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2 years ago

Hi anyone able to explain fully about the continental-continental plate convergence ? I’m a little confused , thanks!!

Geography

1 answer:

PilotLPTM [1.2K]2 years ago

4 0

Answer:

1. Continental-continental convergence are both <em>continental crust</em> plates.

2. Buoyancy basically means these plates are light float on top of the Asthenosphere.

Explanation:

1. To go a little more into detail, this type of convergence ocurrs when two plates made of continental crust (not oceanic) collide. In the case of ocean-continent convergence, the denser oceanic crust sinks underneath the lighter continental crust.

In this scenario, we have two plates, neither of which want to sink, so instead of going down they just crumple and fold and keep colliding. This collision creates very high mountain ranges such as the Himalaya and usually doesn't contain any volcanic activity because no magma is capable of penetrating through so much crustal material.

2. As for why continental crust is buoyant and doesn't sink has much to do with its composition. Continental crust is made mostly of silicate material. This is very different from oceanic crust which is made of basaltic rocks and is therefore heavier. The term buoyancy refers to the 'floating action' of the continental crust on top of the uppermost mantle.

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Answer:

The current amount of the Potassium-40 sample is approximately 37.521 grams.

Explanation:

The amount of the sample of the radioactive isotope decays exponentially in time, the amount of mass of the sample as a function of time ( $m (t)$ ), in grams, is described below:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$ (1)

Where:

$m_{o}$ - Initial mass, in grams.

$t$ - Time, in years.

$\tau$ - Time constant, in years.

The time constant can be found from half life ( $t_{1/2}$ ), in years, described in statement:

$\tau = \frac{t_{1/2}}{\ln 2}$ (2)

If we know that $m_{o} = 300\,g$ , $t = 3.9\times 10^{9}\,yr$ and $t_{1/2} = 1.3\times 10^{9}\,yr$ , then the current amount of the sample is:

$\tau = \frac{1.3\times 10^{9}\,yr}{\ln 2}$

$\tau \approx 1.876\times 10^{9}\,yr$

$m = (300\,g)\cdot e^{-\frac{3.9\times 10^{9}\,yr}{1.876\times 10^{9}\,yr} }$

$m\approx 37.521\,g$

The current amount of the Potassium-40 sample is approximately 37.521 grams.

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