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2 years ago

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

Physics

1 answer:

marta [7]2 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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A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o

iVinArrow [24]

There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: $v=15 m/s$
- radius of the hill: $r=100 m$

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car $W=mg$ (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, $m \frac{v^2}{r}$ , so we can write:
$mg-N=m \frac{v^2}{r}$ (1)
By rearranging the equation and substituting the numbers, we find N:
$N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N$

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
$N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N$

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
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3 years ago

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astra-53 [7]

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While the anode is electrode where electron excess ions move to.

So the relationship between Cathode and Anode involves electrons.

C.

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