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Hitman42 [59]
3 years ago
6

Which sphere is NOT a part of the cycling of oxygen through Earths systems ?

Physics
1 answer:
stepan [7]3 years ago
5 0
The answer would be Exosphere because, there are 3 main regions that circulate oxygen through the Earths system, which are the Biosphere, Atmosphere, and the Lithosphere. 
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What is the Genetic Modification Process?<br><br><br>Brainlyist for the best answer
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is the changing of the DNA of an organism

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The Density of this Rectangular Prism would be 7.5, since the Volume of the Rectangular Prism can be determined by using the formula: length times width times height.

5*2*4, or 2*4*5, or 5*4*2= 40, because 5*2= 10, and 10*4= 40.
The Problem hereby, states that the mass of the Rectangular Prism is 300 grams, right?

Since the Formula for calculating the Density of an Object is Mass divided by Volume, 300 divided by 40 gives you 7.5, and we wouldn't make the answer 7.5 cubed, or 7.5 squared, or 7.5 grams cubed, or 7.5 grams squared, since squared would be for finding only the Area for 2-D Shapes, and cubed would be for finding only the Volume for 3-D Shapes, but in this case that we had here, We were finding Density, which includes Mass AND Volume, not just Volume.
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3 years ago
Explain how the sun's position in the sky affects the length of shadows.
Sergeeva-Olga [200]

Answer:

The position of the sun affects the size of a shadow. A person or object blocks more light when the sun is low in the sky. More blocked light makes shadows longer. ... The sky turns dark when your part of Earth spins away from the sun.

Explanation:

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3 years ago
The minimum voltage that is required to generate an action potential is called the _______.
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The answer would be threshold voltage. It is the minimum value of voltage needed to produce an action potential. It is also known as gate voltage. It is the minimum value that would make turn a transistor on. Hope this answers the question.
3 0
3 years ago
7. Water flows trough a horizontal tube of diameter 2.5 cm that is joined to a second horizontal tube of diameter 1.2 cm. The pr
Usimov [2.4K]

To solve this problem it is necessary to apply the concepts related to the continuity of fluids in a pipeline and apply Bernoulli's balance on the given speeds.

Our values are given as

d_1 = 2.5cm \rightarrow r_1=1.25cm=1.25*10^{-2}m

d_2 = 1.2cm \rightarrow r_2 = 0.6cm = 0.6*10^{-2}m

From the continuity equations in pipes we have to

A_1V_1 = A_2 V_2

Where,

A_{1,2} = Cross sectional Area at each section

V_{1,2} = Flow Velocity at each section

Then replacing we have,

(\pi r_1^2) v_1 = (\pi r_2^2) v_2

(1.25*10^{-2})^2 v_1 = ( 0.06*10^{-2})^2 v_2

v_2 = \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1

From Bernoulli equation we have that the change in the pressure is

\Delta P = \frac{1}{2} \rho (v_2^2-v_1^2)

7.3*10^3 = \frac{1}{2} (1000)([ \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1 ]^2-v_1^2)

7300= 8919.01 v_1^2

v_1 = 0.9m/s

Therefore the speed of flow in the first tube is 0.9m/s

6 0
4 years ago
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