Question

In $1889,$ at Jubbulpore, India, a tug-of-war was finally won after $2 \mathrm{~h} 41 \mathrm{~min},$ with the winning team displacing the center of the rope $3.7 \mathrm{~m} .$ In centimeters per minute, what was the magnitude of the average velocity of that center point during the contest?

Name: In 1889, at Jubbulpore, India, a tug-of-war was finally won after 2 h 41 min, with the winning team displacing the center of the rope 3.7 m . In centimeters per minute, what was the magnitude of the average velocity of that center point during the contest?
Uploaded: 2022-07-11T02:32:31-08:00
Duration: 1 min 39 s
Channel: Madhur L
Description: In 1889, at Jubbulpore, India, a tug-of-war was finally won after 2 h 41 min, with the winning team displacing the center of the rope 3.7 m . In centimeters per minute, what was the magnitude of the average velocity of that center point during the contest?

          In $1889,$ at Jubbulpore, India, a tug-of-war was finally won after $2 \mathrm{~h} 41 \mathrm{~min},$ with the winning team displacing the center of the rope $3.7 \mathrm{~m} .$ In centimeters per minute, what was the magnitude of the average velocity of that center point during the contest?

Added by Taylor H.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Madhur L

07/11/2022

Step 1

$2$ hours is equal to $2 \times 60 = 120$ minutes. $41$ minutes is equal to $41$ minutes. So the total time is $120 + 41 = 161$ minutes. Show more…

Show all steps

Thanks for your feedback!

In $1889,$ at Jubbulpore, India, a tug-of-war was finally won after $2 \mathrm{~h} 41 \mathrm{~min},$ with the winning team displacing the center of the rope $3.7 \mathrm{~m} .$ In centimeters per minute, what was the magnitude of the average velocity of that center point during the contest?