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4 years ago

Describe the relationship of the Earth’s lithosphere to the asthenosphere.

2 answers:

8 0

The Earth's soft putty-like Asthenosphere carries the Lithosphere (which is like a giant jig-saw puzzle fit around the Earth), and the giant continents on its back.

Marina CMI [18]4 years ago

6 0

<h2>Answer:</h2>

<u>Lithosphere floats on the top of Asthenosphere</u>

<h2>Explanation:</h2>

The earth crust is made up of tectonic plates and these tectonic plates are always in flow. Together the earths crust and mantle beneath it is called lithosphere. On the other hand the area beneath the mantle is called the asthenosphere. Actually lithosphere is solid in nature and asthenosphere is semi solid or liquid nature. The lithosphere floats on the asthenosphere.

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Which, if any, of the heat engines violate(s) the second law of thermodynamics?

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The second law states that the total entropy can never decrese over time for an isolated system

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Using the graph above, explain what happened on the amount carbon in the atmosphere 1960 to 2020.

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Answer:

No graph...

Explanation:

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3 years ago

Please Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

arsen [322]

Answer:

Given that

speed u=4*10^6 m/s

electric field E=4*10^3 N/c

distance b/w the plates d=2 cm

basing on the concept of the electrostatices

now we find the acceleration b/w the plates to find the horizontal distance traveled by the electron when it hits the plate.

acceleration a=qE/m= $1.6*10^{-19}*4*10^3/9.1*10^{-31} =0.7*10^{15}$ = $7*10^{14}$ m/s

now we find the horizontal distance traveled by electrons hit the plates

horizontal distance

$X=u[2y/a]^{1/2}$

= $4*10^6[2*2*10^{-2}/7*10^{14}]^{1/2}$

= $3*10^{-2}$ = 3 cm

5 0

4 years ago

please help! find magnitude and direction (the counterclockwise angle with the +x axis) of a vector that is equal to a + c

-BARSIC- [3]

Answer:

Option (2)

Explanation:

From the figure attached,

Horizontal component, $A_x=A\text{Sin}37$

$A_x=12[\text{Sin}(37)]$

= 7.22 m

Vertical component, $A_y=A[\text{Cos}(37)]$

= 9.58 m

Similarly, Horizontal component of vector C,

$C_x$ = C[Cos(60)]

= 6[Cos(60)]

= $\frac{6}{2}$

= 3 m

$C_y=6[\text{Sin}(60)]$

= 5.20 m

Resultant Horizontal component of the vectors A + C,

$R_x=7.22-3=4.22$ m

$R_y=9.58-5.20$ = 4.38 m

Now magnitude of the resultant will be,

From ΔOBC,

$R=\sqrt{(R_x)^{2}+(R_y)^2}$

= $\sqrt{(4.22)^2+(4.38)^2}$

= $\sqrt{17.81+19.18}$

= 6.1 m

Direction of the resultant will be towards vector A.

tan(∠COB) = $\frac{\text{CB}}{\text{OB}}$

= $\frac{R_y}{R_x}$

= $\frac{4.38}{4.22}$

m∠COB = $\text{tan}^{-1}(1.04)$

= 46°

Therefore, magnitude of the resultant vector will be 6.1 m and direction will be 46°.

Option (2) will be the answer.

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3 years ago

The pulse site located at the point where the upper leg bends is called the

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3 years ago