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Lubov Fominskaja [6]
3 years ago
5

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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11. You want to calculate the displacement of an object thrown over a bridge. Using -10 m/s2 for acceleration due to gravity, wh
IrinaVladis [17]

Answer:

The displacement was 320 meters.

Explanation:

Assuming projectile motion and zero initial speed (i.e., the object was dropped, not thrown down), you can calculate the displacement using the kinematic equation:

d = \frac{1}{2}gt^2=\frac{1}{2}10\frac{m}{s^2}\cdot 8^2 s^2=320 m

The displacement was 320 meters.

7 0
3 years ago
i throw a tennis ball straight down from the roof of a building 400 meters tall.id it leaves my hand at 10 m/s, how fast will it
Sidana [21]

consider the motion of the tennis ball in downward direction

Y = vertical displacement = 400 m

a = acceleration = acceleration due to gravity = 9.8 m/s²

v₀ = initial velocity of the ball at the top of building = 10 m/s

v = final velocity of the ball when it hits the ground = ?

using the kinematics equation

v² = v²₀ + 2 a Y

inserting the values

v² = 10² + 2 (9.8) (400)

v = 89.11 m/s

6 0
3 years ago
Can you please answers these for me please today is the last day to turn in work and I need this to pass please I’m begging than
Ymorist [56]

Answer:

1.   <u>F = ma</u>  <em>F = 0.2kg * 20m/s² = 4Kg * m/s² =</em> 4N

2.  <u>F = ma</u>  <em>F - 18Kg * 3m/s² = 54Kg * m/s² =</em> 54N

3.  <u>F = ma</u>  <em>F = 0.025Kg * 5m/s² =</em> 0.125N

4.  <u>F = ma</u>  <em>F = 50Kg * 4m/s² =</em> 200N

5.  <u>F = ma</u>  <em>F = 70Kg * 4m/s² =</em> 280N

6.  <u>F = ma</u>  <em>F = 9Kg * 9.8m/s² =</em> 88.2N

Explanation:

Hope this helps ! ^^

8 0
2 years ago
What is the term for the overall change in position?
mart [117]
That's "displacement". It only depends on the beginning and ending locations, and doesn't care about the route between them.
6 0
3 years ago
Read 2 more answers
In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.
leonid [27]

Answer:

(a): F_e = 8.202\times 10^{-8}\ \rm N.

(b): F_g = 3.6125\times 10^{-47}\ \rm N.

(c): \dfrac{F_e}{F_g}=2.27\times 10^{39}.

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053\times 10^{-9} m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges q_1 and q_2 respectively is given by

F_e = \dfrac{k|q_1||q_2|}{r^2}

where,

  • k = Coulomb's constant = 9\times 10^9\ \rm Nm^2/C^2.
  • r = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, q_1 = +1.6\times 10^{-19}\ C.

The charge on the electron, q_2 = -1.6\times 10^{-19}\ C.

These two are separated by the distance, r = 0.053\times 10^{-9}\ m.

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.

Part (b):

The gravitational force of attraction between two objects of masses m_1 and m_1 respectively is given by

F_g = \dfrac{Gm_1m_2}{r^2}.

where,

  • G = Universal Gravitational constant = 6.67\times 10^{-11}\ \rm Nm^2/kg^2.
  • r = distance of separation between the masses.

For the given system,

The mass of proton, m_1 = 1.67\times 10^{-27}\ kg.

The mass of the electron, m_2 = 9.11\times 10^{-31}\ kg.

Distance between the two, r = 0.053\times 10^{-9}\ m.

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.

The ratio \dfrac{F_e}{F_g}:

\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.

6 0
3 years ago
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