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3 years ago

What is an organism's specific role in an ecosystem

Physics

2 answers:

KengaRu [80]3 years ago

8 0

The answer you are looking for is a Niche. I know since I studied this.

Hope this helped!

Harman [31]3 years ago

7 0

<span>Niche - </span><span>An organism's particular role in an ecosystem, or how it makes its living.</span>

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At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m². A fan is 30 m from the loudspeak

Klio2033 [76]

To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.

The expression for sound power is,

$P = AI$

Here,

A = Area

I = Intensity

P = Power

At the same time the area can be written as,

$A = \frac{\pi d^2}{4}$

Now the intensity is inversely proportional to the square of the distance from the source, then

$I \propto \frac{1}{r^2}$

The expression for the intensity at different distance is

$\frac{I_1}{I_2}= \frac{r^2_2}{r_1^2}$

Here,

$I_1$ = Intensity at distance 1

$I_2$ = Intensity at distance 2

$r_1$ = Distance 1 from light source

$r_2$ = Distance 2 from the light source

If we rearrange the expression to find the intensity at second position we have,

$I_2 = I_1 (\frac{r_1^2}{r_2^2})$

If we replace with our values at this equation we have,

$I_2 = (0.10W/m^2)(\frac{1.0m^2}{30.0m^2})$

$I_2 = 1.11*10^{-4} W/m^2$

Now using the equation to find the area we have that

$A = \frac{\pi (8.4*10^{-3}m)^2}{4}$

$A = 5.5*10^{-5}m^2$

Finally with the intensity and the area we can find the sound power, which is

$P = AI$

$P = (5.5*10^{-5}m^2)(1.11*10^{-4}W/m^2)$

$P = 6.1*10^{-9}J/s$

Power is defined as the quantity of Energy per second, then

$E = 6.1*10^{-9}J$

8 0

3 years ago

Why is it not suitable to determine the volume of an irregular charcoal using displacement method

AleksandrR [38]

Answer:

Displacement method of volume measurement is no suitable

Explanation:

Displacement method of volume measurement is no suitable for the objects that do not get immersed into the water completely because of the hindrance in accuracy of the measurement.

4 0

3 years ago

During the first 5 seconds after a ball is dropped from rest, how far will it fall?

agasfer [191]

The answer is 125metre

5 0

3 years ago

Find the required angular speed, ω, of an ultracentrifuge for the radial acceleration of a point 1.80 cm from the axis to equal

Arturiano [62]

Answer:

The required angular speed ω of an ultra-centrifuge is:

ω = 18074 rad/sec

Explanation:

Given that:

Radius = r = 1.8 cm

Acceleration due to g = a = 6.0 x 10⁵ g

Sol:

We know that

Angular Acceleration = Angular Radius x Speed²

a = r x ω ²

Putting the values

6 x 10⁵ g = 1.8 cm x ω ²

Converting 1.8 cm to 0.018 m, also g = 9.8 ms⁻²

6 x 10⁵ x 9.8 = 0.018 x ω ²

ω ² = (6 x 10⁵ x 9.8) / 0.018

ω ² = 5880000 / 0.018

ω ² = 326666667

ω = 18074 rad/sec

7 0

3 years ago

Is it true or false When objects collide, some momentum is lost.

never [62]

Answer:

True.

Explanation:

5 0

2 years ago

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