Answer:
it impacts the tree form.
Wildfire o Northwest California oak woodlands would naturally be subjected to ... o Fire does not necessarily have a positive impact on growth or recruitment of.
Explanation:
Many species of mature oaks are relatively tolerant to fire. Oaks might replace pines, or drought-tolerant shrubland might take the place ... “Areas with low tree density could maintain droughts for some time because of the cyclical effect of ... diseased plants and removing them from the flora population.
Answer:
The first law of thermodynamics doesn't actually specify that matter can neither be created nor destroyed, but instead that the total amount of energy in a closed system cannot be created nor destroyed (though it can be changed from one form to another).
Explanation:
The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is
1/2 • (10.0 m/s) • (4.0 s) = 20.00 m
Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as
<em>v</em> (ave) = ∆<em>x</em> / ∆<em>t</em>
and under constant acceleration,
<em>v</em> (ave) = (<em>v</em> (final) + <em>v</em> (initial)) / 2
According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so
∆<em>x</em> / (4.0 s) = (10.0 m/s) / 2
∆<em>x</em> = ((4.0 s) • (10.0 m/s)) / 2
∆<em>x</em> = 20.00 m
The main requirement for a good conductor of electricity is to have a lot of valence electrons. Valence electrons are the electrons of the outer shells of atoms not bound with other atoms (for example through covalent bounds). These electrons are "free to escape" as soon as an electric field with enough intensity is applied to the material, and therefore these electrons will be free to move in the material producing an electric current.
Answer:
i = 0.477 10⁴ B
the current flows in the counterclockwise
Explanation:
For this exercise let's use the Ampere law
∫ B . ds = μ₀ I
Where the path is closed
Let's start by locating the current vines that are parallel to the z-axis, so it must be exterminated along the x-axis and as the specific direction is not indicated, suppose it extends along the y-axis.
From BiotSavart's law, the field must be perpendicular to the direction of the current, so the magnetic field must go in the x direction.
We apply the law of Ampere the segment parallel to the x-axis is the one that contributes to the integral, since the other two have an angle of 90º with the magnetic field
Segment on the y axis
L₀ = (y2-y1)
L₀ = 3-0 = 3 cm
Segment on the point x = 2 cm
L₁ = 3-0
L₁ = 3cm
B L = μ₀ I
B 2L = μ₀ I
i = 2 L B /μ₀
i= 2 0.03 / 4π 10⁻⁷ B
i = 4.77 10⁴ B
The current is perpendicular to the magnetic field whereby the current flows in the counterclockwise
