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gayaneshka [121]

3 years ago

7

Describe the forest ecosystem with examples

2 answers:

Andreyy893 years ago

6 0

A forest ecosystem describes the community of plants, animals, microbes and all other organisms in interaction with the chemical and physical features of their environment: Specifically, a terrestrial environment dominated by trees growing in a closed canopy — a forest, in other words.

Hope this helps you buddy! :)

horsena [70]3 years ago

5 0

Answer:

A forest ecosystem describes the community of plants, animals, microbes and all other organisms in interaction with the chemical and physical features of their environment: Specifically, a terrestrial environment dominated by trees growing in a closed canopy — a forest, in other words.

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PERIOD OF THE LEG The period of the leg can be approximated by treating the leg as a physical pendulum, with a period of Equatio

Lesechka [4]

Answer:

Explanation:

We can see from the question that

$T 2\pi\sqrt{\frac{I}{mgh} }$

and $I = \frac{ml^2}{3}$

$m = 16$ % of $67kg$

=> $m =10kg$

from the question $l =48$ % of $1.83m$

Substituting this into the equation

$I = \frac{10.72 * (0.8784)^2}{3}$

=> $I = 2.7571 \ kg m^2$

$h = 0.5 * L$

$h = 0.5 * 0.8784$

$h = 0.4392m$

From the equation above

$T = 2 \pi \sqrt{\frac{2.7571}{10 .72 * 9.81 * 0.4392} }$

$T = 1,534\ sec$

4 0

3 years ago

Which of these is true about a magnetic field?

dimulka [17.4K]

The answer is <span>It points in the same direction as it would make a compass point. In a magnetic field, t</span>he direction of the magnetic field at any location is described as the direction in which the north pole of a compass needle points at that location.

6 0

4 years ago

A 6.3 μC electric charge is placed in an Electric Field with a magnitude of 5.0 x 105 charge due to the Electric Field? N/C. Wha

EastWind [94]

Answer:

F = 3.15 N

Explanation:

Given electric charge, q = 6.3 μC

The magnitude of electric field, $E=5\times 10^5\ N/C$

We need to find the electric force on the charge due to the electric field. The electric force is given by :

F = qE

Putting all the values,

$F=6.3\times 10^{-6}\times 5\times 10^5\\\\F=3.15\ N$

So, the required force on the charge is 3.15 N.

7 0

3 years ago

We have two solenoids: solenoid 2 has twice the diameter, half the length, and twice as many turns as solenoid 1. The current in

leva [86]

Answer:

the field at the center of solenoid 2 is 12 times the field at the center of solenoid 1.

Explanation:

Recall that the field inside a solenoid of length L, N turns, and a circulating current I, is given by the formula:

$B=\mu_0\, \frac{N}{L} I$

Then, if we assign the subindex "1" to the quantities that define the magnetic field ( $B_1$ ) inside solenoid 1, we have:

$B_1=\mu_0\, \frac{N_1}{L_1} I_1$

notice that there is no dependence on the diameter of the solenoid for this formula.

Now, if we write a similar formula for solenoid 2, given that it has :

1) half the length of solenoid 1 . Then $L_2=L_1/2$

2) twice as many turns as solenoid 1. Then $N_2=2\,N_1$

3) three times the current of solenoid 1. Then $I_2=3\,I_1$

we obtain:

$B_2=\mu_0\, \frac{N_2}{L_2} I_2\\B_2=\mu_0\, \frac{2\,N_1}{L_1/2} 3\,I_1\\B_2=\mu_0\, 12\,\frac{N_1}{L_1} I_1\\B_2=12\,B_1$

5 0

4 years ago

A World-class sprinter can reach a top speed of about 11.5 m/s in the first 18.0 m of a race. What is the average acceleration o

irina [24]

Answer

a = 3.674 m / s ^ 2

t = 3.13 s

Using the kinematic equations for the movement we have:

$h(t) = P_{0} + Vot + \frac{1}{2}at ^ 2$ (1)

$V_{f} = V_{0} + at$ (2)

Where:

$P_{0}$ = initial position

$V_{0}}$ = initial velocity

a = acceleration

t = time in seconds

$V_{f}$ = final speed

We know:

$P_{0}=0$

$V_{0}= 0$

h = 18 m

$V_{f} = 11.5\frac{m}{s}$

So:

From (2) we have that: $t =\frac{V_{f}}{a}$

$t =\frac{11.5}{a}$

From (1) we have to:

$h (t) = 0.5at ^ 2\\h = 18 = 0.5at ^ 2$

Then we clear "a" to find the acceleration.

$\frac{36}{t^2} = a\\a = \frac{36}{(\frac{11.5}{a})^2} \\\\a =\frac{11.5^2}{36}\\a = 3.674 m / s ^2$

Then, the time it takes to reach this speed is:

$t =\frac{V_{f}}{a}\\t =\frac{11.5}{3.674}\\t = 3.13 s$

4 0

4 years ago

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