fluid.
-- Measure its mass.
-- Measure its volume.
-- Divide its mass by its volume.
   This gives you the density of the second fluid.

You want their average ?
OK

-- Add  (Density of the first fluid) + (Density of the second fluid).
-- Divide the sum by  2 .

   Now you have the average of the two densities.

Note:
That's NOT necessarily the density of a mixture when you
pour some of fluid-1 and fluid-2 into a jar.  The density of the
fluid in the jar is going to depend on how much of each fluid is
in there. 
I started to calculate how much of each one has to be there in order
for the density of the mixture to be equal to the average of their two
densities. But then I sat up straight, asked myself "Why ? !" . 
Then I stopped, and went into the kitchen and ate some meatloaf.
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levacccp [35]

Answer:

i. 43.5 mH ii.  16 Ω. In phasor form Z = (8.33 + j13.66) Ω iii 58.64°

Explanation:

i. The resistance , R of the non-inductive load R = 125 V/15 A = 8.33 Ω

The reactance X of the inductor is X = 2πfL where f = frequency = 50 Hz.

So, x = 2π(50)L = 100πL Ω = 314.16L Ω

Since the current is the same when the 240 V supply is applied, then

the impedance Z = √(R² + X²) = 240 V/15 A

√(R² + X²) = 16 Ω

8.33² + X² = 16²

69.3889 + X² = 256

X² = 256 - 69.3889

X² = 186.6111

X = √186.6111

X = 13.66 Ω

Since X = 314.16L = 13.66 Ω

L = 13.66/314.16

= 0.0435 H

= 43.5 mH

ii. Since the same current is supplied in both circuits, the impedance Z of the circuit is Z = 240 V/15 A = 16 Ω.

So in phasor form Z = (8.33 + j13.66) Ω

iii. The phase difference θ between the current and voltage is  

θ = tan⁻¹X/R

= tan⁻¹(314.16L/R)

= tan⁻¹(314.16 × 0.0435 H/8.33 Ω)

= tan⁻¹(13.66/8.33)

= tan⁻¹(1.6406)

= 58.64°

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