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2 years ago

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What is the result of seafloor spreading?

1 answer:

Over [174]2 years ago

6 0

Answer:

Seafloor spreading occurs at divergent plate boundaries. As tectonic plates slowly move away from each other, heat from the mantle's convection currents makes the crust more plastic and less dense. The less-dense material rises, often forming a mountain or elevated area of the seafloor. Eventually, the crust cracks.

Explanation:

eventually the crust cracks.

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All but two gaps within a set of Venetian blinds have been blocked off to create a double-slit system. These gaps are separated

Alecsey [184]

Answer:

Explanation:

separation between two gaps, d = 5 cm

angle between central and second order maxima, θ = 0.52°

use

d Sinθ = n λ

n = 2

0.05 x Sin 0.52° = 2 x λ

λ = 2.27 x 10^-4 m

λ = 226.9 micro metre

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2 years ago

A ball is thrown upward at 4 meters per second starting from ground level. How long does it take for the ball to return to the g

nata0808 [166]

Answer:

it takes 2 min to return

Explanation:

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2 years ago

Can I answer my own question

Andre45 [30]

Answer:

<em>Well</em><em> </em><em>okay </em><em>do </em><em>it </em><em>what's</em><em> </em><em>your </em><em>question</em><em> </em><em>?</em><em> </em>

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2 years ago

Read 2 more answers

What are the chemical symbols for carbon aluminium hydrogen oxygen and sodium<br>

Elis [28]

El unico que se es el del oxígeno: o2

4 0

3 years ago

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 27 ft/s. Its height

Ahat [919]

Answer:

$V_{3.01}=-93.2m/s$

$V_{3.005}=-93.1m/s$

$V_{3.002}=-93.04m/s$

$V_{3.001}=-93.02m/s$

$V_{3}=-93m/s$

Explanation:

To calculate average velocity we need the position for both instants t0 and t1.

Now we will proceed to calculate all the positions we need:

$Y_{3}=-99m/s$

$Y_{3.01}=-99.932m/s$

$Y_{3.005}=-99.4655m/s$

$Y_{3.002}=-99.18608m/s$

$Y_{3.001}=-99.09302m/s$

Replacing these values into the formula for average velocity:

$V_{3-3.01}=\frac{Y_{3.01}-Y_{3}}{3.01-3}=-93.2m/s$

$V_{3-3.005}=\frac{Y_{3.005}-Y_{3}}{3.005-3}=-93.1m/s$

$V_{3-3.002}=\frac{Y_{3.002}-Y_{3}}{3.005-3}=-93.04m/s$

$V_{3-3.001}=\frac{Y_{3.001}-Y_{3}}{3.001-3}=-93.02m/s$

To know the actual velocity, we derive the position and we get:

$V=27-40t = -93m/s$

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3 years ago