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vovikov84 [41]
3 years ago
5

What kind of energy involves the flow of positive charges?

Physics
2 answers:
Neporo4naja [7]3 years ago
8 0
Electrical would be the energy that involves the flow of positive charges.
uranmaximum [27]3 years ago
5 0
C is your answer i just answered this for someone else

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Fre.e points ...........
hjlf

Answer:

yooooooooooooooo thanks for the points bestie

7 0
3 years ago
I don’t know this one I need help
nikdorinn [45]
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Using the kinetic energy 1/2mv^2 formula
5*10^5 is the answer
4 0
2 years ago
you are about to work some magic with matter. you have a rectangular cube of matter in its solid state. it measures 3 in x 4 in
Elenna [48]

In solid and liquid the matter can occupy the 90 in³ and 157.1 in³ volume.

The matter in gaseous state can be expanded to occupy the volumes of the container.

<h3>Volume of each of the container</h3>

The volume of each of the container is calculated as follows;

<h3>Volume of the rectangular container</h3>

V = 5 in x 6 in x 3 in

V = 90 in³

<h3>Volume of the cylindrical container</h3>

V = πr²h

V = (π)(2.5 in)²(8 in)

V = 157.1 in³

<h3>Volume of the matter</h3>

Vm = 3 in x 4 in x 5 in

Vm = 60 in³

<h3>Matter in solid and liquid state</h3>

Matter has fixed volume in solid and liquid state.

In solid and liquid the matter can occupy the 90 in³ and 157.1 in³ volume.

<h3>Matter in gaseous state</h3>

Matter has no definite volume in gaseous state.

The matter in gaseous state can be expanded to occupy the volumes of the container.

Learn more about states of matter here:

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6 0
1 year ago
Ice has a specific heat of 2090 J/(kg C) and water has a specific heat of 4186 J/(kg C). Water has a latent heat of fusion of 3.
zepelin [54]

Answer:

3.1 × 10⁷ J

Explanation:

The total heat required is the sum of the heats required in each stage.

1) Solid: from -60°C to 0°C.

Q₁ = c(s) × m × ΔT = (2090 J/kg.°C) × 10 kg × (0°C - (-60°C)) = 1.3 × 10⁶ J

where,

c(s): specific heat of the solid

m: mass

ΔT: change in the temperature

2) Solid to liquid at 0°C

Q₂ = Qf × m = (3.3 × 10⁵ J/kg) × 10 kg = 3.3 × 10⁶ J

where,

Qf: latent heat of fusion

3) Liquid: from 0°C to 100°C

Q₃ = c(l) × m × ΔT = (4186J/kg.°C) × 10 kg × (100°C - 0°C) = 4.2 × 10⁶ J

where,

c(l): specific heat of the liquid

4) Liquid to gas at 100 °C

Q₄ = Qv × m = (2.26 × 10⁶ J/kg) × 10 kg = 2.26 × 10⁷ J

where,

Qv: latent heat of vaporization

Total heat

Q₁ + Q₂ + Q₃ + Q₄

1.3 × 10⁶ J + 3.3 × 10⁶ J + 4.2 × 10⁶ J + 2.26 × 10⁷ J = 3.1 × 10⁷ J

3 0
3 years ago
Help fast please physics
Helga [31]

Answer:

10

Explanation:

7 0
3 years ago
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