Marina is creating a flow chart of the mechanics of mantle convection and how it relates to tectonic plate movement.

1 answer:

8 0

Evidence of the existence of large-scale convection in the mantle from the top to the bottom of the mantle is the density of the material in the lower mantle according shock compression silicates. Calculations showed that under appropriate pressure and temperature distribution of the density in both parts of the mantle are well distributed density oceanic lherzolites - breeds raised in the transform faults of the ocean floor. This is evidence of the homogeneity of the chemical composition of the mantle at the same time is an indirect indication of the existence of large-scale convection in it, constantly stirring its substanceAnd this convection drives plate tectonics

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Which statement is true for particles of the medium of an earth quake p wave

katen-ka-za [31]

I am pretty sure that the only statement which is true for particles of the medium of an earthquake P-wave is being shown in the option : b)vibrate parallel to the wave, forming compressions and rarefactions. As you know, it can be formed in two ways : from alternating compressions and rarefactions or primary wave. I bet you will agree with me.

8 0

4 years ago

A 4.00-m long rod is hinged at one end. The rod is initially held in the horizontal position, and then released as the free end

Natalka [10]

Answer:

The angular acceleration α = 14.7 rad/s²

Explanation:

The torque on the rod τ = Iα where I = moment of inertia of rod = mL²/12 where m =mass of rod and L = length of rod = 4.00 m. α = angular acceleration of rod

Also, τ = Wr where W = weight of rod = mg and r = center of mass of rod = L/2.

So Iα = Wr

Substituting the value of the variables, we have

mL²α/12 = mgL/2

Simplifying by dividing through by mL, we have

mL²α/12mL = mgL/2mL

Lα/12 = g/2

multiplying both sides by 12, we have

Lα/12 × 12 = g/2 × 12

αL = 6g

α = 6g/L

α = 6 × 9.8 m/s² ÷ 4.00 m

α = 58.8 m/s² ÷ 4.00 m

α = 14.7 rad/s²

So, the angular acceleration α = 14.7 rad/s²

5 0

3 years ago

You observe a hockey puck of mass 0.12 kg, traveling across the ice at speed 18.3 m/sec. The interaction of the puck and the ice

Galina-37 [17]

The stopping distance is 143.1 m

Explanation:

First of all, we have to find the acceleration of the hockey puck. This can be done by using Newton's second law of motion:

$\sum F =ma$

where

$\sum F = F_f = -0.14 N$ is the net force acting on the puck (the force of friction, negative because it acts in a direction opposite to the direction of motion)