Question

The aorta is a major artery, rising upward from the left ventricle of the heart and curving down to carry blood to the abdomen and lower half of the body. The curved artery can be approximated as a semicircular arch whose diameter is 4.3 cm. If blood flows through the aortic arch at a speed of 0.36 m/s, what is the magnitude (in m/s2) of the blood's centripetal acceleration?

Answer 1

Answer on Question #58557, Physics / Mechanics | Relativity |

The aorta is a major artery, rising upward from the left ventricle of the heart and curving down to carry blood to the abdomen and lower half of the body. The curved artery can be approximated as a semicircular arch whose diameter is $4.3\mathrm{cm}$. If blood flows through the aortic arch at a speed of $0.36\mathrm{m/s}$, what is the magnitude (in $\mathrm{m/s}^2$) of the blood's centripetal acceleration?

Solution:

The equation for centripetal acceleration is:

$$a _ {r} = \frac {v ^ {2}}{r}$$

where $a_{r}$ is centripetal acceleration, $v$ is velocity in meters per second, and $r$ is the radius of the circle in meters.

In our case,

$$a _ {r} = \frac {0 . 3 6 ^ {2}}{0 . 5 \cdot 0 . 0 4 3} = 6. 0 3 \mathrm {m / s ^ {2}}$$

Answer: $6.03\mathrm{m / s}^2$

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