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UkoKoshka [18]
3 years ago
10

Two animals that can be found at the Grand Canyon include the Abert’s and Kaibab squirrels. The Abert’s squirrel is found at the

south rim of the canyon and in a few surrounding states. The Kaibab squirrel, however, can only be found at the canyon’s north rim, mainly on the Kaibab Plateau. The Kaibab squirrel is not found anywhere else in the world. Scientists believe that the squirrels are actually closely related and have developed separately as a result of geographic isolation. It is possible that the squirrels were separated by plateau erosion that separated the forests and isolated the Kaibab population. Some scientists are studying if the squirrels are now two separate species or if the Kaibab squirrel is simply a subspecies of the Abert’s squirrel. Separate species are unable to reproduce with each other. Which evidence would be used to support the claim that the squirrels are two separate species? The Abert’s squirrel is found at the south rim of the canyon, but the Kaibab squirrel is found at the north rim of the canyon. Geologists prove that plateau erosion is the main cause of forest separation in the region. Scientists observe that Abert’s squirrels are slightly larger and have bushier tails than Kaibab squirrels do. When placed together in pairs for study, no pair of Abert’s squirrel and Kaibab squirrel result in offspring.
Physics
1 answer:
Tom [10]3 years ago
6 0
D. <span>When placed together in pairs for study, no pair of Abert's squirrel and Kaibab squirrel result in offspring.</span>
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A 12.0 μF capacitor is charged to a potential of 50.0 V and then discharged through a 265 Ω resistor. A)How long does the capaci
larisa [96]

(a) The time for the capacitor to loose half its charge is 2.2 ms.

(b) The time for the capacitor to loose half its energy is 1.59 ms.

<h3>Time taken to loose half of its charge</h3>

q(t) = q₀e-^(t/RC)

q(t)/q₀ = e-^(t/RC)

0.5q₀/q₀ = e-^(t/RC)

0.5 = e-^(t/RC)

1/2 =  e-^(t/RC)

t/RC = ln(2)

t = RC x ln(2)

t = (12 x 10⁻⁶ x 265) x ln(2)

t = 2.2 x 10⁻³ s

t = 2.2 ms

<h3>Time taken to loose half of its stored energy</h3>

U(t) = Ue-^(t/RC)

U = ¹/₂Q²/C

(Ue-^(t/RC))²/2C = Q₀²/2Ce

e^(2t/RC) = e

2t/RC = 1

t = RC/2

t = (265 x 12 x 10⁻⁶)/2

t = 1.59 x 10⁻³ s

t = 1.59 ms

Thus, the time for the capacitor to loose half its charge is 2.2 ms and the time for the capacitor to loose half its energy is 1.59 ms.

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6 0
2 years ago
A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p
777dan777 [17]

Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s

Now, we can calculate the angular acceleration  (w0=0rad/s)

\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}

\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}

with this value we can compute the angular velocity

\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}

and the tangential velocity of point B, and then the acceleration of point B:

v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}

hope this helps!!

6 0
3 years ago
Read 2 more answers
How much water will flow in 30 secs through 200 mm of capillary tube of 1.50 mm in diameter, if the pressure difference across t
Paladinen [302]

The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

Qo=1.6 \times 10^{2} \mathrm{~mL}

<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}

Generally, the equation for Rate of flow of Liquid is  mathematically given as

\\$$Q=\frac{\pi r^{4} \times \Delta P}{8 \mu L}

$$

Where dP is pressure difference r is the radius

\mu is the viscosity of water

L is the length of the pipe

Q=\frac{\pi \times\left(\frac{1.5 \times 10^{-3}}{2}\right)^{4} \times 6660}{8 \times 8.01 \times 10^{-4} \times 200 \times 10^{-3}}

Q=5.2 \mathrm{~mL} / \mathrm{s}

In $30s the quantity that flows out of the tube

&Qo=5.2 \times 30 \\&Qo=1.6 \times 10^{2} \mathrm{~mL}

In conclusion, the quantity that flows out of the tube

Qo=1.6 \times 10^{2} \mathrm{~mL}

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5 0
1 year ago
In 2-3 complete sentences, explain why the needle on a compass always points in the direction of magnetic north.
Alex17521 [72]
The needle on a compass always points in the direction of magnetic north because of the magnetic poles of earth. the compass is essentially a magnet itself, so the southern pole of the compass is attracted to the northern pole of earth.
8 0
3 years ago
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In a swimming meet, the swimmers swim a total of 8 laps of a 50-meter-long swimming pool. What is the distance traveled by a swi
N76 [4]

Answer:

The swimmer has a distance traveled of 800 meters.

The final displacement of the swimmer is 0 meters.

Explanation:

A lap is a round trip made by a swimmer in the pool, so that the distance traveled by swimmer is sixteen times the length of the swimming pool. That is:

s = \left(2\,\frac{travels}{lap} \right)\cdot \left(8\,laps \right)\cdot \left(50\,\frac{m}{travel}\right)

s = 800\,m

A swimmer has a distance traveled of 800 meters.

The displacement is the distance between swimmer and a reference point, let suppose that reference point is located at the beginning of the first lap. Hence, the final displacement of the swimmer is 0 meters.

7 0
2 years ago
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