<span>The cell has 1% concentration of the salt. The external environment is highly concentrated with 25% saline solution. This will lead to release of water outside the cell, by passive diffusion from a region of high conentration of solvent to lower concentration. Thus, the cell will shrink.</span>
Segmentation is a squeezing motion of the circular layer of smooth muscle in the small intestine.
<h3>What is segmentation in the small intestine?</h3>
Segmentation, which mostly affects the small intestine, is made up of localized contractions of the Gastro-Intestinal tract's circular muscle. These contractions separate out little portions of the intestine, allowing their contents to move back and forth while being continuously divided, broken up, and mixed.
Our intestines' circular muscles contract during segmentation to churn food back and forth, rather like a washing machine. This churning helps break down food into tiny bits for digestion by allowing it to mix with gastric secretions in the intestines. By bringing chyme into contact with the intestinal walls during segmentation, the technique also helps to increase absorption.
Learn more about segmentation here:
brainly.com/question/2133778
#SPJ4
Dr. Charles is using fMRI when she uses strong magnets to track changes in the brain's blood-oxygen levels.
Functional magnetic resonance imaging or (fMRI) is a techique used to measure brain activity by detecting changes associated with blood flow. When an area of the brain is in use, blood flow to that region will increase.
Given data:
● Angular velocity of the turntable ω = 33 rev/min
Therefore,
ω = 33 rev/min
ω = 33rev/min × 2π rad/rev × 1 min/60 sec = 3.45 rad/s
● The distance of the watermelon seed from the axis of rotation r = 7.8 cm = 0.078 m
● μs = the coefficient of static friction
Section a:
The seed is undergoes circular motion and it is been effected by centripetal acceleration.
ac = rω^2
ac = 0.078 × 3.45^2
ac = 0.9284 m/s^2
Therefore,
the centripetal acceleration of the seed is 9.274 m/s^2
Section b:
If the seed is observed not to slip at the course of the circular motion, then the supplied frictional force given by the seed and surface of turntable would at least be equivalent to the centripetal force working on the seed.
Centripetal Force = Frictional Force
mrω^2 = μsmg
μs = rω^2 /g
μs = 0.078 × 3.45^2
------------
9.81
μs = 0.09464
Thus,
the coefficient of static friction is 0.09464.
