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3 years ago

How do convection currents influence plate tectonics

Physics

1 answer:

jarptica [38.1K]3 years ago

5 0

Answer and Explanation:

Convection happens when particles with a great deal of warmth vitality in a fluid or gas move and replace particles with less warmth vitality.

Lighter (less thick), warm material ascents while heavier (progressively thick) cool material sinks. It is this development that makes course examples known as convection flows in the environment, in water, and in the mantle of Earth.

Huge amount of pressure and heat inside the earth causes the hot magma to stream in convection ebbs and flows. These flows cause the development of the structural plates that make up the world's outside.

Plates at our planet's surface move in view of the extreme warmth in the Earth's center that causes liquid shake in the mantle layer to move.
It moves in an example considered a convection cell that structures when warm material ascents, cools, and in the long run sink down. As the chilled material sinks off, it is warmed and rises once more.

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A man stands at the midpoint between two speakers that are broadcasting an amplified static hiss uniformly in all directions. Th

Illusion [34]

Answer:

Explanation:

The power of each of the speakers is 0.535 W. At a distance d intensity of sound can be found by the following formula

Intensity of sound = Power / 4π d²

= .535 / 4 x 3.14 x (27.3/2)²

= 2.286 x 10⁻⁴ J m⁻² s⁻¹

Intensity of sound due to other source = 5.715 x 10⁻⁵J m⁻² s⁻¹

Total intensity = 2 x 2.286 x 10⁻⁴J m⁻² s⁻¹

= 4.57 x 10⁻⁴J m⁻² s⁻¹

b ) In this case, man is standing at distances 18.15 m and 9.15 m from the sources .

The total intensity of sound reaching him is as follows

0.535 / (4 π x18.15² ) + 0.535 / (4 π x9.15² )

= 1.293 x 10⁻⁴ + 5.087 x 10⁻⁴

= 6.38 x 10⁻⁴J m⁻² s⁻¹

5 0

2 years ago

The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete

oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\$

make shear stress (τ) the subject of the formula

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}$

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of 0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2$

Part (b) shear stress at a distance of 0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0}{2*12} =0$

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3 years ago

Un tubo metálico tiene 100 metros de longitud cuando está a 0 oC y 100,13 metros cuando se calienta hasta 100 oC. ¿Cuál es el co

Artyom0805 [142]

Answer:

α= 1.3 10-5 ºC⁻¹

Explanation:

La dilatación termica de los cuerpos esta dada por la relación

ΔL = L₀ α ( T -T₀)

en este caso nos piden el coeficiente de dilatación térmica

α =DL/L₀ DT

calculemos

α = ( 100,13 -100)/[100 (100 – 0)]

α = 1,3 10-5 ºC⁻¹

Traduction

The thermal expansion of bodies is given by the relationship

ΔL = L₀ α (T -T₀)

in this case they ask us for the coefficient of thermal expansion

α = ΔL / L₀ ΔT

let's calculate

α = (100,13 -100) / [100 (100 - 0)]

α= 1.3 10-5 ºC⁻¹

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2 years ago

A parallel-plate capacitor is formed from two 1.6 cm -diameter electrodes spaced 2.8 mm apart. The electric field strength insid

lidiya [134]

Answer:

12 nC

Explanation:

Capacity of the parallel plate capacitor

C = ε₀ A/d

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Area of plate = π r²

= 3.14 x (0.8x 10⁻²)²

= 2 x 10⁻⁴

C = ( 8.85 X 10⁻¹² x 2 x 10⁻⁴ ) / 2.8 x 10⁻³

= 7.08 x 10⁻¹³

Potential difference between plate = field strength x distance between plate

= 6 x 10⁶ x 2.8 x 10⁻³

= 16.8 x 10³ V

Charge on plate = CV

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11.9 X 10⁻⁹ C

12 nC .

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3 years ago

Which is the fundamental law of energy?

Mkey [24]

Energy cannot be created nor be destroyed

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3 years ago

Read 2 more answers