Question

- The large intestine has an average length of 150 cm and
diameter 7.6 cm. For the purposes of these questions, ignore the
presence of villi. Assuming normal quiet activity, if it takes 120
action potentials (spikes) to contract a 5-cm
^{2}segment of colon smooth muscle, how many action potentials would be required to eventually contract the entire colon?

- Given a 6-oz bolus (~ 170 g) that must be moved has an average speed of 8.60 cm/hr, how much work would it take to move it through the entire length of the colon? Assume that the mass of the bolus does not significantly change during its trip through the colon. How long would it take?

Some useful equations:

Area of a cylinder: Area = 2(pie)(r) (L + r)

Force = ma (Newton’s Equation)

Work = force x distance

Mass (in kg) = weight (in Newtons)/9.8 because mass =/= weight

Answer #1

**Answer :)**

Area of the colon = 2πrl

R = radius = 7.6/2 = 3.8 cm

L = length = 150 cm

Area of the colon = 2 x 3.14 x 3.8 x 150 cm2

Area of the colon = 3579.6 cm2

There are 120 action potentials in 5 cm2 colon.

Therefore,

Total number of action potentials = 120/5 x 3579.6

Total number of action potentials = 85910.4, or

Total number of action potentials = 85910 approximately

Mass of the bolus = 170 g

Mass of the bolus = 170/1000 Kg

Mass of the bolus = 0.17 Kg

Force drive by the bolus = 0.17 x 9.8

Force drive by the bolus = 1.67 N

Work to drive the bolus = Force x distance

Work to drive the bolus = 1.67 x 150/100 m

Work to drive the bolus = 2.5 J

The time is taken by the bolus to move out the colon = Distance/ Speed

The time is taken by the bolus to move out the colon = 150/ 8.6

The time is taken by the bolus to move out the colon = 17.44 hours

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