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3 years ago

Does the elf owl's small size allow it to live in cacti

1 answer:

3 0

Habitat
-In the Sonoran Desert region, elf owls are found mainly in riparian habitats (places where there is water), or in areas where saguaro cactus are plentiful.

Range
-Elf Owls are found from the southwest USA to Central Mexico and Baja California. Northern populations winter in Central Mexico and on the Pacific slope north to Sinaloa, Mexico.

Wild Status
-The most important threat to the elf owl is habitat loss both of its riparian forest habitat and desert-scrub habitats. In Arizona, the elf owl is not uncommon, but it's numbers are decreasing in California and Texas.

-So I mean like ya it's possible that they could live in cacti. Sorry that I couldn't give you a straight answer.

couldlive in

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Estimate the distance (in cm) between the central bright region and the third dark fringe on a screen 5.00 m from two double sli

Artist 52 [7]

Answer:

y = 1.75 cm

Explanation:

In the double-slit experiment the equation for destructive interference is

d sin tea = (m + ½)

let's use trigonometry to find the angle

tan θ = y / L

as all the experiment does not occur at small angles

tan θ = sin θ / cos θ = sin θ = y / L

we substitute

y = (m + 1/2 ) λ L / d

we calculate

y = (3 + ½) 500 10⁻⁹ 5.00 / 0.5 10⁻³

y = 1.75 10⁻² m

y = 1.75 cm

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3 years ago

Why should we invest more money in renewable energy resources

oksian1 [2.3K]

Answer:

It is the smarter option.

Explanation:

Renewable energy is the cheapest source of new power generation for more than two-thirds of the world and has no fuel costs. It can reduce the economic burden of energy bills by eliminating fuel charges — especially when coupled with energy-efficiency upgrades in our homes and businesses.

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3 years ago

An electron is projected with an initial speed vi = 4.60 × 105 m/s directly toward a very distant proton that is at rest. Becaus

frutty [35]

Answer:

$2.99\times 10^{-19}\ m$

Explanation:

<u>Given:</u>

$u$ = initial velocity of the electron = $4.60\times 10^5\ m/s$

$v$ = final velocity of the electron = $3u$

$x$ = initial position of the electron from the proton = very distant = $\infty$

<u>Assume:</u>

$m$ = mass of an electron = $9.1\times10^{-31}\ kg$

$e$ = magnitude of charge on an electron = $1.6\times10^{-19}\ C$

$p$ = magnitude of charge on an proton = $1.6\times10^{-19}\ C$

$k$ = Boltzmann constant = $9\times 10^9\ Nm^2/C^2$

$y$ = final position of the electron from the proton

$\Delta K$ = change in kinetic energy of the electron

$W$ = work done by the electrostatic force

$F$ = electrostatic force

$r$ = instantaneous distance of the electron from the proton

Let us first calculate the work done by the electrostatic force.

$W=\int Fdr\\\Rightarrow W = \int \dfrac{kep}{r^2}dr\\\Rightarrow W = kep\int \dfrac{1}{r^2}dr\\\Rightarrow W = kep\left | \dfrac{1}{r} \right |_{y}^{x}\\\Rightarrow W = kep\left ( \dfrac{1}{x}-\dfrac{1}{y} \right )\\\Rightarrow W = kep\left ( \dfrac{1}{x}-\dfrac{1}{\infty} \right )\\\Rightarrow W =\dfrac{kep}{x}$

Using the principle of the work-energy theorem,

As only the electrostatic force is assumed to act between the two charges, the kinetic energy change of the electron will be equal to the work done by the electrostatic force on the electron due to proton.

$\therefore \Delta K = W\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kep}{x}\\\Rightarrow \dfrac{1}{2}m((3u)^2-u^2)= \dfrac{kep}{x}\\\Rightarrow \dfrac{1}{2}m(8u^2)= \dfrac{kep}{x}\\\Rightarrow x= \dfrac{2kep}{8mu^2}\\\Rightarrow x= \dfrac{2\times 9\times 10^9\times 1.6\times10^{-19}\times 1.6\times10^{-19}}{8\times 9.1\times10^{-31}\times (4.60\times 10^5)^2}\\\Rightarrow x=2.99\times 10^{-10}\ m\leq$

Hence, the electron is at a distance of $2.99\times 10^{-10}\ m$ when the electron instantaneously has speed of three times the initial speed.

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4 years ago

Two objects of the same size will always have the same mass ? Is this statement correct

vlabodo [156]

Answer: The statement is false , if Jupiter and Venus somehow became the same size, Jupiter would still have more mass due to its density and composition and rocks, minerals, and etc.

Explanation:

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3 years ago

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In the oscillating spring ball system, where is the velocity of the ball the greatest?

nlexa [21]

The speed of an object in a mass-spring system is given under the function

$v = \pm \sqrt{\frac{k}{m}(A^2-x^2)}$