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Annette [7]
3 years ago
7

Which of the following woud be expected to have the greatest impact on primary productivity?

Physics
1 answer:
leonid [27]3 years ago
3 0

Answer: C solar energy

Solar energy has the greatest  impact on primary productivity.

Explanation:

Primary productivity  is the amount of organic compounds synthesized from carbon dioxide. Primary production occurs by the process of photosynthesis. Since the process of photosynthesis uses sunlight as the energy source solar energy has the greatest impact on primary productivity.

The synthesis can occur in atmosphere as well as aquatic environment. The organisms that perform the function of primary production are called producers .In terrestrial regions plants are the producers and in aquatic systems algae are the  primary producers.

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A force of 90N is applied to each cart below, which one will accelerate the fastest?
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Answer:

5kg

Explanation:

The cart with the least amount of mass will result in the fastest acceleration.

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Daniel has a sample of pure copper. Its mass is 89.6 grams (g), and its volume is 10 cubic centimeters (cm3). What’s the density
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The orbital period of a satellite is 2 × 106 s and its total radius is 2.5 × 1012 m. The tangential speed of the satellite, writ
LenaWriter [7]

The orbital period of the satellite[T] is given as 2*10^{6} S.

The radius of the satellite is given [R] 2.5*10^{12} m.

we are asked here to calculate the tangential speed of the satellite.

Before going to get the solution first we have understand the tangential speed.

The tangential speed of a satellite is given as the speed required to keep the satellite along the orbit. If satellite speed is less than tangential speed,there is the chance of it falling down towards earth. If it is more,then it will deviate from it orbit and can't stick to the orbit further.In a simple way  the tangential speed is the linear speed of an object in a circular path.

Now we have to calculate the tangential speed [V].

Mathematically the tangential speed [V]   written as -

                                V=\frac{2\pi R}{T}

where T is the time period of the satellite and R is the radius of the satellite.

                        V=\frac{2*3.14*10^{12} }{2*10^{6} }

                               = 7.85*10^{6} m/s

There is also another way through which we can get  the solution as explained below-

We know that the tangential speed of a satellite V=\sqrt{\frac{GM}{R^{2} } }

where G is the gravitational constant and M is the mas of central object.

But we know that g=\frac{GM}{R^{2} }

                               ⇒GM=gR^{2}  where g is the acceleration due to gravity of that central object.


Hence    V=\sqrt{\frac{gR^{2} }{R} }

               ⇒   V=\sqrt{gR}

By knowing the value of g due to that central object we can also calculate its tangential speed.

                           

 




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<em><u>i</u></em><em><u>e(</u></em><em><u>p</u></em><em><u>=</u></em><em><u>m×</u></em><em><u>v</u></em><em><u>)</u></em>

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