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2 years ago

What happens at a convergent boundary?

Physics

2 answers:

LenaWriter [7]2 years ago

8 0

Convergent boundaries form earthquakes, which forms mountains and islands.

SIZIF [17.4K]2 years ago

6 0

Answer:

Convergent plate boundaries are usually characterized by the convergent motion of the tectonic plates and the formation of a subduction zone, deep sea-oceanic trenches.

When the plates move towards each other, there forms a subduction zone where the denser plate goes below the lighter one. This subduction zone is basically identified by the presence of a narrow V-shaped type of depression, which is commonly known as the trench.

As the denser plate subducts into the layer of the asthenosphere, the rocks undergo partial melting and mixes with the magma. This magma is then brought upward into the over-riding plate, giving rise to the formation of volcanoes.

This type of plate boundary is responsible for the occurrence of deep-focus earthquakes.

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How many photons with 10 ev are required to produce 20 joules of energy?

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One electron Volt (eV) is equal to 1.6 x 10^-19 Joules. Therefore, 10 eV is equal to 1.6 x 10^-18 Joules. In order to produce 20 Joules of energy from 10 eV photons, we would require 20 x 1/(1.6 x 10^-18) = 1.25 x 10^19 particles. This demonstrates that in the world of particle physics, the Joule is a massive energy unit relative to the commonly used electron Volt.

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3 years ago

Question 2 (ID=81813)

zaharov [31]

Answer:

The answer is B

Explanation:

5/2=2.5

2.5x2=5

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2 years ago

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Which type of electromagnetic radiation may damage central nervous system

Furkat [3]

Answer:

microwaves

Explanation:

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3 years ago

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Enter your answer in the provided box. The mathematical equation for studying the photoelectric effect is hν = W + 1 2 meu2 wher

siniylev [52]

Answer:

$v = 4.44 \times 10^5 m/s$

Explanation:

By Einstein's Equation of photoelectric effect we know that

$h\nu = W + \frac{1}{2}mv^2$

here we know that

$h\nu$ = energy of the photons incident on the metal

$W$ = minimum energy required to remove photons from metal

$\frac{1}{2}mv^2$ = kinetic energy of the electrons ejected out of the plate

now we know that it requires 351 nm wavelength of photons to just eject out the electrons

so we can say

$W = \frac{hc}{351 nm}$

here we know that

$hc = 1242 eV-nm$

now we have

$W = \frac{1242}{351} = 3.54 eV$

now by energy equation above when photon of 303 nm incident on the surface

$\frac{1242 eV-nm}{303 nm} = 3.54 eV + \frac{1}{2}(9.1 \times 10^{-31})v^2$

$4.1 eV = 3.54 eV + (4.55 \times 10^{-31}) v^2$

$(4.1 - 3.54)\times 1.6 \times 10^{-19}) = (4.55 \times 10^{-31}) v^2$

$8.96 \times 10^{-20} = (4.55 \times 10^{-31}) v^2$

$v = 4.44 \times 10^5 m/s$

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3 years ago

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nexus9112 [7]

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