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RideAnS [48]
2 years ago
7

What will happen in the particles of liquid after 12 hours inside the freezer?​

Physics
1 answer:
snow_tiger [21]2 years ago
3 0

Answer:

When a liquid (like water) is frozen, all of the molecules start sticking to each other and holding on very tightly. ... If you were to put that water in a closed container in the freezer, then it would still get bigger.

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Why is it dangerous to connect a battery to itself. what is the term used to describe that
Yuki888 [10]
In the circuit outside of the battery the electrons have to expend all of their energy on the internal resistance of the battery which causes heating
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3 years ago
A solid non-conducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a radius r (r <
Svet_ta [14]

Answer:  

Hence the answer is E inside = KQr_{1} /R^{3}.

Explanation:  

E inside = KQr_{1} /R^{3}  

so if r1 will be the same then  

E  \begin{bmatrix}Blank Equation\end{bmatrix} proportional to 1/R3  

so if R become 2R  

E becomes 1/8 of the initial electric field.

8 0
2 years ago
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Why is the coefficient of friction independent of the mass?
Svet_ta [14]
Because the coefficient of friction depends on the surface
3 0
3 years ago
Maggie completed a 10000-m race at an average speed of 160
Gala2k [10]

Answer: 200m/min

Explanation:

Divide 10000m by 160m/min, you will get the answer 62.5. You then subtract 12.5 from 62.5 to understand what you will need your answer for the other person’s speed will be. 10000m divided by 50min is 200m/min.

3 0
3 years ago
In which of the two situations described is more energy transferred?
Furkat [3]

Answer:

More energy is transferred in situation A

Explanation:

Each of the situations are analyzed as follows;

Situation A

The temperature of the cup of hot chocolate = 40 °C

The temperature of the interior of the freezer in which the chocolate is placed = -20 °C

We note that at 0°C, the water in the chocolate freezes

The energy transferred by the chocolate to the freezer before freezing is given approximately as follows;

E₁ = m×c₁×ΔT₁

Where;

m = The mass of the chocolate

c₁ = The specific heat capacity of water = 4.184 kJ/(kg·K)

ΔT₁ = The change in temperature from 40 °C to 0°C

Therefore, we have;

E₁ = m×4.184×(40 - 0) = 167.360·m kJ

The heat the coffee gives to turn to ice is given as follows;

E₂ = m·H_f

Where;

H_f = The latent heat of fusion = 334 kJ/kg

∴ E₂ = m × 334 kJ/kg = 334·m kJ

The heat required to cool the frozen ice to -20 °C is given as follows;

E₃ = m·c₂·ΔT₂

Where;

c₂ = The specific heat capacity of ice = 2.108 kJ/(kg·K)

Therefore, we have;

E₃ = m × 2.108 ×(0 - (-20)) = 42.16

E₃ = 42.16·m kJ/(kg·K)

The total heat transferred = (167.360 + 334 + 42.16)·m kJ/(kg·K) = 543.52·m kJ/(kg·K)

Situation B

The temperature of the cup of hot chocolate = 90 °C

The temperature of the room in which the chocolate is placed = 25 °C

The heat transferred by the hot cup of coffee, E, is given as follows;

E = m×4.184×(90 - 25) = 271.96

∴ E = 271.96 kJ/(kg·K)

Therefore, the total heat transferred in situation A is approximately twice the heat transferred in situation B and is therefore more than the heat transferred in situation B

Energy transferred in situation A = 543.52 kJ/(kg·K)

Energy transferred in situation B = 271.96 kJ/(kg·K)

Energy transferred in situation A ≈ 2 × Energy transferred in situation B

∴ Energy transferred in situation A > Energy transferred in situation B.

3 0
2 years ago
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