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3 years ago

8

What is the difference of dispersed phase and continuous phase?

1 answer:

OverLord2011 [107]3 years ago

4 0

Answer:

The phase existing as small droplets is called the dispersed phase and the surrounding liquid is known as the continuous phase. Emulsions are commonly classified as oil-in-water (O/W) or water-in-oil (W/O) depending on whether the continuous phase is water or oil.

Explanation:

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tester [92]

Answer:

nopeeeeeeeeee. false

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3 years ago

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. The p.d. at the terminals of a battery is 25V when no load is connected and 24V when a load taking 10A is connected. Determine

makvit [3.9K]

Answer:

the internal resistance of the cell is 0.1 ohm.

Explanation:

Given;

p.d at the terminals of a battery at no load, E₁ = 25 V

p.d at the terminals of a battery at a load, E₂ = 24 V

current through the circuit, I = 10 A

The potential drop across the circuit, V = E₁ - E₂

= 25 V - 24 V

= 1 V

The internal resistance of the cell is calculated as follows;

r = V/I

r = 1 / 10

r = 0.1 ohm

Therefore, the internal resistance of the cell is 0.1 ohm.

6 0

3 years ago

neptune is an average distance of 4.5×10^12m from the sun. Estimate the length of the Neptunian year.

Vikentia [17]

As per Kepler's third law we know that

$\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}$

now here we know that

$T_1$ = year of Neptune

$T_2$ = year of Earth

$R_1$ = distance of Neptune from Sun

$R_2$ = Distance of Earth from Sun

so now we will have

$\frac{T_1^2}{1} = \frac{(4.5 \times 10^{12})^3}{(1.5 \times 10^11)^3}$

$T_1^2 = 27000$

$T_1 = 164.3 years$

so length of year of Neptune is 164.3 years

6 0

3 years ago

1. On one of the shelves in your physics lab is displayed an antique telescope. A sign underneath the instrument says that the t

Ira Lisetskai [31]

Answer:

a) $f_{e}$ = 3,375 cm , b) f₀ = 77.625 cm

Explanation:

The magnification of a telescope is, to see at the far point of vision (infinity image)

m = - f₀ / $f_{e}$

The length of the tube is

L = f₀ + $f_{e}$

a) The focal length of the eyepiece

L = - m $f_{e}$ + $f_{e}$

L = $f_{e}$ (1-m)

$f_{e}$ = L / (1-m)

Let's calculate

$f_{e}$ = 81.0 / (1 - (-23.0)

$f_{e}$ = 3,375 cm

b) the focal length of the target

f₀ = -m $f_{e}$

f₀ = 23 (3.68)

f₀ = 77.625 cm

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3 years ago

The gravity of Neptune is about 1.1 times the gravity of earth. How will the mass of an object on Neptune compare with its mass

Roman55 [17]

I think the gravity doesn't affect the mass of an object. Only it's weight can be compared

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