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MariettaO [177]
4 years ago
15

When you put water in a freezer it will turn into ice. This is because as the water _____ the movement of its particles ______.

Physics
2 answers:
Marta_Voda [28]4 years ago
7 0
<span>A) cools down, slows down. When you put water in the freezer, it cools down and then the particles reduce their speed until they eventually form a solid by freezing.</span>
Finger [1]4 years ago
4 0


I think the answer is a


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How condensation affects the temperature of the area where liquids forms
NISA [10]
Condensation does not effect the temperature of the area, condensation is the product the temperature difference between an object and the area it is in
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3 years ago
Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1.0 mm.
pantera1 [17]

Answer:

R_a/R_b=3

Explanation:

The resistance in terms of the area and the length of the wire is given by:

R=pL/A

if we have two wires, the first one is a solid wire with a diameter of d_A = 1 x 10^-3 m, and the second one is a hollow wire with inner diameter of d_B,i = 1 x 10^-3 m and outer diameter of d_B,σ= 2 x 10^-3 m, so the cross sectional area of the first wire is:  

A_a=πr^2_a

A_a=πd^2_A/4

hence the resistance is:  

R_a=(4*p*L_a)/π*d^2_A                                     (1)

the area of the second wire is:  

A_b=π*r^2_B,σ-π*r^2_B,i

A_b=π/4(d^2_B,σ-d^2_B,i)

hence the resistance is:  

R_b=(4*p*L_b)/π(d^2_B,σ-d^2_B,i)                   (2)

To find the ratio between the resistances R_a/R_b, we divide (1) over (2) to get:  

R_a/R_b=(d^2_B,σ-d^2_B,i)*L_a/(d^2_a*L_b)

but the wires have the same length, therefore:  

R_a/R_b=(d^2_B,σ-d^2_B,i)/(d^2_a)

substitute with the given values to get:

R_a/R_b=3

6 0
3 years ago
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That the moon has soil within its shadowy craters rich and useful material
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3 years ago
72) What is the freezing point (°C) of a solution prepared by dissolving 11.3 g of Ca(NO3)2 (formula weight = 164 g/mol) in 115
Rina8888 [55]

<u>Answer:</u> The freezing point of solution is -3.34°C

<u>Explanation:</u>

Vant hoff factor for ionic solute is the number of ions that are present in a solution. The equation for the ionization of calcium nitrate follows:

Ca(NO_3)_2(aq.)\rightarrow Ca^{2+}(aq.)+2NO_3^-(aq.)

The total number of ions present in the solution are 3.

  • To calculate the molality of solution, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

Where,

m_{solute} = Given mass of solute (Ca(NO_3)_2) = 11.3 g

M_{solute} = Molar mass of solute (Ca(NO_3)_2) = 164  g/mol

W_{solvent} = Mass of solvent (water) = 115 g

Putting values in above equation, we get:

\text{Molality of }Ca(NO_3)_2=\frac{11.3\times 1000}{164\times 115}\\\\\text{Molality of }Ca(NO_3)_2=0.599m

  • To calculate the depression in freezing point, we use the equation:

\Delta T=iK_fm

where,

i = Vant hoff factor = 3

K_f = molal freezing point depression constant = 1.86°C/m.g

m = molality of solution = 0.599 m

Putting values in above equation, we get:

\Delta T=3\times 1.86^oC/m.g\times 0.599m\\\\\Delta T=3.34^oC

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

\Delta T=\text{freezing point of water}-\text{freezing point of solution}

\Delta T = 3.34 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

3.34^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-3.34^oC

Hence, the freezing point of solution is -3.34°C

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4. Period of intense asteroid bombardment.

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6. Appearance of continents.

7. Start of plate tectonics.

8. Appearance of photosynthetic bacteria.

9. Change in atmosphere due to photosynthetic bacteria.

10. Widespread appearance of complex organisms.

11. Adaptation of plant and animal life to land.

12. Periods of intense, long-lasting volcanic activity.

13. Re-positioning of Earth's continents and oceans, and its

resultant effect on weather patterns and ocean circulation.

14. Changes in Earth's magnetic fields.

15. Changes in Earth's orientation to the sun.

16. Random devastating meteorite impacts.

17. Glaciations, floods, earthquakes.

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8 0
4 years ago
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