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AURORKA [14]
4 years ago
7

PLEASE HELP! What is slab pull?

Physics
1 answer:
skad [1K]4 years ago
3 0

Explanation:

Slab pull is that part of the motion of a tectonic plate caused by its subduction. In 1975 Forsyth and Uyeda showed using inverse theory methods that of the many likely driving forces of plates slab pull was the strongest.[1] Plate motion is partly driven by the weight of cold, dense plates sinking into the mantle at oceanic trenches.[2][3] This force and slab suction account for almost all of the force driving plate tectonics. The ridge push at rifts contributes only 5 to 10%.[4]

Carlson et al. (1983)[5] in Lallemandet al. (2005)[6] defined the slab pull force as:

{\displaystyle F_{sp}=K\times \Delta \rho \times L\times {\sqrt {A}}}F_{sp} = K \times \Delta\rho \times L \times \sqrt{A}

Where:

K is 4.2g (gravitational acceleration = 9.81 m/s2) according to McNutt (1984);[7]

Δρ = 80 kg/m3 is the mean density difference between the slab and the surrounding asthenosphere;

L is the slab length calculated only for the part above 670 km (the upper/lower mantle boundary);

A is the slab age in Ma at the trench.

The slab pull force manifests itself between two extreme forms:

The aseismic back-arc extension as in the Izu–Bonin–Mariana Arc.

And as the Aleutian and Chile tectonics with strong earthquakes and back-arc thrusting.

Between these two examples there is the evolution of the Farallon Plate: from the huge slab width with the Nevada, the Sevier and Laramide orogenies; the Mid-Tertiary ignimbrite flare-up and later left as Juan de Fuca and Cocos plates, the Basin and Range Province under extension, with slab break off, smaller slab width, more edges and mantle return flow.

Some early models of plate tectonics envisioned the plates riding on top of convection cells like conveyor belts. However, most scientists working today believe that the asthenosphere does not directly cause motion by the friction of such basal forces. The North American Plate is nowhere being subducted, yet it is in motion. Likewise the African, Eurasian and Antarctic Plates. Ridge push is thought responsible for the motion of these plates.

The subducting slabs around the Pacific Ring of Fire cool down the Earth and its Core-mantle boundary. Around the African Plate upwelling mantle plumes from the Core-mantle boundary produce rifting including the African and Ethiopian rift valleys.

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The bonds of oxygen molecules are broken by sunlight. The minimum energy required to break the oxygen-oxygen bond is 495 kJ/mol.
makvit [3.9K]

Answer:

The wavelength of sunlight that can cause this bond breakage is 242 nm

Explanation:

The minimum energy of the sunlight that'll break Oxygen-oxygen bond must match 495 KJ/mol

But 1 mole of any molecule contains 6.02 × 10²³ molecules/mol

Each molecule of Oxygen will require (495 × 10³)/(6.02 × 10²³) = 8.22 × 10⁻¹⁹ J

E = hf

v = fλ

f = v/λ

f = frequency of the sunlight

λ = wavelength of the sunlight

v = speed of light = 3.0 × 10⁸ m/s

E = hv/λ

λ = hv/E

h = Planck's constant = 6.63 × 10⁻³⁴ J.s

λ = (6.63 × 10⁻³⁴)(3 × 10⁸)/(8.22 × 10⁻¹⁹)

λ = 2.42 × 10⁻⁷ m = 242 nm.

5 0
3 years ago
Una rueda que tiene 15 cm de radio, realiza 64 vueltas en 16 seg. Calcula: Periodo Frecuencia Velocidad angular Velocidad lineal
liraira [26]

Answer:

i) El período de la rueda es de 0,25 segundos.

ii) La frecuencia de la rueda es 4 Hertz

iii) La velocidad angular es aproximadamente 25.133

iv) La velocidad lineal es de aproximadamente 3,77 m / s

Explanation:

El radio de la rueda, r = 15 cm = 0,15 m

El número de vueltas que hace la rueda = 64 vueltas

El tiempo que tarda el volante en dar 64 vueltas = 16 segundos

i) El período = El tiempo que tarda la rueda en dar 1 vuelta

∴ El período de la rueda, T = 16 segundos/(64 vueltas) = 0,25 segundos

El período de la rueda, T = 0,25 segundos

ii) La frecuencia = El número de vueltas por segundo

∴ La frecuencia de la rueda, f = 64 vueltas /(16 segundos) = 4 Hertz

1 vuelta = 2 · π radianes

La frecuencia de la rueda, f = 4 Hertz

iii) Velocidad angular = La medida del ángulo girado por segundo

∴ La velocidad angular, ω = 64 × 2 × π/16 segundos ≈ 8 · π rad/segundos ≈ 25.133 rad/seg

La velocidad angular, ω ≈ 25.133

iv) La velocidad lineal, v = r × ω

∴ v = 0,15 m × 8 · π rad / segundos ≈ 3,77 m/s

La velocidad lineal, v ≈ 3.77 m/s

7 0
3 years ago
Why is the escape speed for a spacecraft independent of the spacecraft's mass?
scoray [572]
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7 0
4 years ago
How much heat is needed to raise the temperature of 9g of water by 17oC?
andre [41]
Well the heat that is needed to raise the temperature of 10g of water by 17oC is 7
6 0
3 years ago
I cant solve this problem, and our teacher said that this would be in the test we'll have tomorrow, can someone help me?
Ad libitum [116K]

Answer:

d = 11.1 m

Explanation:

Since the inclined plane is frictionless, this is just a simple application of the conservation law of energy:

\frac{1}{2} m {v}^{2}  = mgh

Let d be the displacement along the inclined plane. Note that the height h in terms of d and the angle is as follows:

\sin(15)  =  \frac{h}{d}  \\ or \: h = d \sin(15)

Plugging this into the energy conservation equation and cancelling m, we get

{v}^{2}  = 2gd \sin(15)

Solving for d,

d =  \frac{ {v}^{2} }{2g \sin(15) }  =  \frac{ {(7.5 \:  \frac{m}{s}) }^{2} }{2(9.8 \:  \frac{m}{ {s}^{2} })(0.259)}   \\ = 11.1 \: m

3 0
3 years ago
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