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3 years ago

Where does the heat come from that drives this convection current in the mantle

Physics

1 answer:

marta [7]3 years ago

3 0

Answer:

Earth's interior (Core)

Explanation:

The earth is comprised of 3 distinct layers namely the Core, the Mantle and the Crust, which are divided based on their composition as well as density.

The core of the earth is extremely very hot where the inner core remains solid and outer core acts a liquid. It is mainly comprised of iron, nickel and other siderophile elements.

A large amount of heat (energy) is radiated from this core region towards the surface of the earth. Due to this, the mantle rocks forms magma that creates the convection currents, where the hot and less dense magma rises upward and the cool and denser magma sinks to the bottom. This occurs continuously, as a result of which the lithospheric plates are forced to move over the less dense layer of asthenosphere.

Thus, the heat energy that drives the convection current in the mantle is provided from the interior (core) of the earth.

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La tension que se transmite en la cuerda BD es de 75 lb. Calcula el momento de fuerza generada por la cuerda respecto al punto C

Bas_tet [7]

Answer:

Mc = 1920[lb*in]

Explanation:

Para poder solucionar este problema debemos realizar un análisis estático, por tal motivo lo primero es realizar un diagrama de cuerpo libre con las respectivas fuerzas actuando sobre la barra ABC. DE igual manera calcular la geometría de la configuración mostrada.

El diagrama de cuerpo libre se puede ver en la imagen adjunta, con la solución de este problema.

Lo primero es determinar el angulo t, el cual por medio de las propiedades del triangulo rectángulo se puede determinar.

Con este angulo (t) ya determinado, fijamos la atención en el triangulo BCD, este triangulo no es rectángulo, pero por medio de la ley de senos podemos determinar el angulo omega.

Después de determinar el angulo omega, restamos el angulo (t) para poder determinar el angulo (a).

Seguidamente realizamos una sumatoria de momentos alrededor del punto C, utilizado las respectivas fuerzas con los ángulos descompuestos.

El momento en el punto C es de 1920 [Lb*in].

Nota: ya que no se menciona la fuerza en el punto A, esta se desprecia y no se tiene en cuenta en los calculos. En la imagen adjunta se puede ver el procedimiento desarrollado.

7 0

3 years ago

Help pls, see picture. Will mark Brainliest

Pani-rosa [81]

Answer:

a ) option 2 is correct

b) -ve acceleration for upward motion ,0 acceleration at top point ,+ve acceleration on downward motion ...

Explanation:

mark me as brainliest ❤️

5 0

2 years ago

Read 2 more answers

Two blocks A and B with mA = 2.2 kg and mB = 0.84 kg are connected by a string of negligible mass. They rest on a frictionless h

madreJ [45]

Answer:

(a) a = 1.875 m/s²

(b)T = 1.575 N

(c)T increase

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

We define the x-axis in the direction parallel to the movement of the blocks on the horizontal surface and the y-axis in the direction perpendicular to it.

Forces acting on the block A

WA: Weight of of the A block : In vertical direction downaward (-y)

NA : Normal force of the A block :In vertical direction upaward (+y)

F= 5.7 N In in the direction parallel to the movement of the blocks (+x)

T : tension of the string: In in the direction (-x)

Forces acting on the block B

WB: Weight of the B block: In vertical direction downaward (-y)

NB : Normal force of the B block :In vertical direction upaward (+y)

T : tension of the string: In in the direction (+x)

Data

mA = 2.2 kg

mB = 0.84 kg

(a) Magnitude of the acceleration of the blocks

Newton's second law to B block:

∑Fx = m*a

T = (0.84)*a Equation (1)

Newton's second law to A block:

∑Fx = m*a

5.7 - T = (2.2)*a We replace T of the Equation (1)

5.7 - (0.84)*a = (2.2)*a

5.7 = (2.2)*a + (0.84)*a Equation (2)

5.7 =(3.04)*a

a =5.7 / (3.04)

a = 1.875 m/s²

(b)Tension (in N) in the string connecting the two blocks

We replace data in the Equation (1)

T = (0.84)*a

T = (0.84)*(1.875)

T = 1.575 N

(c) How will the tension in the string be affected if mA is decreased?

We observed Equation (2)

5.7 = (2.2)*a + (0.84)*a

5.7 = (mA)*a + (0.84)*a

5.7 =a*( mA+ 0.84)

if mA decrease , then, the acceleration increase and T increase

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Answer:

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