Because evaporation requires heat (the latent heat of evaporation) and condensation releases heat (the latent heat of condensation).<span>
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Answer:
A. change in abiotic factors
Explanation:
Abiotic factors are those factors that have no life, the most important of which can be mentioned: water, temperature, light, pH, soil, humidity, oxygen and nutrients.
Answer:

Explanation:
The relationship between the current and the electric field inside a wire is given by the following equation:

where ρ is the resistivity, and J is the current density. The current density is the current per area:

The resistivity is the reciprocal of conductivity. Therefore,

Finally, the electric field can be found:

The final velocity of the rock before it touches the ground is 28 m/s.
Answer:
Explanation:
As the rock is thrown down, this means the acceleration due to gravity will be exerting on the rock. So the rock will be exhibiting a free fall motion. Thus, the acceleration of the rock will be equal to the magnitude of acceleration due to gravity. Then using the third equation of motion, we can determine the final velocity of the rock provided the values for initial velocity, displacement and acceleration is given in the problem itself.
So the acceleration is equal to 9.8 m/s² due to its free fall motion and displacement will be equal to the height of the tower which is given as 30 m. And the initial speed of the rock is stated as 14 m/s. The initial speed is represented as u, final speed is represented as v, displacement is represented as s and acceleration is represented as a.

Then, 2 × 9.8 × 30 = v²-(14)²
v²=784
v= 28 m/s
So the final velocity of the rock before it touches the ground is 28 m/s.
Answer:
the maximum current is 500 A
Explanation:
Given the data in the question;
the B field magnitude on the surface of the wire is;
B = μ₀i / 2πr
we are to determine the maximum current so we rearrange to find i
B2πr = μ₀i
i = B2πr / μ₀
given that;
diameter d = 2 mm = 0.002 m
radius = 0.002 / 2 = 0.001 m
B = 0.100 T
we know that permeability; μ₀ = 4π × 10⁻⁷ Tm/A
so we substitute
i = (0.100)(2π×0.001 ) / 4π × 10⁻⁷
i = 500 A
Therefore, the maximum current is 500 A