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Ulleksa [173]
3 years ago
14

In Alaska, salmon feed on tiny plants called phytoplankton. During late summer, salmon gather in rivers to reproduce. Grizzly be

ars visit these rivers to feed on the large number of salmon. Phytoplankton are . Both the salmon and the grizzly bears are . Grizzly bears are .
Physics
2 answers:
Daniel [21]3 years ago
6 0

Answer:

correct answer below

Explanation:

d1i1m1o1n [39]3 years ago
5 0

Answer:

-->Phytoplankton are AUTOTROPHS

--> Both the Salmon and the grizzly bears are CONSUMERS.

--> Grizzly bears are SECONDARY CONSUMERS.

Explanation:

In an ecosystem, which is the basic functioning unit of nature, the biotic components is made up of all the living organisms in it. These organisms are divided into two main groups which includes:

--> The AUTOTROPHS: These are organisms that are able to use sunlight or chemical energy to manufacture their own food from simple inorganic substances. Since Autotrophs are the only organisms that can produce food in an ecosystem, they are also known as PRIMARY PRODUCERS. Example of autotrophs includes Phytoplankton( all green plants).

-->HETEROTROPHS: These organisms cannot manufacture their own food rather they feed on ready-made food which comes from the tissues of organisms in their environment. In an ecosystem, Heterotrophs are known as the CONSUMERS. Example of Heterotrophs includes all animals (salmon and the grizzly bears).

These biotic components of the ecosystem forms a feeding pathway through which energy and nutrients are transferred step by step among organisms. This is called a FOOD CHAIN. This feeding pathway follows a certain pattern:

--> it begins with a primary producer (Phytoplankton)

--> the primary producer is eaten by a primary consumer (salmon)

--> the primary consumer is eaten by a secondary consumer (grizzly bears)

That is the reason both the Salmon and the grizzly bears are CONSUMERS but while salmon is a primary consumer, grizzly bears are SECONDARY consumers.

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A playground merry-go-round of radius r = 1 m has a moment of inertia of i = 240 kg*m^2. and is rotating at a rate of ω = 8 rev/
Ghella [55]
When the child jumps onto the merry-go-around the moment of inertia of the system changes. If we consider the child to be point-like mass then its moment of inertia would be:
I_{ch}=mr^2
We get the new moment of inertia by simply adding the child's moment of inertia to the old moment of inertia.
I_{new}=I_{old}+I_{ch}=240+35(1)^2=275 $kgm^2
Since there is no force mention we must assume that angular momentum is conserved.
L=const.\\ L=I_{old}\omega_0=I_{new}\omega'\\ \omega'=\frac{I_{old}\omega_0}{I_{new}}
When we plug in all the numbers we get:
\omega'=\frac{240\cdot8}{275}=6.98 \ \frac{rev}{min}

3 0
3 years ago
Light of wavelength 656 nm and 410 nm emitted from a hot gas of hydrogen atoms strikes a grating with 5300 lines per centimeter.
padilas [110]

Answer:

20.32^{\circ} and 44.08^{\circ}

12.56^{\circ} and 25.77^{\circ}

Explanation:

\lambda = Wavelength

\theta = Angle

m = Order

Distance between grating is given by

d=\dfrac{1}{5300}\\\Rightarrow d=0.0001886\ \text{cm}

\lambda=656\ \text{nm}

We have the relation

d\sin\theta=m\lambda\\\Rightarrow \theta=\sin^{-1}\dfrac{m\lambda}{d}

m = 1

\theta=\sin^{-1}\dfrac{1\times 656\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=20.35^{\circ}

m = 2

\theta=\sin^{-1}\dfrac{2\times 656\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=44.08^{\circ}

The first and second order angular deflection is 20.32^{\circ} and 44.08^{\circ}

\lambda=410\ \text{nm}

m = 1

\theta=\sin^{-1}\dfrac{1\times 410\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=12.56^{\circ}

m = 2

\theta=\sin^{-1}\dfrac{2\times 410\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=25.77^{\circ}

The first and second order angular deflection is 12.56^{\circ} and 25.77^{\circ}.

4 0
3 years ago
An aluminum bar has a mass of 9 kg in the air. Calculate its volume. Now imagine that you submerge the aluminum bar in water han
yan [13]

Answer:

55.7 N

Explanation:

The density of aluminum is 2710 kg/m³.  So its volume is:

V = (9 kg) / (2710 kg/m³)

V = 0.00332 m³

The apparent weight is the actual weight minus the buoyant force.

N = mg − B

N = mg − ρVg

N = g (m − ρV)

N = (9.8 m/s²) (9 kg − (1000 kg/m³) (0.0332 m³))

N = 55.7 N

7 0
3 years ago
Light with a wavelength of 494 nm in vacuo travels from vacuum to water. Find the wavelength of the light inside the water in nm
Mamont248 [21]

Answer:

370.6 nm

Explanation:

wavelength in vacuum = 494 nm

refractive index of water with respect to air = 1.333

Let the wavelength of light in water is λ.

The frequency of the light remains same but the speed and the wavelength is changed as the light passes from one medium to another.

By using the definition of refractive index

n = \frac{wavelength in air}{wavelength in water}

where, n be the refractive index of water with respect to air

By substituting the values, we get

1.333 = \frac{494}{\lambda }

λ = 370.6 nm

Thus, the wavelength of light in water is 370.6 nm.

8 0
3 years ago
What happens to the forces acting on a falling object at terminal velocity?
xenn [34]

Answer: Objects falling through a fluid eventually reach terminal velocity. At terminal velocity, the object moves at a steady speed in a constant direction because the resultant force acting on it is zero.

8 0
3 years ago
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