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3 years ago

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A forest was completely destroyed by a volcanic eruption. All living organisms were killed. Then, due to the arrival of propagul

es carried by birds from nearby areas, certain plant species started growing in the forest. These species have started colonizing the forest because they could survive the harsh conditions of the forest. They also facilitated the growth of other species. What stage of succession is taking place in the forest?

1 answer:

IRISSAK [1]3 years ago

7 0

I took the test on Plato and got it correct. The answer is ecesis.

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Sound waves travel at the rate of 343 m/s at 20°C. If a man standing 450 meters away from the wall of a canyon yells, “Hello,” h

nekit [7.7K]

Answer:

2.62seconds

Explanation:

Speed is defined as the ratio of the distance covered by a body with respect to time.

Speed v = Distance (s)/Time (t)

For a traveling sound wave, the distance between the source of a sound and the reflector is '2S'.

Speed v = 2 × distance (S)/Time (T)

V = 2S/t

2S = vt

Given speed of the wave = 342m/s

Distance covered = 450m

t = 2S/v

t = (2×450)/343

t = 900/343

t = 2.62seconds

It will take him 2.62seconds for him to hear his own voice echo off of the wall.

5 0

3 years ago

Read 2 more answers

The distance between the earth and sun is 1.5 x 108 kilometers and the speed of light is 3.00 x 108 meters per second. Calculate

butalik [34]

Answer:

time = 8.3333 minutes.

Explanation:

distance between earth and sun = 1.5 * $10^{8}km$

speed of light = 3* $10^{8}m/s$

convert the distance unit from km to m so we can have uniform units.

distance between earth and sun = 1.5 * $10^{8}*1000$ m

distance between earth and sun = 1.5 * $10^{11}$ m

speed = distance /time

time = distance / speed

time = $\frac{1.5*10^{11} }{3*10^{8} }$

$= 0.5*10^{3}$

time =500 sec

time = 500/60 minutes

time = 8.3333 minutes.

3 0

3 years ago

Consider an insulating sphere of radius 6 cm surrounded by a conducting sphere of inner radius 18 cm and outer radius 26 cm. Fur

dimulka [17.4K]

Answer:

$-1.7908787542\times 10^{-9}\ C$

$3.4260289211\times 10^{-9}\ C$

$1.7908787542\times 10^{-9}\ C$

$1.6351501669\times 10^{-9}\ C$

Explanation:

r = Radius

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Electric field is given by

$E=-\dfrac{kq}{r^2}\\\Rightarrow q=-\dfrac{Er^2}{k}\\\Rightarrow q=-\dfrac{1610\times 0.1^2}{8.99\times 10^9}\\\Rightarrow q=-1.7908787542\times 10^{-9}\ C$

The charge is $-1.7908787542\times 10^{-9}\ C$

$Q+q=\dfrac{Er^2}{k}\\\Rightarrow Q=\dfrac{Er^2}{k}-q\\\Rightarrow Q=\dfrac{120\times 0.35^2}{8.99\times 10^9}-(-1.7908787542\times 10^{-9})\\\Rightarrow Q=3.4260289211\times 10^{-9}\ C$

The charge is $3.4260289211\times 10^{-9}\ C$

The charge inside will have the polarity changed

$q=+1.7908787542\times 10^{-9}\ C$

Outside the charge will be

$3.4260289211\times 10^{-9}-1.7908787542\times 10^{-9}\\ =1.6351501669\times 10^{-9}\ C$

3 0

4 years ago

Writers periodically compete to see who can write the longest sentence in literature. James Joyce long held the English record w

Scorpion4ik [409]

Answer:

Explanation:

Writers periodically compete to see who can write the longest sentence in literature. James Joyce long held the English record with a 4,391 word sentence in Ulysses. Jonathan Coe one-uped him in 2001 with a 13,955 word sentence in The Rotter’s Club. More recently, a single-sentence, 469,375 word novel appeared.Writers periodically compete to see who can write the longest sentence in literature. James Joyce long held the English record with a 4,391 word sentence in Ulysses. Jonathan Coe one-uped him in 2001 with a 13,955 word sentence in The Rotter’s Club. More recently, a single-sentence, 469,375 word novel appeared.

Writers periodically compete to see who can write the longest sentence in literature. James Joyce long held the English record with a 4,391 word sentence in Ulysses. Jonathan Coe one-uped him in 2001 with a 13,955 word sentence in The Rotter’s Club. More recently, a single-sentence, 469,375 word novel appeared.

Writers periodically compete to see who can write the longest sentence in literature. James Joyce long held the English record with a 4,391 word sentence in Ulysses. Jonathan Coe one-uped him in 2001 with a 13,955 word sentence in The Rotter’s Club. More recently, a single-sentence, 469,375 word novel appeared.

Writers periodically compete to see who can write the longest sentence in literature. James Joyce long held the English record with a 4,391 word sentence in Ulysses. Jonathan Coe one-uped him in 2001 with a 13,955 word sentence in The Rotter’s Club. More recently, a single-sentence, 469,375 word novel appeared.

Writers periodically compete to see who can write the longest sentence in literature. James Joyce long held the English record with a 4,391 word sentence in Ulysses. Jonathan Coe one-uped him in 2001 with a 13,955 word sentence in The Rotter’s Club. More recently, a single-sentence, 469,375 word novel appeared.

8 0

3 years ago

A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with

Julli [10]

Explanation:

Given that,

Capacitor $C=1.66\ \mu F$

Resistor $R=80.0\ \Omega$

Peak voltage = 5.10 V

(A). We need to calculate the crossover frequency

Using formula of frequency

$f_{c}=\dfrac{1}{2\pi R C}$

Where, R = resistor

C = capacitor

Put the value into the formula

$f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}$

$f_{c}=1198.45\ Hz$

(B). We need to calculate the $V_{R}$ when $f = \dfrac{1}{2f_{c}}$

Using formula of $V_{R}$

$V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})$

Put the value into the formula

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=2.280\ Volt$

(C). We need to calculate the $V_{R}$ when $f = f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=3.606\ Volt$

(D). We need to calculate the $V_{R}$ when $f = 2f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=4.561\ Volt$

Hence, This is the required solution.

8 0

4 years ago