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3 years ago

A shallow groove on the surface of the cortex is called a ________.

Physics

1 answer:

Juli2301 [7.4K]3 years ago

3 0

Answer:

sulcus

Explanation:

A sulcus is an indentation or depression in the brain that causes it to look like it ridges or folds

Cerebral sulci and fissures are grooves between the adjacent gyri on the surface of the cerebral hemispheres.

Sulci can be basically can be divided into three basic function

limiting sulcus: This happens to develop between areas differing in structure and function, for example central sulcus

axial sulcus: This develops along the axis of a rapidly growing/developing area (e.g. calcarine sulcus)

operculated sulcus: a sulcus may be between two structurally-different areas and a third sulcus may lie in its wall and does not appear on the surface (e.g. lunate sulcus)

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2 years ago

A massless string connects a 10.00 kg mass to a 13.00 kg cart which is resting on a frictionless horizontal surface. The mass ha

ch4aika [34]

The cart's acceleration to the right after the mass is released is determined as 7.54 m/s².

<h3>Acceleration of the cart</h3>

The acceleration of the cart is determined from the net force acting on the mass-cart system.

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Learn more about acceleration here: brainly.com/question/14344386

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2 years ago

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2 years ago

Read 2 more answers

Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infini

amid [387]

Answer:

Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:

$E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}$

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

$E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}$

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

$V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2} } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r$

= $\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2} }{2R^{3} } ]$

∴NOTE: Graph is attached

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3 years ago